Bỏ dấu ngoặc rồi thu gọn biểu thức :
a, \(A=\left(a-2b+c\right)-\left(a-2b-c\right)\)
b, \(B=\left(-x-y+3\right)-\left(-x+2-y\right)\)
c, \(C=2.\left(3a+b-1\right)-3\left(2a+b-2\right)\)
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30 tháng 4 2020
\(a^2b^2c^2+\left(a+1\right)\left(1+b\right)\left(1+c\right)\ge a+b+c+ab+bc+ca+3\)
\(\Leftrightarrow\left(abc\right)^2+abc-2\ge0\Leftrightarrow\left(abc+2\right)\left(abc-1\right)\ge0\Leftrightarrow abc\ge1\)
Áp dụng BĐT Cosi ta có:
\(\frac{a^3}{\left(b+2c\right)\left(2c+3a\right)}+\frac{b+2c}{45}+\frac{2c+3a}{75}\ge3\sqrt[3]{\frac{a^3}{\left(b+2c\right)\left(2c+3b\right)}\cdot\frac{b+2c}{45}\cdot\frac{2c+3a}{75}}=\frac{a}{5}\left(1\right)\)
Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(c+2a\right)\left(2a+3b\right)}+\frac{c+2a}{45}+\frac{2a+3b}{75}\ge\frac{b}{5}\left(2\right)\\\frac{c^3}{\left(a+2b\right)\left(2b+3c\right)}+\frac{a+2b}{45}+\frac{2b+3c}{75}\ge\frac{c}{5}\left(3\right)\end{cases}}\)
Từ (1)(2)(3) ta có:
\(P+\frac{2\left(a+b+c\right)}{15}\ge\frac{a+b+c}{5}\Leftrightarrow P\ge\frac{1}{15}\left(a+b+c\right)\)
Mà \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow S\ge\frac{1}{5}\)
Dấu "=" xảy ra <=> a=b=c=1
20 tháng 4 2017
Bài giải:
a) (a + b)2 – (a – b)2 = (a2 + 2ab + b2) – (a2 – 2ab + b2)
= a2 + 2ab + b2 – a2 + 2ab - b2 = 4ab
Hoặc (a + b)2 – (a – b)2 = [(a + b) + (a – b)][(a + b) – (a – b)]
= (a + b + a – b)(a + b – a + b)
= 2a . 2b = 4ab
b) (a + b)3 – (a – b)3 – 2b3
= (a3 + 3a2b + 3ab2 + b3) – (a3 – 3a2b + 3ab2 – b3) – 2b3
= a3 + 3a2b + 3ab2 + b3 – a3 + 3a2b - 3ab2 + b3 – 2b3
= 6a2b
Hoặc (a + b)3 – (a – b)3 – 2b3 = [(a + b)3 – (a – b)3] – 2b3
= [(a + b) – (a – b)][(a + b)2 + (a + b)(a – b) + (a – b)2] – 2b3
= (a + b – a + b)(a2 + 2ab + b2 + a2 – b2 + a2 – 2ab + b2) – 2b3
= 2b . (3a2 + b2) – 2b3 = 6a2b + 2b3 – 2b3 = 6a2b
c) (x + y + z)2 – 2(x + y + z)(x + y) + (x + y)2
= x2 + y2 + z2+ 2xy + 2yz + 2xz – 2(x2 + xy + yx + y2 + zx + zy) + x2 + 2xy + y2
= 2x2 + 2y2 + z2 + 4xy + 2yz + 2xz – 2x2 – 4xy – 2y2 – 2xz – 2yz = z2
T
7 tháng 7 2019
2) Để sau đi (em chưa nghĩ ra)
3) \(A=\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)^2\left(x-y\right)+\left(y+z\right)^2\left(y-z\right)+\left(z+x\right)^2\left(z-x\right)\)
Đặt x - y = a; y - z = b => z - x = -(a+b)
\(A=\left(x+y\right)^2a+\left(y+z\right)^2b-\left(z+x\right)^2a-\left(z+x\right)^2b\)
\(=a\left[\left(x+y\right)^2-\left(z+x\right)^2\right]+b\left[\left(y+z\right)^2-\left(z+x\right)^2\right]\)
\(=\left(x-y\right)\left(x+y-z-x\right)\left(x+y+z+x\right)+\left(y-z\right)\left(y+z-z-x\right)\left(y+z+z+x\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(2x+y+z\right)-\left(y-z\right)\left(x-y\right)\left(2z+x+y\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
Em tính sai sót chỗ nào thì thông cảm cho em ạ :>
3 tháng 11 2017
1)
=2(a4+b4+c4-4a2b2-4a2c2-4b2c2)
=2a4+2b4+2c4-4a2b2-4a2c2-4b2c2
=(a4-2a2b2+b4)+(a4-2a2c2+c4)+(b4-2b2c2+c4
18 tháng 7 2017
a) \(\left(x^2-2x+2\right)\left(x-2\right)\left(x^2-2x+2\right)\left(x+2\right)\)
\(=\left(x^3-2x^2-2x^2+4x+2x-4\right)\left(x^3+2^3\right)\)
\(=\left(x^3-4x^2+6x-4\right)\left(x^3+8\right)\)
\(=x^6+8x^3-4x^5-32x^2+6x^4+48x-4x^3-32\)
\(=x^6-4x^5+4x^3-32x^2+48x-32\)
b) \(\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)
\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]+x^3-3x\left(x^2-1\right)\)
\(=2x\left[\left(x^2+2x+1\right)-\left(x^2-1\right)+\left(x^2-2x+1\right)\right]+x^3-\left(3x^3-3x\right)\)
\(=2x\left(x^2+2x+1-x^2+1+x^2-2x+1\right)+x^3-3x^3+3x\)
\(=2x\left(x^2+3\right)+x^3-3x^3+3x\)
\(=2x^3+6x-2x^3+3x\)
\(=9x\)
2 câu kia đợi tí đã nhé!
18 tháng 7 2017
c) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+\left(2a-b\right)^2\)
\(=\left(a^2+b^2+c^2+2ab+2bc+2ca\right)+\left(a^2+b^2+c^2+2ab-2bc-2ca\right)+\left(4a^2-4ab+b^2\right)\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ca+4a^2-4ab+b^2\)
\(=6a^2+3b^2+2c^2\)
d) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ca+2a^2+2ab+b^2\)
\(=4a^2+4b^2+2c^2+6ab.\)
a) \(A=\left(a-2b+c\right)-\left(a-2b-c\right)\)
\(A=a-2b+c-a+2b+c=2c\)
b) \(B=\left(-x-y+3\right)-\left(-x+2-y\right)\)
\(B=-x-y+3+x-2+y=1\)
c) \(C=2\left(3a+b-1\right)-3\left(2a+b-2\right)\)
\(C=6a+2b-2-6a-3b+6=4-b\)
a. \(A=\left(a-2b+c\right)-\left(a-2b-c\right)=a-2b+c-a+2b+c=0\)
b. \(B=\left(-x-y+3\right)-\left(-x+2-y\right)=-x-y+3+x-2+y=1\)
c. \(C=2\left(3a+b-1\right)-3\left(2a+b-2\right)=6a+2b-2-6b-3b+6=4-3b\)