b ) 3 - ( x - 5 ) = - 7 + ( - 8 )

     3 - ( x - 5 ) = - 15

            x - 5   = 3 - ( - 15 )

            x - 5   = 18

                 x   = 18 + 5

                 x   = 23

Vậy x = 23

\(a,\) Vì \(2x⋮x\Rightarrow3⋮x\Rightarrow x\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(b,\left(8x+4\right)⋮\left(2x-1\right)\\ \Rightarrow\left[\left(8x-4\right)+8\right]⋮\left(2x-1\right)\\ \Rightarrow\left[4\left(2x-1\right)+8\right]⋮\left(2x-1\right)\)

\(Vì.4\left(2x-1\right)⋮\left(2x-1\right)\Rightarrow8⋮\left(2x-1\right)\Rightarrow\left(2x-1\right)\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

Ta có bảng:

Vậy \(x\in\left\{0;1\right\}\)

\(c,\left(x^2-x+7\right)⋮\left(x-1\right)\\ \Rightarrow\left[x\left(x-1\right)+7\right]⋮\left(x-1\right)\)

\(Vì.x\left(x-1\right)⋮\left(x-1\right)\Rightarrow7⋮\left(x-1\right)\Rightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

Ta có bảng:

Vậy \(x\in\left\{-6;0;2;8\right\}\)

a. \(\left|x-3\right|-7=13\Leftrightarrow\left|x-3\right|=20\)

TH1: \(x-3=20\Leftrightarrow x=23\)

TH2: \(x-3=-20\Leftrightarrow x=-17\)

b. \(72-3.\left|x+1\right|=9\Leftrightarrow\left|x+1\right|=21\)

TH1: \(x+1=21\Leftrightarrow x=20\)

TH2: \(x+1=-21\Leftrightarrow x=-22\)

c. \(17-\left(43-\left|x\right|\right)=45\Leftrightarrow\left|x\right|=71\)

TH1: \(x=71\)

TH2: \(x=-71\)

d. \(3\left|x-1\right|-5=7\Leftrightarrow\left|x-1\right|=4\)

TH1: \(x-1=4\Leftrightarrow x=5\)

TH2: \(x-1=-4\Leftrightarrow x=-3\)

e. \(-12\left(x-5\right)+7\left(3-x\right)=5\Leftrightarrow-12x+60+21-7x=5\Leftrightarrow-19x=-76\Leftrightarrow x=4\)

a.

\(\left|x-3\right|-7=13\)

\(\left|x-3\right|=13+7\)

\(\left|x-3\right|=20\)

\(x-3=\pm20\)

• \(x-3=20\)

               \(x=20+3\)

               \(x=23\)

• \(x-3=-20\)

               \(x=-20+3\)

               \(x=-17\)

Vậy x = 23 hoặc x = - 17.

b.

\(72-3\times\left|x+1\right|=9\)

\(3\times\left|x+1\right|=72-9\)

\(3\times\left|x+1\right|=63\)

\(\left|x+1\right|=63\div3\)

\(\left|x+1\right|=21\)

\(x+1=\pm21\)

• \(x+1=21\)

               \(x=21-1\)

               \(x=20\)

• \(x+1=-21\)

               \(x=-21-1\)

               \(x=-22\)

Vậy x = 20 hoặc x = - 22.

c.

\(17-\left(43-\left|x\right|\right)=45\)

\(43-\left|x\right|=17-45\)

\(43-\left|x\right|=-28\)

\(\left|x\right|=43-\left(-28\right)\)

\(\left|x\right|=71\)

\(x=\pm71\)

Vậy x = 71 hoặc x = - 71.

d.

\(3\times\left|x-1\right|-5=7\)

\(3\times\left|x-1\right|=7+5\)

\(3\times\left|x-1\right|=12\)

\(\left|x-1\right|=12\div3\)

\(\left|x-1\right|=4\)

\(x-1=\pm4\)

• \(x-1=4\)

               \(x=4+1\)

               \(x=5\)

• \(x-1=-4\)

               \(x=-4+1\)

               \(x=-3\)

Vậy x = 5 hoặc x = - 3.

e.

\(-12\times\left(x-5\right)+7\times\left(3-x\right)=5\)

\(-12\times x+\left(-12\right)\times\left(-5\right)+7\times3-7\times x=5\)

\(-12\times x+60+21-7\times x=5\)

\(-19\times x+81=5\)

\(-19\times x=5-81\)

\(-19\times x=-76\)

\(x=\frac{-76}{-19}\)

\(x=4\)

Chúc bạn học tốt

Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

a) \(2x^2-4x=0\)

\(2x\left(x-2\right)=0\)

TH1:2x=0⇒x=0

TH2:x-2=0⇒x=2

\(a,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow x-4=0\Leftrightarrow x=4\)

mình cần gấp

a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)

\(\Leftrightarrow2x^2-2x-2x^2=-6\)

\(\Leftrightarrow x=3\)

b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

a)(2x-3)(x+5)=0

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

Vậy x=3/2 hoặc x=-5

a) \(\left(2x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};-5\right\}\)

b) \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{2;\dfrac{7}{2}\right\}\)

c) \(5x\left(2x-3\right)-6x+9=0\)

\(\Leftrightarrow5x\left(2x-3\right)-3\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\5x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};\dfrac{3}{5}\right\}\)

a. -6x=18

\(\Rightarrow x=18:\left(-6\right)\)

\(\Rightarrow x=-3\)

Vậy x=-3

b,2x-(-3)=7

\(\Rightarrow2x=7+\left(-3\right)\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=4:2\)

\(\Rightarrow x=2\)

Vậy x=2

c,(x-5)(x+6)=0

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0+5\\x=0-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)

- 6x = 18

x = 18 : ( - 6 )

x = -  3