Làm ơn có ai giúp mih câu hỏi này với

b. (2:x+1)^2x=5^2x

    \(\Rightarrow\)2:x+1=5

            2:x     =5-1

            2:x     =4

            x        =2:4

           x         =1/2

a)\(^{\left(7x+2\right)^{-1}=3^{-2}}\)

            1/(7x+2)=1/3^2

             1/(7x+2)=1/9

       \(\Rightarrow\)7x+2  = 9

                 7x      =9-2

                 7x      = 7

                   x      =7:7

                  x        =1

Nếu đúng thì *** cho mình nhaaa!!!@@

a. \(5^{4-x}+1=26\)

\(\Leftrightarrow5^{4-x}=26-1=25\)

\(\Leftrightarrow5^{4-x}=5^2\)

\(\Leftrightarrow4-x=2\)

\(\Leftrightarrow x=2\)

b. \(\left(\frac{2}{x}+1\right)^{2x}=5^{2x}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{x}+1=5\\\frac{2}{x}+1=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{x}=4\\\frac{2}{x}=-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{3}\end{cases}}\)

c. \(\left(1-2x\right)^4-\left(1-2x\right)^6=0\)

\(\Leftrightarrow\left(1-2x\right)^4.\left[1-\left(1-2x\right)^2\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(1-2x\right)^4=0\\1-\left(1-2x\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}1-2x=0\\\left(1-2x\right)^2=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=1\\2x=0hoac2x=-2\end{cases}}\)

\(\Leftrightarrow x=\frac{1}{2},x=0,x=-1\)

\(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2017}\)

\(-5B=\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2018}\)

\(-5B-B=\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2018}-\)\(\left[\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2017}\right]\)

 \(-6B=\left(-5\right)^0-\left(-5\right)^{2018}\)

\(B=\left(5^{2018}-1\right):6\)

a)\(f\left(-1\right)=\left(-1\right)^2+5\cdot\left(-1\right)=1+\left(-5\right)=-4\)

\(f\left(-2\right)=\left(-2\right)^2+5\cdot\left(-2\right)=4+\left(-10\right)=-6\)

\(f\left(0\right)=0^2+5\cdot0=0\)

b)\(f\left(x\right)=-6\Leftrightarrow x^2+5x=-6\)

\(x^2+5x-\left(-6\right)=0\)

\(x^2+5x+6=0\)

\(x^2+2x+3x+6=0\)

\(x\left(x+2\right)+3\left(x+2\right)=0\)

\(\left(x+2\right)\left(x+3\right)=0\)

\(\Rightarrow x+2=0\) hoặc x+3=0

\(\Rightarrow\)x=-2 hoặc -3

a) f(-1) = (-1)² + 5(-1) = -4 =y

tuong tu

b) x² + 5x = -6

x² +5x +6 = 0 => x² +3x +2x +6 = 0

(x+3)(x+2) = 0

x = -3; x = -2

( chiều yên tâm đi học r)

minh co gang lam tung buoc nho cho ban hieu.

\(4-\left(x-\frac{1}{2}\right)-3\left(5-x\right)=0\)

{nhan pp pha ngoac

\(4-x+\frac{1}{2}-3.5+3.x=0\)

{gom cac so hang, hang so chuyen ben tay phai}

\(\left(3x-x\right)=3.5-4-\frac{1}{2}=15-4-\frac{1}{2}=11-\frac{1}{2}=\frac{21}{2}\)

\(2x=\frac{21}{2}\Rightarrow x=\frac{21}{2}:\frac{2}{1}=\frac{21}{2}.\frac{1}{2}=\frac{21}{4}\)

\(x=\frac{21}{4}\)

x=-7,75

\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

_Tần vũ_

\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)

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

\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)

\(\Leftrightarrow3x=\frac{1}{6}\)

\(\Leftrightarrow x=\frac{1}{18}\)

_Tần Vũ_

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{3^2}{5^2}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow\hept{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=\frac{3}{5}-\frac{3}{5}\\2x=-\frac{3}{5}-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=0\\2x=\frac{-6}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0:2\\x=-\frac{6}{5}:2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=0-\frac{1}{9}\)

\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}:3\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)

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

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

\(\Rightarrow3x-\frac{1}{2}=-\frac{1}{3}\)

\(\Rightarrow3x=-\frac{1}{3}+\frac{1}{2}\)

\(\Rightarrow3x=\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{6}:3\)

\(\Rightarrow x=\frac{1}{18}\)

a) => 4x + 2/3 = 0 hoặc 2/3x - 1 =0 

4x= -2/3 hoặc 2/3x= 1

x = -2/3 . 1/4 hoặc x = 1.3/2

x = -1/6 hoặc x = 3/2 

b) x+2 / x -1 = 5/2 

=> 2(x+2) = 5(x-1)

2x + 4 = 5x - 5

5x - 2x= 4+5

3x = 9

=> x= 3

a) (4x+\(\frac{2}{3}\)) . ( \(\frac{2}{3}\)x-1)=0

\(\Rightarrow\)\(\orbr{\begin{cases}4x+\frac{2}{3}=0\\\frac{2}{3}x-1=0\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=\\x=\end{cases}}\)........

Tới đây bn tự giải nha

còn câu c nữa các bạn ơi ,giúp mình với