x=0

x=0.

I15x-7I=I15x+7I

TH1: 15x-7=15x+7 => -7=7 ( vô nghiệm)

TH2: 15x-7=-15x-7 <=> 30x=0 <=> x=0

thử lại thấy thỏa mãn

KL: vạy x=0

7/5 x X + 3/8 x X= 9

X x (7/5 +3/8)=9

X x 71/40 = 9

X             = 9:71/40

X             = 360/71

     \(\frac{7}{15}X+\frac{3}{8}X=9\)

     \(\left(\frac{7}{15}+\frac{3}{8}\right)X=9\) 

\(\left(\frac{56}{120}+\frac{45}{120}\right)X=9\)

                  \(\frac{101}{120}X=9\)

                          \(X=9:\frac{101}{120}\)

                          \(X=9x\frac{120}{101}\)

                          \(X=\frac{1080}{101}\)

b, \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)

\(x.0=-1\)

\(\Rightarrow x\in rỗng\)

loz mẹ làm như cái loz

[15x-30]*[x-7]=0

15x*[x-7]-30*[x-7]=0

=>15x*[x-7]=30*[x-7]

15x=30

x=30/15=2

theo de bai =>

TH1 15*x-30=0

        15*x     =30+0

         x          =30:15

         x          =2

TH2  x-7=0

        x    =0+7

       x     =7

CHHUC HOC TOT

\(\left[15x-30\right].\left[x-7\right]=0\)

\(\orbr{\begin{cases}15x-30=0\\x-7=0\end{cases}}\)

\(TH1:15x-30=0\)                                                     \(TH2:x-7=0\)

\(15x=30\)                                                                                       \(x=0+7\)

\(x=2\)                                                                                           \(x=7\)

                                   Vậy   \(\orbr{\begin{cases}x=2\\x=7\end{cases}}\)

\(15x-13x+5x=49:7\)

\(\Leftrightarrow7x=7\Leftrightarrow x=1\)

\(15x-13x+5x=49:7\)

\(\Leftrightarrow2x+5x=7\)

\(\Leftrightarrow7x=7\)

\(\Leftrightarrow x=1\)

\(15x-13x+5x=49:7\)

\(\Leftrightarrow7x=7\)

\(\Leftrightarrow x=1\)

-5(3x-7) - ( -15x+3) - ( 12 -x ) =-4

=>-15x+35+15x-3-12+x=-4

=>20+x=-4

=>x=-4-20

=>x=-24

-3(4x-2) - ( -12+8) - (-x)=0

=>-12x+6+12-8+x=0

=>-11x+10=0

=>-11x=0-10

=>-11x=-10

=>11x=10

=>x=10:11=\(\frac{10}{11}\)

( 2. l x l - 1) .(7 - 3x ) =0                                        ( x² + 1 ). ( 1/2 - 3x ) <0                                  ( l x l + 1 ) . ( 15x - 1 ) = 0

=> 2 . l x l - 1 = 0 hoặc 7 - 3x = 0                           => x²+1 hoặc 1/2 -3x < 0                              => l x l + 1 hoặc 15x - 1 =0

+  2 . l x l - 1 = 0 => 2 . l x l =1 => x = 1/2             + x² +1< 0 => x không tồn tại                       + l x l - 1 = 0 => l x l = 1 => thuộc 1 : -1

+ 7 - 3x = 0 => 3x = 7 => x = 7/3                           + 1/2 - 3x < 0 => 3x > 1/2 => x > 1                 + 15x - 1 = 0 => 15x =1 => x = 1/15