|3x-5|=3/5
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a) 3/2.|x - 5/3| - 4/5 = 4/3.|x - 5/3| + 1
<=> 3/2.|x - 5/3| = 4/3.|x - 5/3| + 1 + 4/5
<=> 3/2.|x - 5/3| = 9/5 + 4|x - 5/3|/3
<=> 3/2.|x - 5/3| - 4.|x - 5/3|/3 = 9/5
<=> |x - 5/3|/6 = 9/5
<=> |x - 5/3| = 9/5.6
<=> |x - 5/3| = 54/5
<=> x - 5/3 = 54/5 hoặc x - 5/3 = -54/5
x = 54/5 + 5/3 x = -54/5 - 5/3
x = 187/15 x = -137/15
b) 2.|3x + 1| = 1/3.|3x + 1| + 5
<=> 2.|3x + 1| - 1/3.|3x + 1| = 5
<=> 5/3.|3x + 1| = 5
<=> 5.|3x + 1| = 5.3
<=> 5.|3x + 1| = 15
<=> |3x + 1| = 15 : 5
<=> |3x + 1| = 3
3x + 1 = 3 hoặc 3x + 1 = -3
3x = 3 - 1 3x = -3 - 1
3x = 2 3x = -4
x = 2/3 x = -4/3
=> x = 2/3 hoặc x = -4/3
c) làm tương tự câu a) mình hơi lời
Làm câu c) cho
\(\frac{1}{4}-\frac{5}{2}\left|3x-\frac{1}{5}\right|=\frac{2}{3}\left|3x-\frac{1}{5}\right|-\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{4}+\frac{2}{3}=\frac{2}{3}\left|3x-\frac{1}{5}\right|+\frac{5}{2}\left|3x-\frac{1}{5}\right|\)
\(\Leftrightarrow\frac{3}{12}+\frac{8}{12}=\left|3x-\frac{1}{5}\right|\left(\frac{2}{3}+\frac{5}{2}\right)\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|\left(\frac{4}{6}+\frac{15}{6}\right)=\frac{11}{12}\)
\(\Leftrightarrow\frac{19}{6}\left|3x-\frac{1}{5}\right|=\frac{11}{12}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{12}.\frac{6}{19}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{38}\)
\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{5}=\frac{11}{38}\\3x-\frac{1}{5}=-\frac{11}{38}\end{cases}}\)
Giải tiếp nha

2: 12-10x=25-30x
=>20x=13
=>x=13/20
3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)
=>6x+9-8x+10=10x+21
=>10x+21=-2x+19
=>12x=-2
=>x=-1/6
4: \(\Leftrightarrow25x-15-6x+12=11-5x\)
=>19x-3=11-5x
=>24x=14
=>x=7/12
5: \(\Leftrightarrow8-12x-5+10x=4-6x\)
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
6: \(\Leftrightarrow32x-24-6+9x=13-40x\)
=>41x-30=13-40x
=>81x=43
=>x=43/81
7: \(\Leftrightarrow10x-5+20x=5x-11\)
=>30x-5=5x-11
=>25x=-6
=>x=-6/25

mk giúp bạn câu cuối nhé:
3|x+2|-5=16
3|x+2|=16+5
3|X+2|=21
|x+2|=21:3
|x+2|=7
=>x+2=7 hoặc x+2=-7
+) với x+2=7 +) với x+2= -7
x=5. x=-9
vậy x€{5,-9}
nếu có TGian mk sẽ giải cho bạn mấy câu trên
cam ơn bạn nhé bạn có giup mình not câu trên trong vong ngay ko

\(A=x^2+6x+9-4x-1-2x-x^2=9\\ B=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ C=\left(3x+5-3x+5\right)^2=100\)
a: \(A=x^2+6x+9-4x-1-2x-x^2=8\)
b: \(B=2x^2+3x-10x-15-2x^2+6x+x+7=-8\)

a) \(2\chi-3=3\left(\chi+1\right)\)
\(\Leftrightarrow2\chi-3=3\chi+3\)
\(\Leftrightarrow2\chi-3\chi=3+3\)
\(\Leftrightarrow\chi=-6\)
Vậy phương trình có tập nghiệm S= \(\left\{-6\right\}\)
\(3\chi-3=2\left(\chi+1\right)\)
\(\Leftrightarrow3\chi-3=2\chi+2\)
\(\Leftrightarrow3\chi-2\chi=2+3\)
\(\Leftrightarrow\chi=5\)
Vậy phương trình có tập nghiệm S= \(\left\{5\right\}\)
b) \(\left(3\chi+2\right)\left(4\chi-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3\chi+2=0\\4\chi-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3\chi=-2\\4\chi=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\chi=\dfrac{-2}{3}\\\chi=\dfrac{5}{4}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S= \(\left\{\dfrac{-2}{3};\dfrac{5}{4}\right\}\)
\(\left(3\chi+5\right)\left(4\chi-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3\chi+5=0\\4\chi-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3\chi=-5\\4\chi=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\chi=\dfrac{-5}{3}\\\chi=\dfrac{1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S= \(\left\{\dfrac{-5}{3};\dfrac{1}{2}\right\}\)
c) \(\left|\chi-7\right|=2\chi+3\)
Trường hợp 1:
Nếu \(\chi-7\ge0\Leftrightarrow\chi\ge7\)
Khi đó:\(\left|\chi-7\right|=2\chi+3\)
\(\Leftrightarrow\chi-7=2\chi+3\)
\(\Leftrightarrow\chi-2\chi=3+7\)
\(\Leftrightarrow\chi=-10\) (KTMĐK)
Trường hợp 2:
Nếu \(\chi-7\le0\Leftrightarrow\chi\le7\)
Khi đó: \(\left|\chi-7\right|=2\chi+3\)
\(\Leftrightarrow-\chi+7=2\chi+3\)
\(\Leftrightarrow-\chi-2\chi=3-7\)
\(\Leftrightarrow-3\chi=-4\)
\(\Leftrightarrow\chi=\dfrac{4}{3}\)(TMĐK)
Vậy phương trình có tập nghiệm S=\(\left\{\dfrac{4}{3}\right\}\)
\(\left|\chi-4\right|=5-3\chi\)
Trường hợp 1:
Nếu \(\chi-4\ge0\Leftrightarrow\chi\ge4\)
Khi đó: \(\left|\chi-4\right|=5-3\chi\)
\(\Leftrightarrow\chi-4=5-3\chi\)
\(\Leftrightarrow\chi+3\chi=5+4\)
\(\Leftrightarrow4\chi=9\)
\(\Leftrightarrow\chi=\dfrac{9}{4}\)(KTMĐK)
Trường hợp 2: Nếu \(\chi-4\le0\Leftrightarrow\chi\le4\)
Khi đó: \(\left|\chi-4\right|=5-3\chi\)
\(\Leftrightarrow-\chi+4=5-3\chi\)
\(\Leftrightarrow-\chi+3\chi=5-4\)
\(\Leftrightarrow2\chi=1\)
\(\Leftrightarrow\chi=\dfrac{1}{2}\)(TMĐK)
Vậy phương trình có tập nghiệm S=\(\left\{\dfrac{1}{2}\right\}\)

* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)

a. Kiểm tra lại mẫu số vế phải, \(7-5x\) hay \(7-3x\)
b. ĐKXĐ: \(x\ne-\dfrac{5}{3}\)
\(\dfrac{3x+5}{12}=\dfrac{3}{5+3x}\)
\(\Leftrightarrow\dfrac{\left(3x+5\right)^2}{12\left(3x+5\right)}=\dfrac{36}{12\left(3x+5\right)}\)
\(\Rightarrow\left(3x+5\right)^2=36=6^2\)
\(\Rightarrow\left[{}\begin{matrix}3x+5=6\\3x+5=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{11}{3}\end{matrix}\right.\) (thỏa mãn)

a) 12 ( 2x - 6 ) - 43= 53
12 ( 2x - 6 ) - 43= 125
12 ( 2x - 6 ) =125+43
12 ( 2x - 6 ) =168
2x - 6=168:12
2x - 6=14
2x=14+6
2x=20
x=20:2
x=10
b) 7 ( 25 - 3x ) = 34 - 25
7 ( 25 - 3x ) = 49
25 - 3x =49:7
25 - 3x =7
3x=25-7
3x=18
x=18:3
x=6
c) ( 3x - 5 ) + 34 + 60 = 53
( 3x - 5 ) + 81 + 1 = 125
( 3x - 5 ) + 81=125-1
( 3x - 5 ) + 81=124
3x - 5=124-81
3x - 5=43
3x=43+5
3x=48
x=48:3
x=16
d) 3x + 2x ( 23 . 5 - 32 . 4 ) + 52 = 44
3x + 2x ( 8 . 5 - 9 . 4 ) + 25 = 256
3x + 2x.4 + 25 = 256
3x + 2x.4=256-25
3x + 2x.4=231
(3+2).x.4=231
5x.4=231
5x=231:4
5x=57,75
x=57,75:5
x=11,55
e) 720 : [ 41 - (2x - 5 ) ] = 23 .5
720 : [ 41 - ( 2x - 5 ) ] =40
41 - ( 2x - 5 )=720:40
41 - ( 2x - 5 )=18
2x - 5=41-18
2x - 5=23
2x=23+5
2x=28
x=28:2
x=14
Ta có: \(\left|3x-5\right|=\frac{3}{5}\)
\(\Rightarrow\left[\begin{matrix}3x-5=\frac{3}{5}\\3x-5=-\frac{3}{5}\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=\frac{28}{15}\\x=\frac{22}{15}\end{matrix}\right.\)
Vậy: \(x=\frac{28}{15}\) hoặc \(\frac{22}{15}\)
Ta có:
\(\left|3x-5\right|=\frac{3}{5}\)
=> \(\left[\begin{matrix}3x-5=\frac{3}{5}\\3x-5=\frac{-3}{5}\end{matrix}\right.\)
=> \(\left[\begin{matrix}3x=\frac{3}{5}+5\\3x=\frac{-3}{5}+5\end{matrix}\right.\)
=> \(\left[\begin{matrix}3x=\frac{28}{5}\\3x=\frac{22}{5}\end{matrix}\right.\)
=> \(\left[\begin{matrix}x=\frac{28}{5}:3\\x=\frac{22}{5}:3\end{matrix}\right.\)
=> \(\left[\begin{matrix}x=\frac{28}{15}\\x=\frac{22}{15}\end{matrix}\right.\)
Vậy \(\left[\begin{matrix}x=\frac{28}{15}\\x=\frac{22}{15}\end{matrix}\right.\).