\(\left(x-1\right)^{x+1}-\left(x-1\right)^{x+12}=0\\ \Leftrightarrow\left(x-1\right)^{x+1}\left[1-\left(x-1\right)^{11}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+11}=0\\\left(x-1\right)^{11}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) (3x – 2)(4x + 5) = 0

 ⇔ 3x – 2 = 0 hoặc 4x + 5 = 0

 ⇔ 3x = 2 hoặc 4x  = -5

 ⇔ x = \(\dfrac{2}{3}\) hoặc x  = \(\dfrac{-5}{4}\)

 Vậy tập nghiệm là S = {\(\dfrac{2}{3}\); \(\dfrac{-5}{4}\)}

b) 2x(x – 3) + 5(x – 3) = 0

 ⇔ (x – 3)(2x + 5) = 0

 ⇔ x – 3 = 0 hoặc 2x + 5 = 0

 ⇔ x = 3 hoặc 2x = \(-5\)

 ⇔ x = 3 hoặc x = \(\dfrac{-5}{2}\)

 Vậy tập nghiệp là S = {3; \(\dfrac{-5}{2}\)}

a) |x-3|+5=-7

   |x+3|=-7-5

   |x-3|=-12( vô lí) 

Vậy không có giá trị nào của x thỏa mãn đề bài 

b)||x-7|-6|=3

suy ra |x-7|-6=3 hoặc |x-7|-6 = -3

          |x-7|=3+6        |x-7|=-3+6  

          |x-7|=9           |x-7|=3

suy ra x-7=9 hoặc x-7=-9 hoặc x-7=3 hoặc x-7=-3

TH1 ; x-7=9        TH2 : x-7=-9      TH3 : x-7=3        TH4 : x-7=-3 

          x=9+7                x= -9+7             x=3+7                x=-3+7 

         x=16                  x=-2                  x=10                  x=4

Vậy x thuộc {10;-2;10;4}

c) |x-1|=5-x 

suy ra x-1=5-x hoặc x-1=-(5-x)

         x+x=1+5        x-1=-5+x 

         2x=6             x-x=1-5|(vô lí) 

          x=6:2          

         x=3

Vậy x=3 

d) |1-x|+x+3=6

   |1-x|+x=6-3

   |1-x|+x=3

   |1-x|=3-x

suy ra 1-x=3-x hoặc 1-x=-(3-x) 

TH1 : 1-x=3-x                                 TH2 : 1-x=(-3-x)

        -x+x=-1+3 (vô lí )                   1-x=-3+x

                                                    -x-x=-1-3

                                                   -2x=-4

                                                    x= -4:(-2)

                                                   x=2

vậy x=2

a, l x-3 l +5 = 7                   

    l x-3 l =2 

=> x-3=2 hoac x-3=-2

     x=5   hoac x=1

Ta có : x + 4 > x - 9 

\(\left\{{}\begin{matrix}x+4>0\\x-9< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-4\\x< 9\end{matrix}\right.\)<=> -4 < x < 9 

\(\left(x+4\right)\left(x-9\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+4>0\\x-9< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+4< 0\\x-9>0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-4\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x< -4\\x>9\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-4< x< 9\)

a) |x| + |x-2| = 2 

TH1: \(x< 0\Rightarrow-x-x+2=2\Leftrightarrow-2x=0\Rightarrow x=0\)

TH2: \(0\le x< 2\Rightarrow x-x+2=0\Rightarrow0x=-2\)(PT vô nghiệm)

TH3: \(x>2\Rightarrow x+x-2=0\Rightarrow2x=2\Rightarrow x=1\)(loại vì 1<2 không thỏa mãn ĐK)

b) |x-1| + |x-4| = 3x

TH1: \(x-1\le0\Rightarrow x\le1\)

\(-x+1-x+4=3x\Leftrightarrow5-2x=3x\Leftrightarrow5x=5\Rightarrow x=1\)

TH2: \(1\le x\le4\)

\(x-1-x+4=3x\Leftrightarrow3x=3\Rightarrow x=1\)

TH3: \(x\ge4\)

\(x-1+x-4=3x\Leftrightarrow2x-5=3x\Rightarrow x=-5\)(loại vì -5<4 không thỏa mãn với ĐK)

Em cảm ơn anh ạ :))

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\Leftrightarrow-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

=> x \(\in\) {-1;0;1;2;3;4;5;6}

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\Leftrightarrow\)\(\frac{9-10}{12}\le\frac{x}{12}< 1-\left(\frac{8-3}{12}\right)\)

\(\Leftrightarrow\)\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

\(\Leftrightarrow-1\le x< 7\)

Mà x nguyên

=>x={-1;0;1;2;3;4;5;6}

Ta có:\(\frac{x+1}{11}+\frac{x+2}{10}=\frac{x+3}{9}+\frac{x+4}{8}\)

\(\Rightarrow1+\frac{x+1}{11}+1+\frac{x+2}{10}=1+\frac{x+3}{9}+1+\frac{x+4}{8}\)

\(\Rightarrow\frac{x+12}{11}+\frac{x+12}{10}=\frac{x+12}{9}+\frac{x+12}{8}\)

\(\Rightarrow\frac{x+12}{11}+\frac{x+12}{10}-\frac{x+12}{9}-\frac{x+12}{8}=0\)

\(\Rightarrow\left(x+12\right)\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)

Mà \(\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)>0\)

\(\Rightarrow x+12=0\Rightarrow x=-12\)

\(\frac{x+1}{11}+\frac{x+2}{10}=\frac{x+3}{9}+\frac{x+4}{8}\)

<=> \(\frac{x+1}{11}+\frac{x+2}{10}-\frac{x+3}{9}-\frac{x+4}{8}=0\)

<=> \(\left(\frac{x+1}{11}+1\right)+\left(\frac{x+2}{10}+1\right)-\left(\frac{x+3}{9}+1\right)-\left(\frac{x+4}{8}+1\right)=0\)<=> \(\frac{x+12}{11}+\frac{x+12}{10}-\frac{x+12}{9}-\frac{x+12}{8}=0\)

<=> \(\left(x+12\right)\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)

<=> x + 12 = 0.Vì \(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\ne0\)

<=> x = -12

\(\frac{x+1}{11}+\frac{x+2}{10}=\frac{x+3}{9}+\frac{x+4}{8}\)

\(\Rightarrow\left(\frac{x+1}{11}+1\right)+\left(\frac{x+2}{10}+1\right)=\left(\frac{x+3}{9}+1\right)+\left(\frac{x+4}{8}+1\right)\)

\(\Rightarrow\frac{x+12}{10}+\frac{x+12}{11}=\frac{x+12}{9}=\frac{x+12}{8}\)

\(\Rightarrow\frac{x+12}{11}+\frac{x+12}{10}-\frac{x+12}{9}-\frac{x+12}{8}=0\)

\(\Rightarrow\left(x+12\right)\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)

Mà \(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\ne0\)

\(\Rightarrow x+12=0\)

\(\Rightarrow x=-12\)

Vậy \(x=-12\)

sai đề tích nha

Đề sai nha! Đăng lại nhé!

ung ho mih nha cac bn

\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(\frac{-1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)

\(\hept{\begin{cases}\frac{1}{7}x-\frac{2}{7}=0\\\frac{-1}{5}x+\frac{3}{5}=0\\\frac{1}{3}x+\frac{4}{3}=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=3\\x=-4\end{cases}}}\)

KL

b, \(\left|\frac{5}{3}x\right|=\left|\frac{-1}{6}\right|\)

\(\left|\frac{5}{3}x\right|=\frac{1}{6}\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}x=\frac{1}{6}\\\frac{5}{3}x=\frac{-1}{6}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}\\x=\frac{-1}{10}\end{cases}}}\)

KL

c, \(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\left|\frac{-3}{4}\right|\)

\(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\frac{3}{4}\)

\(\Rightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|=\frac{3}{2}\)

\(\Rightarrow\orbr{\begin{cases}\frac{3}{4}x-\frac{3}{4}=\frac{3}{2}\\\frac{3}{4}x-\frac{3}{4}=\frac{-3}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{-3}{4}\end{cases}}}\)

KL

lang nhang qua