giúp mik với

\(\frac{2-x}{x+3}=\frac{6}{5}\)

<=>\(5\left(2-x\right)=6\left(x+3\right)\)

<=>\(10-5x=6x+18\)

<=>\(\left(-5x\right)-6x=18-10\)

<=>\(-11x=8\)

<=>\(x=\frac{-8}{11}\)

cảm ơn bn

a) |x - 1,7| = 2,3

Xét 2 trường hợp:

TH1: x - 1,7 = -2,3

         x         = -2,3 +1,7

         x         = -0,6

TH2: x - 1,7 = 2,3

         x         = 2,3 + 1,7

         x         = 4

Vậy: Tự kl :<

c)

+)x<1=>/x-1/=1-x=2x-3=>1-x-(2x-3)=0=>4-3x=0=>x=4/3 (loại)

+)x>=1=>x-1=2x-3=>2x-x-3+1=0=>x-2=0=>x=2(t/m)

Vậy: x=2 haizz

Ê thằng kia!Ai cho đạo lại tên t hả cục shit?Hay m muốn t đạo lại m?À mà không sao,t chỉ việc báo ad một cái thôi mà,là níc m bay ngay=)

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)

\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2-\left(\frac{1}{4}\right)^2=0\)

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

\(\Leftrightarrow\left(\frac{1}{x}-\frac{5}{12}\right)\left(\frac{1}{x}-\frac{11}{12}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{5}{12}=0\\\frac{1}{x}-\frac{11}{12}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{5}{12}\\\frac{1}{x}=\frac{11}{12}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}\)

Vậy....

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)

\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)

\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2\)

\(\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{x}=\frac{11}{12}\)

\(\Rightarrow x=\frac{11}{12}\)

\(=>\left(\frac{1}{x}-\frac{2}{4}\right)^2=\frac{1}{16}\)

=> \(\orbr{\begin{cases}\frac{1}{x}-\frac{2}{4}=\frac{1}{4}\\\frac{1}{x}-\frac{2}{4}=-\frac{1}{4}\end{cases}}\)

=>\(\orbr{\begin{cases}\frac{1}{x}=\frac{3}{4}\\\frac{1}{x}=\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{4}{3}\\x=4\end{cases}}\)

học tốt

x^2-7/9x=0

x(x-7/9)=0

x=0, x-7/9=0

x=0,x=7/9. Vậy x=0, x=7/9

X²-\(\frac{7}{9}\)X=0   <=>  X(X-\(\frac{7}{9}\))=0

    =>   x=0 hoặc x-\(\frac{7}{9}\)=0

            x-\(\frac{7}{9}\)=0 <=>X=0+\(\frac{7}{9}\)=\(\frac{7}{9}\)

        =>  X=0 hoặc \(\frac{7}{9}\)

B1:

a)x=-3/5*9/25 =>x=-27/125

b)x=(4/7)⁶:(4/7)⁴ =>x=(4/7)²=16/49

c)(x/4)²=4:(x/2)

    (x/4)²=8/x

     x²/16=8/x2

     x³=128

x=5,039

B2

M=2^3.10+2^2.10/2^3.4+2^2.11

⁼2³⁰+2²⁰/2¹²+2²²

⁼2³⁰+2⁸+2²²

=2⁸(2²²+1+2¹⁴)

=2

Bài 1:

\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)

\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)

Bài 2:

\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)

<=>  \(\frac{7}{8}-x=\frac{27}{40}\)

<=>  \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)

Vậy...

bài 2 mình tính sai, sửa

.......

<=>  \(\frac{7}{8}-x=\frac{37}{40}\)

<=>  \(x=\frac{7}{8}-\frac{37}{40}=\frac{-1}{20}\)

Vậy....

\(\frac{4}{5}-\frac{-1}{8}=\frac{7}{8}-x\)

\(\Leftrightarrow x=\frac{7}{8}-\frac{4}{5}-\frac{1}{8}\)

\(\Leftrightarrow x=\frac{-1}{20}\)