x=15 và x=-9 bạn nhé

a: =>5x+25-3x+6=25+18

=>2x+41=43

=>2x=2

=>x=1

b: =>4x+8=3x+3+17

=>4x+8=3x+20

=>x=12

a: =>5x+25-3x+6=25+18

=>2x+41=43

=>2x=2

=>x=1

b: =>4x+8=3x+3+17

=>4x+8=3x+20

=>x=12

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

\(a.\dfrac{3}{2}+\dfrac{-1}{3}< \dfrac{x}{6}< \dfrac{1}{9}+\dfrac{31}{18}\)

\(\Leftrightarrow\dfrac{7}{6}< \dfrac{x}{6}< \dfrac{11}{6}\)

\(\Leftrightarrow7< x< 11\)

\(\Leftrightarrow x\in\left\{8;9;10\right\}\)

\(b.\dfrac{-5}{12}+\dfrac{7}{12}+\dfrac{-1}{12}< \dfrac{x}{12}< \dfrac{2}{15}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{12}< \dfrac{x}{12}< \dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{1}{12}< \dfrac{x}{12}< \dfrac{4}{12}\)

\(\Leftrightarrow1< x< 4\)

\(\Leftrightarrow x\in\left\{2;3\right\}\)

Bạn làm chi tiết chỗ 1/9+31/18 đc hông~

b: =>x/23=1+3/4+4/7=65/28

=>x=23*65/28=1495/28

c: =>3/5:x=3/5-1/4-1/2=9/40

=>x=3/5:9/40=8/3

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

giải phần còn lại giúp mình được ko?

Câu 1: 

a) Ta có: x-3 là ước của 13

\(\Leftrightarrow x-3\inƯ\left(13\right)\)

\(\Leftrightarrow x-3\in\left\{1;-1;13;-13\right\}\)

hay \(x\in\left\{4;2;16;-10\right\}\)(thỏa mãn)

Vậy: \(x\in\left\{4;2;16;-10\right\}\)

b) Ta có: \(x^2-7\) là ước của \(x^2+2\)

\(\Leftrightarrow x^2+2⋮x^2-7\)

\(\Leftrightarrow x^2-7+9⋮x^2-7\)

mà \(x^2-7⋮x^2-7\)

nên \(9⋮x^2-7\)

\(\Leftrightarrow x^2-7\inƯ\left(9\right)\)

\(\Leftrightarrow x^2-7\in\left\{1;-1;3;-3;9;-9\right\}\)

mà \(x^2-7\ge-7\forall x\)

nên \(x^2-7\in\left\{1;-1;3;-3;9\right\}\)

\(\Leftrightarrow x^2\in\left\{8;6;10;4;16\right\}\)

\(\Leftrightarrow x\in\left\{2\sqrt{2};-2\sqrt{2};-\sqrt{6};\sqrt{6};\sqrt{10};-\sqrt{10};2;-2;4;-4\right\}\)

mà \(x\in Z\)

nên \(x\in\left\{2;-2;4;-4\right\}\)

Vậy: \(x\in\left\{2;-2;4;-4\right\}\)

Câu 2: 

a) Ta có: \(2\left(x-3\right)-3\left(x-5\right)=4\left(3-x\right)-18\)

\(\Leftrightarrow2x-6-3x+15=12-4x-18\)

\(\Leftrightarrow-x+9+4x+6=0\)

\(\Leftrightarrow3x+15=0\)

\(\Leftrightarrow3x=-15\)

hay x=-5

Vậy: x=-5

mk mới lớp 6 thui, sao bn lại giải phần b câu 1 kiểu này

a: =>2x=-18+5=-13

=>x=-13/2

b: =>3^x-1=81

=>x-1=4

=>x=5

c: =>4(5-x)=24

=>5-x=6

=>x=-1