a, \(\left(\dfrac{1}{2}+\dfrac{4}{7}\right):x=\dfrac{-3}{4}\)

\(\dfrac{15}{14}:x=\dfrac{-3}{4}\)

=> x= \(\dfrac{-7}{10}\)

b, 0,5:x-\(1\dfrac{3}{4}\)= 25%

0,5:x-\(\dfrac{7}{4}=\dfrac{1}{4}\)

0,5:x = 2

=> x = \(\dfrac{1}{4}\)

a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)

=>x=-3; y=28; z=-60

b: 5/12=x/-72

=>x=-72*5/12=-6*5=-30

c: =>x+3=-5

=>x=-8

a: \(\left(\dfrac{3}{4}x+2\dfrac{1}{2}\right)\cdot\dfrac{-2}{3}=\dfrac{1}{8}\)

=>\(\left(\dfrac{3}{4}x+\dfrac{5}{2}\right)=\dfrac{1}{8}:\dfrac{-2}{3}=\dfrac{-3}{16}\)

=>\(\dfrac{3}{4}x=-\dfrac{3}{16}-\dfrac{5}{2}=-\dfrac{3}{16}-\dfrac{40}{16}=-\dfrac{43}{16}\)

=>\(x=-\dfrac{43}{16}:\dfrac{3}{4}=\dfrac{-43}{16}\cdot\dfrac{4}{3}=\dfrac{-43}{12}\)

b: \(\dfrac{1}{3}\cdot x-0,5x=0,75\)

=>\(x\left(\dfrac{1}{3}-\dfrac{1}{2}\right)=0,75\)

=>\(x\cdot\dfrac{-1}{6}=0,75\)

=>\(x=-0,75\cdot6=-4,5\)

\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{13}{24}\)

\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{6}{5}\)

\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)

\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)

\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)

\(=\dfrac{1}{5}\)

a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)

            \(x=\dfrac{-3}{8}-\dfrac{1}{6}\)

           \(x=\dfrac{-13}{24}\)

vậy x =....

b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=\dfrac{17}{12}\)

                    \(x=\dfrac{3}{4}-\dfrac{17}{12}\)

                   \(x=\dfrac{-2}{3}\)

vậy x =....

a, ĐKXĐ:\(x\ge1\)

\(\sqrt{x-1}=3\\ \Rightarrow x-1=9\\ \Rightarrow x=10\)

\(b,x^2-64=0\\ \Rightarrow\left(x-8\right)\left(x+8\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\\ c,x^2+16=25\\ \Rightarrow x^2=9\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ d,ĐKXĐ:x\ge0\\ \left|\sqrt{x}-3\right|+3=9\\ \Rightarrow\left|\sqrt{x}-3\right|=6\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x}-3=-6\\x-3=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x}=-3\left(vô.lí\right)\\x=9\left(tm\right)\end{matrix}\right.\)

a) \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Rightarrow3x-6+2x-6=5\)

\(\Rightarrow5x=17\Rightarrow x=\dfrac{17}{5}\)

b) \(\left(2x-8\right)^2-16=0\)

\(\Rightarrow\left(2x-8-4\right)\left(2x-8+4\right)=0\)

\(\Rightarrow\left(2x-12\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=12\\2x=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)

c) \(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=3\)

\(\Rightarrow4x^2-4x+1-4x^2+12x-x+3=3\)

\(\Rightarrow7x=-1\Rightarrow x=-\dfrac{1}{7}\)

a: Ta có: \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Leftrightarrow3x-6+2x-6=5\)

\(\Leftrightarrow5x=17\)

hay \(x=\dfrac{17}{5}\)

b: Ta có: \(\left(2x-8\right)^2-16=0\)

\(\Leftrightarrow\left(2x-4\right)\left(2x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

a. \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Leftrightarrow3x-6+2x-6=5\)

\(\Leftrightarrow5x=17\)

\(\Leftrightarrow x=\dfrac{17}{5}\)

b. \(\left(2x-8\right)^2-16=0\)

\(\Leftrightarrow\left(2x-8-4\right)\left(2x-8+4\right)=0\)

\(\Leftrightarrow4\left(x-6\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)

c. \(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=3\)

\(\Leftrightarrow4x^2-4x+1-4x^2+11x+3-3=0\)

\(\Leftrightarrow7x+1=0\)

\(\Leftrightarrow x=-\dfrac{1}{7}\)

a)

\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)                

Vậy \(x = \frac{{ - 3}}{2}\).

b)

\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)

Vậy \(x = \frac{{ - 5}}{6}\).

c)

\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)      

Vậy \(x = \frac{4}{5}\)

d)

\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)

Vậy \(x = \frac{{ - 2}}{5}\).

Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.

a) \(\left(x-4\right)^2-\left(x-4\right)=0\)

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

\(\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

b) \(5x^2\left(x-7\right)+7\left(x-7\right)=0\)

\(\left(x-7\right)\left(5x^2+7\right)=0\)

\(\left[{}\begin{matrix}x-7=0\\5x^2+7=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=7\\x^2=\dfrac{-7}{5}\end{matrix}\right.\)

\(x=7\)

c) \(x^2\left(x-3\right)-\left(x-3\right)=0\)

\(\left(x-3\right)\left(x^2-1\right)=0\)

\(\left[{}\begin{matrix}x=3\\x=\pm1\end{matrix}\right.\)

a) (x - 4)^2=(x - 4)

(x - 4) (x -4)=(x -4 )

(x - 4) (x - 4)-(x - 4)=0

(x-4) (x-4-1)=0

(x-4) (x-5)=0

TH1:x-4=0                          TH2:x-5=0

            x=4                                      x=5