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a) M = 8ab;
b) N = [ ( 3 a + + 2 ) + ( 1 – 2 b ) ] 2 = ( 3 a – 2 b + 3 ) 2 .
`Answer:`
`a)`
`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`
`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`
`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`
`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`
`=>A=-2x^2+28x-6`
`b)`
`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`
`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`
`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`
`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`
Thay `x=-7` vào ta được:
`B=10(-7)^2-2(-7)^3-7(-7)-6`
`=>B=10.49-2(-343)+49-6`
`=>B=490+686+49-6`
`=>B=1219`
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 \(x\in\left\{3;1;5;-1\right\}\)
a: \(A=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\cdot\left(x+1\right)\cdot x+\dfrac{1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{\left(x^2+1\right)\left(x+1\right)+x-1}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{x^3+x^2+x+1+x-1}{\left(x-1\right)}\cdot\dfrac{x+1}{2x+1}\)
\(=\dfrac{x^3+x^2+2x}{x-1}\cdot\dfrac{x+1}{2x+1}=\dfrac{x\left(x^2+x+2\right)\left(x+1\right)}{\left(x-1\right)\left(2x+1\right)}\)
b: Khi x=1/2 thì \(A=\dfrac{\dfrac{1}{2}\left(\dfrac{1}{4}+\dfrac{1}{2}+2\right)\left(\dfrac{1}{2}+1\right)}{\left(\dfrac{1}{2}-1\right)\left(2\cdot\dfrac{1}{2}+1\right)}=-\dfrac{33}{16}\)
a) \(\left(3x-2\right)^2-\left(2x+3\right)\left(2x-3\right)\)
\(=9x^2-12x+4-4x^2+9\)
\(=5x^2-12x+13\)
b) \(3x\left(5x-2\right)-\left(2x^2-1\right)\left(2-x\right)\)
\(=15x^2-6x-\left(4x^2-2x^3-2+x\right)\)
\(=15x^2-6x-4x^2+2x^3+2-x\)
\(=11x^2-7x+2x^3+2\)
\(=\left(a-1-b+1\right)\left(a-1+b-1\right)=\left(a-b\right)\left(a+b-2\right)\)
=(a−1−b+1)(a−1+b−1)=(a−b)(a+b−2)
hằng đẳng thức số 3