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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`
Bài 1:
\(x^3-x^2-x+1=0\)
\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy x = 1 hoặc x = -1
Bài 2:
\(2x-2x^2-1=-2\left(x^2-x+\dfrac{1}{2}\right)\)
\(=-2\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4}\right)\)
\(=-2\left(x^2-\dfrac{1}{2}\right)^2-\dfrac{1}{2}< 0\)
\(\Rightarrowđpcm\)
Có: \(x^3-y^3=-3xy\left(y-x\right)\)
\(\Leftrightarrow x^3-y^3=-3xy^2+3x^2y\)
\(\Leftrightarrow x^3-3x^2y+3xy^2-y^3=0\)
\(\Leftrightarrow\left(x-y\right)^3=0\)
\(\Leftrightarrow x-y=0\Leftrightarrow x=y\)
Khi đó bt A trở thành:
\(A=\left(2x-y\right)\left(y-2x\right)\left(y-y\right)^2=\left(2x-y\right)\left(y-2x\right)\cdot0=0\)
Q=\(-2\left(X^2-2.X.\frac{5}{2}+\frac{25}{4}\right)-\frac{25}{4}\)
Q=\(-2\left(X-\frac{5}{2}\right)^2-2.\frac{-25}{4}\)
Q=\(-2\left(X-\frac{5}{2}\right)^2+\frac{25}{2}\)
=>\(GTLN\) LÀ 25/2 TẠI X=5/2
N=
\(\frac{2x+2}{\left(x+1\right)\left(x-1\right)}.ĐKXĐ:x\ne\pm1\)
\(=\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{2}{x-1}\)
~~