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a ) \(\left(3x-1\right)^2+\left(3x+1\right)^2+2\left(3x+1\right)\left(1-3x\right)\)
\(=\left(1-3x\right)^2+\left(3x+1\right)^2+2\left(3x+1\right)\left(1-3x\right)\)
\(=\left(1-3x+3x+1\right)^2\)
\(=2^2=4\)
b ) \(\left(2x-3y\right)^2+\left(2x+3y\right)^2+\left(2x-3y\right)\left(2x+3y\right)\)
\(=4x^2-12xy+9y^2+4x^2+12xy+9y^2+4x^2-9y^2\)
\(=\left(4x^2+4x^2+4x^2\right)+\left(9y^2+9y^2-9y^2\right)+\left(12xy-12xy\right)\)
\(=12x^2+9y^2\)
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\(\left(2x+3y\right)\left(2x-3y\right)-\left(2x-1\right)^2+\left(3y-1\right)^2\)
\(=4x^2-9y^2-4x^2+4x-1+9y^2-6y+1=4x-6y\)
Thay x = 1 ; y = -1 ta được :
\(4+6=10\)
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\(C=\left(2x+3y\right)\left(2x-3y\right)-\left(2x-1\right)^2+\left(3y-1\right)^2\)
\(C=4x^2-6xy+6xy-9y^2-4x^2+4x-1+9y^2-6y+1\)
\(C=4x-6y\)
Giá trị cảu biểu thức trên tại x=1 và y=-1 là:
\(C=4.1-6.\left(-1\right)=4+6=10\)
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\(\left(2x^2+5x+3\right):\left(x+1\right)-\left(4x-5\right)\)
\(=\dfrac{2x^2+2x+3x+3}{x+1}-4x+5\)
\(=\dfrac{2x\left(x+1\right)+3\left(x+1\right)}{x+1}-4x+5\)
\(=\dfrac{\left(x+1\right)\left(2x+3\right)}{x+1}-4x+5\)
\(=2x+3-4x+5\)
\(=-2x+8\)
thay x=-2 vào biểu thức ta có:
\(=-2\left(-2\right)+8=4+8=12\)
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a. \(\left(2x^2+y\right)\left(2x^2-y\right)-4x^2+y^2=\left(2x^2\right)^2-y^2-4x^2+y^2\)
\(=4x^4-4x^2\)
b. \(\left(2x^2+y\right)^2-\left(2x^2-y^2\right)=4x^4+4x^2y+y^2-2x^2+y^2\)
\(=4x^4+4x^2y-2x^2+2y^2\)
c. \(\left(2x+1\right)\left(2x-1\right)-4x^2=\left(2x\right)^2-1^2-4x^2=4x^2-1-4x^2=-1\)
d. \(\left(2x^{3y}+y\right)^2-\left(y-2x^{3y}\right)^2\)
\(=\left(2x^{3y}+y+y-2x^{3y}\right)\left(2x^{3y}+y-y+2x^{3y}\right)\)
\(=2y.2.2x^{3y}=4y.2x^{3y}\)
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\(\left(2x+3y\right)^2+2\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)
\(\left(2x+3y\right)^2+2\left(2x+3y\right)+1\)
\(=\left(2x+3y+1\right)^2\)
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\(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy.\left(x^2-y^2-2y-1\right)\)
\(=2xy.[x^2-\left(y^2+2y+1\right)]\)
\(=2xy.[x^2-\left(y+1\right)^2]\)
\(=2xy.\left(x+y+1\right).\left(x-y-1\right)\)
Vậy chọn đáp án A
\(\left(2x+3y\right)^2+2\left(2x+3y\right)+1\)
\(=\left(2x+3y+1\right)^2\)
chú ý:đặt 2x+3y=t thì x2+2x+1=(x+1)2 đơn giản :))