\(\sqrt{-2x^2+6}=x-1\left(đk:\sqrt{3}\ge x\ge1\right)\)

\(\Leftrightarrow-2x^2+6=x^2-2x+1\)

\(\Leftrightarrow3x^2-2x-5=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=\dfrac{5}{3}\left(tm\right)\end{matrix}\right.\)

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

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

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

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

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

Cho g(x) = 0

x + 1 = 0

x = -1

Để f(x) chia hết cho g(x) thì x = -1 cũng là nghiệm của f(x)

Hay f(1) = 0

3.1² + 2.1² - 7.1 - m + 2 = 0

-2 - m + 2 = 0

m = 0

Vậy m = 0 thì f(x) chia hết cho g(x)

Giải chi tiết của em đây :

F(x) = 3x² + 2x² - 7x - m + 2 

F(x) \(⋮\) x + 1 \(\Leftrightarrow\) F(x) \(⋮\) x - (-1)

Theo bezout ta có : F(x) \(⋮\) x - (-1) \(\Leftrightarrow\) F(-1) = 0

\(\Leftrightarrow\) 3(-1)² + 2(-1)² - 7.(-1) - m + 2 = 0

    3 + 2 + 7 - m + 2 =0

              14 - m = 0

                     m = 14

Kết luận với m = 14 thì F(x) chia hết cho x + 1 

làm ơn, mình đang cần rất gấp !!!!!!!!!!!!!

:((((((((((

Do x = -1 là nghiệm của phương trình

⇒ a - b - 1 - 2 = 0

⇒ a - b = 3

Tương tự ta có a + b = 1

Vậy a = 2 ; b = -1 

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)

                                                             Bài giải

a, \(\left|x-0,6\right|< \frac{1}{2}\)

* Nếu \(x-0,6< 0\) thì :

\(-\left(x-0,6\right)< \frac{1}{2}\)

\(-x+\frac{3}{5}< \frac{1}{2}\)

\(-x< \frac{1}{2}-\frac{3}{5}\)

\(-x< -\frac{1}{10}\)

\(x< \frac{1}{10}\)

                                                           Bài giải

a, \(\left|x-0,6\right|< \frac{1}{2}\)

* Nếu \(x-0,6< 0\) thì :

\(-\left(x-0,6\right)< \frac{1}{2}\)

\(-x+\frac{3}{5}< \frac{1}{2}\)

\(-x< \frac{1}{2}-\frac{3}{5}\)

\(-x< -\frac{1}{10}\)

\(x< \frac{1}{10}\)

                                                                         Bài giải

b, \(\left|2x-1\right|>\left|-\frac{3}{4}\right|\)

\(\left|2x-1\right|>\frac{3}{4}\)

* Nếu \(2x-1< 0\)  ta có : 

\(-\left(2x-1\right)>\frac{3}{4}\)

\(-2x+1>\frac{3}{4}\)

\(-2x>\frac{3}{4}-1\)

\(-2x>-\frac{1}{4}\)

\(x>-\frac{1}{4}\text{ : }\left(-2\right)\)

\(x>\frac{1}{8}\)

* Nếu \(2x-1>0\) ta có : 

\(2x-1>\frac{3}{4}\)

\(2x>\frac{3}{4}+1\)

\(2x>\frac{7}{4}\)

\(x>\frac{7}{4}\text{ : }2\)

\(x>\frac{7}{8}\)