Gợi ý thui :v

Giải hơi dài dòng :v

\(a,x-\frac{1}{5}=-2\frac{1}{3}\)

\(\Rightarrow x-\frac{1}{5}=-\frac{7}{3}\)

\(\Leftrightarrow x=-\frac{7}{3}+\frac{1}{5}=-\frac{32}{15}\)

\(b,\frac{1}{4}+\frac{3}{4}:x=\frac{1}{2}\)

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

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

c, Tự làm

a. x - 1/5 = -2 1/3

    x - 1/5 = -7/3

    x         = -2 2/15

b. 1/4 + 3/4 : x = 1/2

             3/4 : x = 1/2 - 1/4

             3/4 : x = 1/4

                     x = 3/4 x 4

                     x = 3

c. -0,75x + 8/11x = 23/44

     (-3/4 + 8/11)x = 23/44

                -1/44x = 23/44

                        x = -23

\(a,x-\frac{1}{5}=-2\frac{1}{3}\)

\(\Rightarrow x-\frac{1}{5}=\frac{-7}{3}\)

\(\Rightarrow x=\frac{-7}{3}+\frac{1}{5}\)

\(\Rightarrow x=\frac{-35}{15}+\frac{3}{15}\)

\(\Rightarrow x=\frac{-32}{15}\)

a \(\dfrac{2}{3}x+\dfrac{1}{3}=\dfrac{1}{5}\\ \dfrac{2}{3}x=\dfrac{1}{5}-\dfrac{1}{3}\\ \dfrac{2}{3}x=\dfrac{-2}{15}\\ x=-\dfrac{2}{15}:\dfrac{2}{3}\\ x=-\dfrac{1}{5}\)   b) \(\dfrac{4}{5}-\dfrac{5}{3}x=-2\\ \dfrac{5}{3}x=\dfrac{4}{5}+2\\ \dfrac{5}{3}x=\dfrac{14}{5}\\ x=\dfrac{14}{5}:\dfrac{5}{3}\\ x=\dfrac{42}{25}\)c) \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\\ \dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}\\ \dfrac{5}{3}:x=\dfrac{3}{10}\\ x=\dfrac{5}{3}:\dfrac{3}{10}\\ x=\dfrac{50}{9}\)d) \(\dfrac{5}{7}:x-3=-\dfrac{2}{7}\\ \dfrac{5}{7}:x=3-\dfrac{2}{7}\\ \dfrac{5}{7}:x=\dfrac{19}{7}\\ x=\dfrac{5}{7}:\dfrac{19}{7}\\ x=\dfrac{5}{19}\)

Bài 1

a) \(x=x^5\)

\(x^5-x=0\)

\(x\left(x^4-1\right)=0\)

\(x=0\) hoặc \(x^4-1=0\)

* \(x^4-1=0\)

\(x^4=1\)

\(x=1\)

Vậy x = 0; x = 1

b) \(x^4=x^2\)

\(x^4-x^2=0\)

\(x^2\left(x^2-1\right)=0\)

\(x^2=0\) hoặc \(x^2-1=0\)

*) \(x^2=0\)

\(x=0\)

*) \(x^2-1=0\)

\(x^2=1\)

\(x=1\)

Vậy \(x=0\); \(x=1\)

c) \(\left(x-1\right)^3=x-1\)

\(\left(x-1\right)^3-\left(x-1\right)=0\)

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

\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)

*) \(x-1=0\)

\(x=1\)

*) \(\left(x-1\right)^2-1=0\)

\(\left(x-1\right)^2=1\)

\(x-1=1\) hoặc \(x-1=-1\)

**) \(x-1=1\)

\(x=2\)

**) \(x-1=-1\)

\(x=0\)

Vậy \(x=0\); \(x=1\); \(x=2\)

Bài 2: 

a: \(\left(6x-39\right):7=3\)

\(\Leftrightarrow6x-39=21\)

hay x=10

a: \(2^x=2^3\)

nên x=3

c: \(11^x=1331\)

nên x=3

d: \(2^x+4=12\)

nên \(2^x=8\)

hay x=3

\(\dfrac{1}{6}+x=\dfrac{5}{12}\)
\(=>x=\dfrac{5}{12}-\dfrac{2}{12}=\dfrac{1}{4}\)
\(\dfrac{3}{4}+\dfrac{1}{4}x=-\dfrac{1}{2}\)
\(=>\dfrac{1}{4}x=-\dfrac{5}{4}\)
\(=>x=-\dfrac{5}{4}.4=-5\)
\(7^{2x}+7^{2x+3}=344\)
\(< =>49^x+49^x.343=344\)
\(=>x=?\)

Bài 1: 

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)

\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)

Suy ra: \(-12x-3=8x-2-6x-8\)

\(\Leftrightarrow-12x-3-2x+10=0\)

\(\Leftrightarrow-14x+7=0\)

\(\Leftrightarrow-14x=-7\)

\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)

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

Lời giải:

a. $2x^2+3(x-1)(x+1)=5x(x+1)$

$\Leftrightarrow 2x^2+3x^2-3=5x^2+5x$

$\Leftrightarrow 5x^2-3=5x^2+5x$
$\Leftrightarrow 5x=-3$

$\Leftrightarrow x=\frac{-3}{5}$

b.

PT $\Leftrightarrow (-5x^2-2x+16)+4(x^2-x-2)=4-x^2$

$\Leftrightarrow -x^2-6x+8=4-x^2$

$\Leftrightarrow -6x+8=4$
$\Leftrightarrow -6x=-4$

$\Leftrightarrow x=\frac{2}{3}$

c.

PT $\Leftrightarrow 4(x^2+4x-5)-(x^2+7x+10)=3(x^2+x-2)$

$\Leftrightarrow 4x^2+16x-20-x^2-7x-10=3x^2+3x-6$

$\Leftrightarrow 3x^2+9x-30=3x^2+3x-6$

$\Leftrightarrow 6x=24$

$\Leftrightarrow x=4$

a x=4

\(a,\dfrac{x}{2}=\dfrac{8}{x}\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\\ b,\dfrac{x+1}{5}=\dfrac{x+1}{5}\left(luôn.đúng\right)\\ c,\dfrac{x+1}{5}=\dfrac{x+3}{10}\\ \Rightarrow\dfrac{2x+2}{10}=\dfrac{x+3}{10}\\ \Rightarrow2x+2=x+3\\ \Rightarrow2x-x=3-2\\ \Rightarrow x=1\\ d,\dfrac{x}{4}=\dfrac{18}{x+1}\\ \Rightarrow x\left(x+1\right)=4.18\\ \Rightarrow x^2+x=72\\ \Rightarrow x^2+x-72=0\\ \Rightarrow\left(x^2+9x\right)-\left(8x+72\right)=0\\ \Rightarrow x\left(x+9\right)-8\left(x+9\right)=0\\ \Rightarrow\left(x-8\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-9\end{matrix}\right.\)