1Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên
2Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên
3Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên
4Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên
a b. a b,
a > b, a − b a b,
a b, a − b a
b, a < b, a − b a
b a b, a −b
a |a| =
a a 0
−a a < 0.
a, b, c n
a > b ⇐⇒ a −b > 0
a > b ⇐⇒ a + c > b + c
a > b ⇐⇒ a
2n+1
> a
2n+1
|a| > |b| ⇐⇒ a
2n
> a
2n
a b ⇐⇒
a=b
a>b.
a > b, c > 0 ⇐⇒ ac > bc
c < 0 ⇐⇒ ac < bc.
a > b, b > c =⇒ a > c.
= a
3
+ 3ab(a + b) + b
3
(a −b)
3
= a
3
− 3ab(a − b) −b
3
.
a
2
− b
2
= (a − b)(a + b).
a
3
− b
3
= (a − b)(a
2
+ ab + b
2
) a
3
+ b
3
= (a + b)(a
2
+ (ay − bx)
2
+ (bz −
cy)
2
+ (cx − az)
2
.
|ab| = |a||b|, |
a
d
| =
|a|
|d|
|a| = |b| a = ±b.
a, b, c, x, y, z d = 0
a
2
+ b
2
2ab.
(a
2
+ b
2
)(x
2
+ y
2
) (ax + by)
)(x
2
+ y
2
) = (ax + by)
2
+ (ay − bx)
2
(ax + by)
2
(a
2
+ b
2
)(x
2
+ y
2
) (ax + by)
2
.
a
x
=
b
y
.
(a
2
+b
+z
2
) (ax+by+cz)
2
.
a
x
=
b
y
=
c
z
.
|a| ±a, |b| ±b. a+b 0 |a+b| = a+b |a|+|b|;
a+b < 0 |a+b| = −a−b |a|+|b|. |a+b| |a|+|b|.
6Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên
|a| = |a +b +(−b)| |a +b|+|−b| = |a + b|+ |b| |a|−|b| |a +b|.
|b| = |a + b + (−a)| |a + b| + | − a| = |a + b| + |a|
|b| −|a| |a + b|. ||a| −|b|| |a + b| |a| + |b|.
a, b, c, x, y, z, u, v, t 0
a + b + c 3
3
√
abc.
3
(a + x)(b + y)(c + z)
3
√
3
√
abc
a + b + c +
3
√
abc 4
3
√
abc a + b + c 3
3
√
abc.
a + x, b + y, c + z a + x = 0,
a = x = 0 a + x, b + y, c + z = 0 :
a
a + x
+
b
b + y
+
c
c + z
(a + x + u)(b + y + v)(c + z + t)
3
(a + x)(b + y)(c + z)+
3
√
uvt
3
(a + x + u)(b + y + v)(c + z + t)
3
√
abc+
3
√
xyz+
3
√
uvt.
a, b, c 0.
1
1 + a
2
+
1
1 + b
2
2
2
+
1
(1 + b)
2
2
1 + ab
ab 1.
1
(1 + a)
2
+
1
(1 + b)
2
+
1
(1 + c)
2
3
1 + abc
a, b, c 1.
7Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên
1
√
1 + a
2
+
1 +
a + b + c
3
2
a, b, c
1
√
2
.
(ab−1)(a−b)
2
0.
1
1 + a
2
+
1
1 + b
2
2
1 + ab
ab 1.
a, b, c 1
1 + bc
2
1 + abc
1
1 + c
2
+
1
1 + a
2
2
1 + ca
2
1 + abc
1
1 + a
2
+
1
1 + b
2
+
1
1 + c
2
3
1
(1 + a)
2
+
1
(1 + b)
2
2
1 + ab
2
1 + abc
1
(1 + b)
2
+
1
(1 + c)
2
2
1 + bc
2
1 + abc
1
(1 + c)
2
x > 0 y
= −x(1+x
2
)
−3/2
< 0. y
y” = 3x
2
(1 + x
2
)
−5/2
− (1 + x
2
)
−3/2
=
2x
2
− 1
(1 + x
2
)
5
0 x
1
+
1
√
1 + c
2
3
1 +
a + b + c
3
2
8Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên
a, b, c
1
√
2
.
ab > 1 a = 2, b = 1
ab = 2 > 1
1
(1 + a)
2
+
1
(1 + b)
2
=
2
1 + ab
.
A B A −B A − B 0.
a, b, c a
2
+ b
2
+ c
2
ab + bc + ca.
a
2
+b
2
+c
2
−ab−bc−ca =
1
2
(a−b)
2
+(b−c)
2
+(c−a)
2
0
a
T =
1 −a
128
1 −a
1
1 −a
.
a, b, c
a
2
+ b
2
+ c
2
< 2(ab + bc + ca).
a
2
> (b −c)
2
, b
2
> (c −a)
2
c
2
> (a −b)
2
a
2
1.
(b
2
+ c
2
)(x
2
+ y
2
) (bx + cy)
2
= a
2
x
2
+ y
2
1.
9Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên
a, b ∈ [1; 2]
a + b
1
a
+
1
b
a + b
2
a
+
2
b
a + b
1
a
+
1
b
9
2
.
a, b, c ∈ [−2; 3] a + b + c = 0
a
2
+ b
2
+ c
2
18.
a, b, c ∈ [−2; 3] (a + 2)(a − 3) 0, (b + 2)(b − 3) 0
(c + 2)(c −3) 0. a
2
+ (b + c)
2
−2bc a
2
+ (6 −a)
2
= 2a
2
−12a + 36.
a
2
+ b
2
+ c
2
2(a
2
−6a + 18) = 2[(a −2)(a −4) + 10] 20.
a = 4, b = 2, c = 0.
a b c a
4
+ b
4
+ c
4
a
3
b + b
3
= (a −b)(a
3
− c
3
) + (b − c)(b
3
− c
3
) 0.
a
4
+ b
4
+ c
4
a
3
b + b
3
c + c
3
a.
a, b, c 0
a
b + c
+
b
c + a
+
c
−
a
b + c
=
a
2
(b + c) − a(b
2
+ c
2
)
(b
2
+ c
2
)(b + c)
.
a
2
b
2
+ c
2
−
a
b + c
=
ab(a −b) + ac(a − c)
(b
2
b + c
+
b
c + a
+
c
a + b
= ab(a −b)
1
(b
2
+ c
2
)(b + c)
−
1
(c
2
+ a
2
)(c + a)
+ bc(b −c)
1
(c
2
+ a
2
)(c + a) (b
2
+ c
2
)(b + c); a > b
(c
2
+ a
2
)(c + a) > (b
2
+ c
2
)(b + c).
ab(a −b)
1
(b
2
+ c
2
)(b + c)
−
1
(c
2
+ a
2
)(c + a)
2
3
=
3
8
(a + b)(a − b)
2
0.
2ab
1
2
(a + b)
2
T = a
3
+ b
3
+ 2 −2ab −a −b
1
4
(a + b)
3
−
1
2
(a + b)
2
− (a + b) + 2 =
1
a
2
a
2
− (b − c)
2
= (a + b − c)(a −b + c)
b
2
b
2
− (c − a)
2
= (a + b − c)(−a + b + c)
c
2
c
2
− (a − b)
2
= (−a + b + c)(a −b + c).
a+b−c < 0 a−b+c > 0, −a+b+c > 0. (a+b−c)(a−b+
c)(−a+b+c) < 0 abc. a+b−c 0,
a−b+c 0 −a+b+c 0 a
2
(1 + x
2
+ y
2
)
2
3(x + x
3
y + y
3
).
4(a
2
+ b
2
+ c
2
− ab − bc −ca)
(a
2
+ b
2
+ c
2
)
2
− 3(a
3
b + b
(a
3
+ b
3
+ c
3
) −(a
2
b + b
2
c + c
2
a) −2(ab
2
+ bc
2
+ ca
2
) + 6abc
2
0.
a = b = c.
a, b, c > 0
1
a
+
1
b
+
2
b
+
a
2
+ b
2
c
2(a + b + c).
b
2
+ c
2
a
+ a +
c
2
+ a
2
b
+ b +
a
2
+ b
2
c
+ c 3(a + b + c)
3(a
2
+b
1
a
+
1
b
+
1
c
(a+b+c)
2
1
a
+
1
b
+
1
c
9(a+b+c).
a, b, c, d > 0.
1
1
a
+
1
b
+
1
d
(a + c)(b + d).
1
1
a
+
1
b
+
1
1
c
+
1
d
1
1
a + c
+
1
b + d
=
(a + c)(b + d)
a + b + c + d
1
1
a
+
2
+
√
1 −b
2
2
1 −
a + b
2
2
(1 −a
2
)(1 −b
2
) 1 −ab.
(1 − a
2
)(1 − b
2
) (1 − ab)
2
(a − b)
2
0.
√
1 −a
+ b + 1 −5a) 0 :
b
2
+ b + 1 a
2
+ a + 1 2a + a + 1 5a.
5abc + c
3
5abc + 1 (c −1)(c
2
+ c + 1 −5ab) 0 :
c
2
+ c + 1 a
2
+ a + 1 5a 5ab.
a = 2, b = c = 1.
a, b, c ∈ [1; 2]
(a + b + c)
1
a
+
1
b
+
1
c
10.
.
ab + bc b
2
+ ac
c
a
+ 1
c
b
+
b
a
.
a
b
+
b
c
+
c
b
+
b
a
a
c
+
c
a
. 1 x 2. (x − 2)(x − 1) 0 x +
2
x
3.
1
2
1
x
x +
1
x
+
1
2
x +
2
x
3. x +
1
x
5
2
.
a
b
+
a
c
2
−a
2
)(c
2
+a
2
−b
2
).
a
2
, b
2
, c
2
(b
2
−(a −c)
2
)(c
2
−(a −b)
2
)(a
2
−(b −c)
2
)
(−a
)(a
2
+ b
2
− c
2
).
b
4
−2b
2
(a −c)
2
+ (a −c)
4
b
4
−(a
2
−c
2
)
2
(a −c)
2
(a
2
−b
2
+ c
2
+ c
2
)(a
2
− c
2
+ b
2
).
a, b, c ∈ [1; 2]
(a + b + c)
1
a
+
1
b
+
1
c
6
a
b + c
+
b
c + a
+
a
b + c
+
b
c + a
+
c
a + b
−3 =
(a −b)
2
(a + c)(b + c)
+
(b −c)
2
(b + a)(c + a)
+
(c −a)
2
(c + b)(a + b)
.
S
a
(b −c)
2
+ S
b
(c −a)
2
S
a
=
1
bc
−
3
(c + a)(c + b)
.
a b c,
S
a
S
b
S
c
. S
b
+ S
c
0,
S
b
+ S
c
=
1
a
1
. a b + c
V P
3
b + c
b + c
2b + c
+
b + c
2c + b
=
3
2b + c
+
3
2c + b
.
1
b
+
1
c
3
2b + c
+
3
2c + b
b
(c −a)
2
+ S
c
(a −b)
2
(S
b
+ S
c
)(a −b)
2
0.
a = b = c a = 2, b = c = 1
a, b, c 0, a + b + c = 3,
2(a
2
b
2
+ b
2
c
2
+ c
2
a
2
) + 3 3(a
2
+ b
2
c
2
+ c
2
a
2
= c
2
(a
2
+ b
2
) + a
2
b
2
= c
2
(2x
2
+ 2y
2
) + (x
2
− y
2
)
2
+ 2y
2
+ c
2
2x
2
+ c
2
.
2x
4
+ 4x
2
c
2
+ 3 3(2x
2
+ c
2
).
c = 3 − 2x 2x
4
+ 4x
2
(3 − 2x)
2
+ 3 6x
2
+ 3(3 − 2x)
2
c
a
− 1
+ c
2
a
b
− 1
0.
a
3
b
2
+ b
3
c
2
+ c
3
a
2
abc(a
2
+ b
2
+ c
2
3
.
T = 2(a
3
b
2
+ b
3
c
2
+ c
3
a
2
) − 2abc(a
2
+ b
2
+ c
2
)
T = a
3
(b
2
− 2bc + c
2
) + b
3
(c
3
(b −c)
2
+ b
3
(c −a)
2
+ c
3
(a −b)
2
− a
2
(b −c)
3
− b
2
(c −a)
3
− c
2
(a −b)
3
= a
2
(b −c)
2
(a −b + c) + b
2
(c −a)
T =
(b −c)(c − a)(a − b)(a + b + c)
(b −c)(c − a)(a − b)
= a + b + c
T 3
3
√
abc
17Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên
a, b, c > 0
a
3
b
3
+
b
3
c
3
+
c
3
a
3
a
b
+
b
c
b
c
+
c
a
a
3
b
3
+
b
3
c
3
+
c
3
a
3
1
9
a
b
+
b
c
+
c
2
3
1 + sin A sin B sin C
.
1
1 + sin
2
A
+
1
1 + sin
2
B
+
1
1 + sin
2
C
6
√
7
π
2
∠A, ∠B,
∠C
.
1 + a
2
2a
a
1 + a
2
1
2
.
b
1 + b
2
1
2
,
c
1 + c
2
1
2
.
a
1 + a
2
+
b
6
x
1 +
z
y
+
x
y
6
y
1 +
x
z
+
y
z
6
z
.
6
1
x
+
1
y
+
1
=
1
x
+
1
y
+
1
z
3
2
a
1 + a
2
+
b
1 + b
2
+
c
1 + c
2
1
1 + a
+
1
1 + b
+
b + c
+
b
c + a
+
c
a + b
> 2.
a
b + c + d
+
b
c + d + a
+
c
d + a + b
+
d
a + b + c
> 2.
2[1 +
a
b + c
+ 1 +
b + c
+
c
c + d
+
d
d + a
+
a
a + b
N =
c
b + c
+
d
c + d
+
a
d + a
+
b
a + b
.
M + S =
a + b
b + c
+
b + c
c + d
+
4(a + c)
a + b + c + d
+
4(b + d)
a + b + c + d
= 4.
19Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên
M + N + 2S 8 S 2.
a
b + c
2a
a + b + c
b
c + a
+
c
a + b
> 2.
b
c + d + a
+
c
d + a + b
+
d
a + b + c
2.
a = b+c+d, b = c+d+a, c = d+a+b, d = a+b+c.
a = b = c = d = 0 :
a
b + c + d
+
b
c + d + a
+
c
d + a + b
+
d
a + b + c
> 2.
a, b, c
.
a = 0 b = 0 c = 0
abc = 0. a
4
b
4
c
4
x =
ab
c
2
, y =
bc
a
2
, z =
ca
b
2
.
1
x
2
+
1
y
2
+
1
y
2
+
1
z
2
+ 3 2(x + y + z).
a, b, c > 0
3 + (a + b + c) + (
1
a
+
1
b
+
1
c
) + (
a
b
+
b
c
+
c
a
)
3(a + 1)(b + 1)(c + 1)
abc + 1
.
2
+
1
(1 + d)
2
1.
1
(1 + a)
2
+
1
(1 + b)
2
1
1 + ab
,
1
(1 + c)
2
+
1
(1 + d)
2
1
1 + cd
T
1
1 + ab
2
c + c
2
b + b
2
a).
c = min{a, b, c}. a = c+u, b =
c + v u, v 0.
V T = 48c
3
+ 48(u + v)c
2
+ (36u
2
+ 36v
2
+ 12uv)c + 12(u
3
+ v
3
)
V P = 48c
3
+ 48(u + v)c
2
+ 16(u + v)
2
c + 10u
2
v + 6uv
+
1
b
+
1
c
9
a + b + c
.
a + b + c 3
3
√
abc
1
a
+
1
b
+
1
c
3
3
√
abc
1
a
+
− b
2
− c
2
2
a
2
+ b
2
+ c
2
+ 2(
√
a +
√
b +
√
c) 9.
a
2
+
√
a +
√
a 3a, b
2
+
√
b +
√
c
(a + b)
2
9
4(a + b + c)
.
(a + b + c)
a
(b + c)
2
+
b
(c + a)
2
+
c
(a + b)
2
a
b + c
+
b
c + a
+
c
2
.
22Số hóa bởi Trung tâm Học liệu – Đại học Thái Nguyên
a, b, c > 0 a + b + c = 3.
T =
a
2
a + 2b
2
+
b
2
b + 2c
2
+
c
2
c + 2a
2
1.
[a
2
(a + 2b
2
) + b
2
(b + 2c
2
) + c
2
+ b
2
c
2
+ c
2
a
2
)
.
(a
2
+ b
2
+ c
2
)
2
a
3
+ b
3
+ c
3
+ 2(a
2
b
2
+ b
2
+ c
3
)(a + b + c) (a
2
+ b
2
+ c
2
)
2
3(a
2
+ b
2
+ c
2
) (a + b + c)
2
= 9 a
3
+ b
3
+ c
3
a
2
+ b
2
+ c
2
3
+ c
3
.
a = b = c = 1.
a, b, c, d > 0 a + b + c + d = abc + bcd +
cda + dab.
abc + bcd + cda + dab
a
2
+ 1
2
+
b
2
+ 1
2
+
c
2
+ 1
2
+
d
2
+ 1
+ 1
d + a
=
(a + b + c + d)
2
a + b + c + d
= a + b + c + d.
2
a + b +c + d
a
2
+ 1
a + b
+
b
2
+ 1
b + c
+
c
2
+ 1
c + d
+
d
2
+ 1
d + a
2
+ 1
2
+
c
2
+ 1
2
+
d
2
+ 1
2
.
a, b, c, d > 0, abcd = 1,
(a + b + c + d)
6
2
8
(a
2
+ 1)(b
2
+ 1)(c
2
+ 1)(d
2
+ 1).
+ 1).
xyzt = 1 (x
2
+ y
2
+ z
2
+ t
2
)
6
2
8
(x
3
+ yzt)(y
3
+ xzt)(z
3
+ xyt)(t
3
+ xyz).
4(x
3
+yzt)(y
3
+xzt) (x
3
+y
.
4(x
2
− xy + y
2
+ zt)(z
2
− zt + t
2
+ xy) (x
2
+ y
2
+ z
2
+ t
2
)
2
(x + y)(z + t) = xz + yt + xt + yz x
2
+ y
2
+ z
2
+ t
2
a, b, c > 0, abc = 1,
a + b
c + ac 5b
c
2
b + c
2
b + c
2
a + c
2
a + ab 5c
2[ab(a + b)+ bc(b +c) + ca(c + a)] 5(a + b+ c) −(ab + bc + ca).
b
2
a + b
2
a + a
2
b + a
2
b + c 5ab
b
2
c + b
2
c + c
2
b + c
2
b + a 5bc
a
a(a + c)
9
2
c
b
+
b
a
+
a
c
+
a
a + b
+
b
b + c
+
c
c + a
9
2
c + b
b
+
b + a
a
+
c
c + a
1.
a + b
4a
+
a
a + b
+
b + c
4b
+
b
b + c
+
c + a
4c
+
c
c + a
3.
3
4
a + b
a
+
b + c
b
+
)(c
2
− ca + a
2
) 12.
a b c.
b
2
−bc+c
2
b
2
, a
2
−ac+c
2
(a+c)
2
, a
2
−ab+b
2
(a+c)
2
−(a+c)b+b
2
.
M = b
2
(a+c)
−x
2
)
3
2
(s
2
−x
2
)(s
2
+ 3x
2
)
4
3
s
2
3
=
27
9
4
M 27 M 12.
c = 0, s
2
= 9x
2