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Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
b/ \(4\left(x-1\right)\left(x+1\right)-5x\left(x-2\right)+x^2\)
= \(4\left(x^2-1\right)-5x^2+10x+x^2\)
= \(4x^2-4-5x^2+10x+x^2\)
= \(10x-4\)
= \(2\left(5x-2\right)\)
c/ \(\left(3-2x\right)\left(x-2\right)+4\left(x-1\right)\left(x-3\right)-2\left(x-2\right)\left(x+2\right)\)
= \(3x-6-2x^2+4x+4\left(x^2-4x-4\right)-2\left(x^2-4\right)\)
= \(3x-6-2x^2+4x+4x^2-16x-16-8x^2-18\)
= \(-9x-12\)
= \(-3\left(3x+4\right)\)
1, ( x + 3 )( x- 4 ) + ( x - 4 ) mũ 2
=x^2+4x+3x-12+x^2-8x+16
=2x^2-x+4
3, x( x -14 ) - 10(x - 1) mũ 2
=x^2-14x-10(x^2-2x+1)
=x^2-14x-10x^2-20x+10
=-9x^2-34x+10
Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
( x - 1 ) . ( x3 + x2 + x + 2 )
= x4 + x3 + x2 + 2x - x3 - x2 - x - 2
= x4 + x - 2
( x + 1 ) . ( x4 - x3 + x2 - x + 1 )
= x5 - x4 + x3 - x2 + x + x4 - x3 + x2 - x + 1
= x5 + 1
Study well
1) (x - 1)(x3 + x2 + x + 2)
= x(x3 + x2 + x + 2) - (x3 + x2 + x + 2)
= x4 + x3 + x2 + 2x - x3 - x2 - x - 2
= x4 + x - 2
2) (x + 1)(x4 - x3 + x2 - x + 1)
= x(x4 - x3 + x2 - x + 1) + x4 - x3 + x2 - x + 1
= x5 - x4 + x3 - x2 + x + x4 - x3 + x2 - x + 1
= x4 + 1