\(x^5-x^3+x^2-1=x^3\left(x^2-1\right)+\left(x^2-1\right)=\left(x^2-1\right)\left(x^3+1\right)=\left(x-1\right)\left(x+1\right)^2\left(x^2-x+1\right)\)

a) Ta có: \(N=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{3\sqrt{x}}{x-\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1-3}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

b) Ta có: \(\left\{{}\begin{matrix}x+3y=9\\2x-5y=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+6y=18\\2x-5y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11y=22\\x+3y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=9-3y=9-3\cdot2=3\end{matrix}\right.\)

GIÚP MÌNH VỚI

Chọn A

a) Thay x=-1 vào g(x),ta được:

\(2\cdot\left(-1\right)^2+5a=0\)

\(\Leftrightarrow5a=-2\)

hay \(a=-\dfrac{2}{5}\)

B

\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\dfrac{4x}{3x-3}\\ =\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}.\dfrac{3x-3}{4x}\\ =\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}.\dfrac{3\left(x-1\right)}{4x}\\ =\dfrac{4x.3\left(x-1\right)}{4x\left(x-1\right)\left(x+1\right)}\\ =\dfrac{3}{x+1}\)

\(\left|x^2-1\right|=2x+1\left(dk:2x+1\ge0\Leftrightarrow2x\ge-1\Leftrightarrow x\le-\dfrac{1}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+1\\x^2-1=-2x-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1-2x-1=0\\x^2-1+2x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-2=0\\x^2+2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-2+3=3\\x.\left(x+2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=3\\\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1^2\right)-\left(\sqrt{3}\right)^2=0\\\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1-\sqrt{3}\right).\left(x-1+\sqrt{3}\right)=0\\\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1-\sqrt{3}\left(loai\right)\\x=1+\sqrt{3\left(loai\right)}\end{matrix}\right.\\\left[{}\begin{matrix}x=0\left(loai\right)\\x=-2\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy x =  -2

\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)

\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)

\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)

hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)

a)Ta có:5/3.x^2-1/2.x^2y

          =(5/3-1/2).x^2y

          = 7/6.x^2y(Bậc 3)

b)Ta có: 7/6.(-2)^2(-1)

             = 7/6.4.(-1)

             = 7/6.(-4)

             =-28/6

a, -  A=\(\dfrac{5}{3}\).x².y-\(\dfrac{-1}{2}\).x².y

      =\(\dfrac{13}{6}\).x².y

- Bậc= 3.

b, A=\(\dfrac{13}{6}\).(-2)².(-1)

      =\(\dfrac{13}{6}\).4.(-1)

      =\(\dfrac{-26}{3}\)

bạn Tt_Cindy_tT làm sai rồi 5/3-1/2=7/6 chứ ko bằng 13/6

DKXD : x khac -1

\(\frac{-x}{x+1}\)+ 3 =\(\frac{2x+3}{x+1}\)

<=> \(\frac{-x}{x+1}\)+\(\frac{3\left(x+1\right)}{x+1}\)= \(\frac{2x+3}{x+1}\)

=>   -x + 3x +3 = 2x +3

<=> 2x -2x =3-3

<=>  0x=0

<=> x=0(TMDK)

`[2-x]/x >= 1`

`<=>[2-x-x]/x >= 0`

`<=>[2-2x]/x >= 0`

`<=>0 < x <= 1`

     `->\bb B`

B

B