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\(A=\left(2x+y\right)^2+\left(2x-y\right)^2+\left(4x^2-y^2\right)+3y\\ =\left(4x^2+4xy+y^2\right)+\left(4x^2-4xy+y^2\right)+\left(4x^2-y^2\right)+3y\\ =4x^2+4x^2+4x^2+4xy-4xy+y^2+y^2-y^2+3y=12x^2+3y-y^2\\ B=\left(x-2\right)\left(x+2\right)-\left(x+2\right)^2\\ =\left(x+2\right)\left(x-2-x-2\right)=-4\left(x+2\right)=-4x-8\\ C=\left(3x-4y\right)^2+\left(3x-4y\right)^2\\ =\left(9x^2-24xy+16y^2\right)+\left(9x^2-24xy+16y^2\right)\\ =18x^2-48xy+32y^2\)
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a,
\(\left(x^2-2xy+y^2\right)\left(x-y\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\left[\left(x^2-2xy+y^2\right)\left(x-y\right)\right]-\left[\left(x-y\right)\left(x^2+xy+y^2\right)\right]\)
\(=\left[\left(x-y\right)^2\left(x-y\right)\right]-\left(x-y\right)^3\)
\(=\left(x-y\right)^3-\left(x-y\right)^3\)
\(=0\)
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a) \(A=4x^2-4x+1+9-4x^2=-4x+10\)
\(=-4.\dfrac{1}{4}+10=9\)
b) \(B=x^3+xy-x^3-8y^3=y\left(x-8y^2\right)\)
\(=\left(-2\right).\left(32-32\right)=0\)
a: Ta có: \(A=\left(2x-1\right)^2+\left(3-2x\right)\left(3+2x\right)\)
\(=4x^2-4x+1+9-4x^2\)
\(=-4x+10\)
\(=-4\cdot\dfrac{1}{4}+10=-1+10=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)
Thay x = 15 vào bt A ta có
A = 9 . 15 = 135
b/ \(B=5x^2-20xy-4y^2+2xy=5x^2-4y^2\)
Thay x = -1/5 ; y = - 1/2 vào bt B ta có
\(B=5.\dfrac{1}{25}-4.\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
c/ \(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)
\(=9x^2y^2-xy^3-8x^3\)
Thay x = 1/2 ; y = 2 vào bt C ta có
\(C=9.4.\dfrac{1}{4}-\dfrac{1}{2}.8-8.\dfrac{1}{8}=9-4-1=4\)
d/ \(D=6x^2+10x-3x-5+6x^2-3x+8x-2\)
\(=12x^2+12x-3\)
\(\left|x\right|=2\Rightarrow x=\pm2\)
Thay x = 2 vào bt D có
\(D=12.4+12.2-3=69\)
Thay x = - 2 vào bt D ta có
\(D=12.4-12.2-3=21\)
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\(1)A=2x\left(x-y\right)-y\left(y-2x\right)\)
\(=2x^2-2xy-y^2+2xy\)
\(=2x^2-y^2=2.\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{1}{3}\right)^2\)
\(=\dfrac{8}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
\(2)B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(=5x^2-20xy-4y^2+20xy\)
\(=5x^2-4y^2=5.\left(-\dfrac{1}{5}\right)^2-4.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
\(3)C=\text{x.(x^2-y^2)-x^2(x+y)+y(x^2-x)}\)
\(=x^3-xy^2-x^3-x^2y+x^2y-xy\)
\(=-xy\left(x+1\right)\)
\(A=2x^2+2xy+x^2-2xy+y^2-4y^2=3x^2-3y^2\)
\(A=2x^2+2xy+x^2-2xy+y^2-4y^2=3x^2-3y^2\)