=>\(=\hept{\begin{cases}x=-4\\x=6\end{cases}}\)

a) \(0,\left(31\right)+x=0,3\left(7\right)\\ \Rightarrow\dfrac{31}{99}+x=\dfrac{17}{45}\\ \Rightarrow x=\dfrac{17}{45}-\dfrac{31}{99}=\dfrac{32}{495}=0,0\left(64\right)\)

Vậy \(x=0,0\left(64\right)\)

b) \(0,\left(4\right)\cdot x=\dfrac{5}{6}\\ \Rightarrow\dfrac{4}{9}\cdot x=\dfrac{5}{6}\\ \Rightarrow x=\dfrac{5}{6}:\dfrac{4}{9}\\ \Rightarrow x=\dfrac{5}{6}\cdot\dfrac{9}{4}\\ \Rightarrow x=\dfrac{15}{8}=1,875\)

Vậy \(x=1,875\)

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

a) 7/x-1=x+1/9
7/x-x-1=1/9
7/x-x=1/9+1
7/x-x=10/9
còn lại thì bạn tự tính nhé
câu b làm tương tự

tick cho mik nha

b: Ta có: \(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)

\(\Leftrightarrow x^2+2x-3=x^2-4\)

\(\Leftrightarrow2x=-1\)

hay \(x=-\dfrac{1}{2}\)

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge-1\\x>1\end{matrix}\right.\)\(\Rightarrow x>1\)

Ta có : \(PT\Leftrightarrow\sqrt{x+1}=2\sqrt{x-1}\)

\(\Leftrightarrow x+1=4x-4\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\dfrac{5}{3}\left(TM\right)\)

Vậy ...

b, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge1\\x>-1\end{matrix}\right.\)\(\Rightarrow x\ge1\)

Ta có : \(PT\Leftrightarrow\sqrt{x-1}=2\sqrt{x+1}\)

\(\Leftrightarrow x-1=4x+4\)

\(\Leftrightarrow3x=-5\)

\(\Leftrightarrow x=-\dfrac{5}{3}\left(L\right)\)

Vậy phương trình vô nghiệm .

a) ĐKXĐ: \(x>1\)

Ta có: \(\dfrac{\sqrt{x+1}}{\sqrt{x-1}}=2\)

\(\Leftrightarrow\sqrt{x+1}=2\sqrt{x-1}\)

\(\Leftrightarrow x+1=4x-4\)

\(\Leftrightarrow x-4x=-4-1\)

\(\Leftrightarrow-3x=-5\)

hay \(x=\dfrac{5}{3}\left(nhận\right)\)

Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)

b) ĐKXĐ: \(\left\{{}\begin{matrix}x>-1\\x\ne1\end{matrix}\right.\)

Ta có: \(\dfrac{\sqrt{x-1}}{\sqrt{x+1}}=2\)

\(\Leftrightarrow\sqrt{x-1}=2\sqrt{x+1}\)

\(\Leftrightarrow x-1=4x+4\)

\(\Leftrightarrow x-4x=4+1\)

\(\Leftrightarrow-3x=5\)

hay \(x=-\dfrac{5}{3}\)(loại)

Vậy: \(S=\varnothing\)

x/0,3 = 7,5/x

=>x^2 = 2,25

=>x^2 = (1,5)^2 hoặc x = (-1,5)^2

=>x = 1,5         hoặc x = -1,5

x/0.3=7.5/x

=>x^2=7.5*0.3

=>x^2=2.25

=>x=+-1.5

a) \(\dfrac{x}{4}\)=\(\dfrac{1}{x}\)→x²=4→x=2 hoặc x=-2
b) \(\dfrac{1}{5}\) = x:4-\(\dfrac{1}{10}\)→x:4=\(\dfrac{1}{5}\)+\(\dfrac{1}{10}\)→x:4=\(\dfrac{2+1}{10}\)→x:4=\(\dfrac{3}{10}\)→x=\(\dfrac{3}{10}\)x4→x=\(\dfrac{12}{10}\)→x=\(\dfrac{6}{5}\)

a) \(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

b) \(\Rightarrow x\left(x-1\right)-2\left(x-1\right)=0\Rightarrow\left(x-1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a x=4

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