rút gọn (x+2)^2 - (x+4)(x+1)
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a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
\(T=\dfrac{2\left(x-1\right)}{\sqrt{x}+1}+\dfrac{x-4}{\sqrt{x}-2}\)
\(T=\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\)
\(T=2\left(\sqrt{x}+1\right)+\left(\sqrt{x}+2\right)\)
\(T=2\sqrt{x}+2+\sqrt{x}+2\)
\(T=3\sqrt{x}+4\)
\(x=4\)
\(\Rightarrow T=3\sqrt{4}+4=3.2+4=10\)
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)^4 - 6(x + 1)^2 - (x^2 - 2)(x^2 + 2)
= (x^2 + 2x + 1)(x^2 + 2x + 1) - 6(x^2 + 2x + 1) - (x^2 - 2)(x^2 + 2)
= x^2.(x^2 + 2x + 1) + 2x.(x^2 + 2x + 1) + x^2 + 2x + 1 - (x^2 - 2)(x^2 + 2)
= x^4 + 2x^3 + x^2 + 2x^3 + 4x^2 + 2x + x^2 + 2x + 1 - 6x^2 - 12x - 6 - x^2 + 2^2
= 4x^3 - 8x - 1
\(\left(x+1\right)^4-6\left(x+1\right)^2-\left(x^2-2\right)\left(x^2+2\right)\)
\(=\left(x^2+2x-5\right)\left(x^2+2x+1\right)-x^4+2\)
\(=x^4+2x^3+x^2+2x^3+4x^2+2x-5x^2-10x-5-x^4+4\)
\(=4x^3-8x-1\)
Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x^2-1\right)-x+2\)
\(=x^3-8-x^3+x-x+2\)
\(=-6\)
a: Ta có: \(\left(x+3\right)^2-\left(x+1\right)\left(x-4\right)\)
\(=x^2+6x+9-x^2+3x+4\)
\(=9x+13\)
b: Ta có: \(x\left(x+2\right)\left(x-2\right)-x\left(x-1\right)^2\)
\(=x\left(x^2-4\right)-x\left(x^2-2x+1\right)\)
\(=x^3-4x-x^3+2x^2-x\)
\(=2x^2-5x\)
`(x+2)^2-(x+4)(x+1)`
`=x^2 +2*x*2+2^2 - (x^2+x+4x+4)`
`=x^2+4x+4-x^2-x-4x-4`
`= (x^2-x^2)+(4x-x-4x)+(4-4)`
`= -x`
thank