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-x - y^2 - x^2 - y
rút gọn: y; ^2
-> -x - x
Theo mik là thế chứ không bt đúng hay sai đâu
Bài 2:
a: ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)\(A=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}\)
\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)
b: Để A>0 thì x-3>0
hay x>3
( x + 1 )2 +3( x - 5)(x+ 5)-( 2x-1)2
=x2 + 2x + 1 + 3(x2 - 25) - 4x2 - 4x + 1
= x2 + 2x + 1 + 3x2 - 75 - 4x2 - 4x + 1
= -2x - 73
k cho mk nhe!!
( x + 1 )2 +3( x - 5)(x+ 5)-( 2x-1)2
=x2+2x+1+3x2-75-4x2+4x-1
=(x2+3x2-4x2)+(2x+4x)-(1-1)-75
=6x-75
Vậy ms đúng bn kia sai r`
\(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\)
\(=\left(x^8+x^4+1\right)\left(x^8-x^4+1\right)\)
\(=x^{16}+x^8+1\)
\(\left(x^2+x+1\right)\left(x^2-x-1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\)
\(=\left(x^4-x^3-x^2+x^3-x^2-x+x^2-x-1\right)\) \(\left(x^{32}-x^{16}+x^4-x^{16}+x^8-x^2+x^8-x^4+1\right)\)
\(=\left(x^4-x^2-2x-1\right)\left(x^{32}-2x^{16}+2x^8-x^2+1\right)\)
\(\frac{2x+2}{\left(x+1\right)\left(x-1\right)}.ĐKXĐ:x\ne\pm1\)
\(=\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{2}{x-1}\)
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