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9) \(\dfrac{x}{4}=\dfrac{9}{x}\)
Theo định nghĩa về hai phân số bằng nhau, ta có:
\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
8)
\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)
7)
\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)
6)
\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)
5)
\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)
4)
\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
3)
\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)
2)
\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)
1)
\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)
a) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
x=\(\dfrac{1}{10}:-\dfrac{2}{3}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\).
b) \(\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)
\(\dfrac{2}{3}:x=-7-\dfrac{1}{3}\)
\(\dfrac{2}{3}:x=-\dfrac{22}{3}\)
\(x=\dfrac{2}{3}:-\dfrac{22}{3}\)
\(x=-\dfrac{1}{11}\)
Vậy \(x=-\dfrac{1}{11}\).
c) \(60\%x=\dfrac{1}{3}\cdot6\dfrac{1}{3}\)
\(60\%x=\dfrac{19}{9}\)
\(\dfrac{3}{5}x=\dfrac{19}{9}\)
\(x=\dfrac{19}{9}:\dfrac{3}{5}\)
\(x=\dfrac{95}{27}\)
Vậy \(x=\dfrac{95}{27}\).
d) \(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\)
\(\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\)
\(\dfrac{2}{3}-x=\dfrac{3}{20}\)
\(x=\dfrac{2}{3}-\dfrac{3}{20}\)
\(x=\dfrac{31}{60}\)
Vậy \(x=\dfrac{31}{60}\).
e) \(-2x-\dfrac{-3}{5}:\left(-0.5\right)^2=-1\dfrac{1}{4}\)
\(-2x-\dfrac{-12}{5}=-1\dfrac{1}{4}\)
\(-2x=-1\dfrac{1}{4}+\dfrac{-12}{5}\)
\(-2x=-\dfrac{73}{20}\)
\(x=-\dfrac{73}{20}:\left(-2\right)\)
\(x=\dfrac{73}{40}\)
Vậy \(x=\dfrac{73}{40}\).
6. \(\dfrac{x}{4}=\dfrac{9}{x}\)
=>x2=4.9=36
=>x\(\in\)\(\left\{-6;6\right\}\)
\((\dfrac{2x}{5}+2):\left(-4\right)=-1\dfrac{1}{2}\)
(\(\dfrac{2x}{5}+2):\left(-4\right)=-\dfrac{3}{2}\)
\(\dfrac{2x}{5}=-\dfrac{3}{2}.\left(-4\right)\)
\(\dfrac{2x}{5}=6\)
\(\dfrac{2x}{5}=\dfrac{30}{5}\)
2x = 30
x = 30 : 2 = 15
a) \(\left(2x-3\right)\left(6-2x\right)=0\)
\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)
\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)
Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)
\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)
\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)
\(-\dfrac{11}{15}=-x\left(x-1\right)\)
\(\Rightarrow x=1.491631652\)
Vậy \(x=1.491631652\)
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\).
e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(x=\dfrac{3\cdot7}{10}\)
\(x=\dfrac{21}{10}\)
Vậy \(x=\dfrac{21}{10}\).
f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)
\(\dfrac{x}{5}=\dfrac{11}{10}\)
\(x=\dfrac{5\cdot11}{10}\)
\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Vậy \(x=\dfrac{11}{2}\).
g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)
Vậy \(x=2\).
h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)
Vậy \(x=14\).
a: \(\Leftrightarrow x\cdot\dfrac{1}{2}-\dfrac{3}{5}x+\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7}{10}x\)
=>3/5x=-4
hay x=-4:3/5=-20/3
b: \(\Leftrightarrow4x-6-9=5-3x-2\)
=>4x-15=-3x+3
=>7x=18
hay x=18/7
Bài 2: a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Leftrightarrow7x-21=5x+25\)
\(\Leftrightarrow7x-5x=21+25\)
\(\Leftrightarrow2x=46\)
\(\Rightarrow x=46:2=23\)
b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Rightarrow x^2=\left(\pm8\right)^2\)
\(\Rightarrow x=8\) hoặc \(x=-8\)
2)a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)
\(7x-21=5x+25\)
\(7x-5x+25=21\)
\(2x+25=21\)
\(2x=-4\Rightarrow x=-2\)
b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x+1\right)\left(x-1\right)\)
\(63=x\left(x-1\right)+1\left(x-1\right)\)
\(63=x^2-x+x-1\)
\(x^2=63+1=64\)
\(x=\left\{\pm8\right\}\)
c) \(\dfrac{x+4}{20}=\dfrac{2}{x+4}\)
\(\Leftrightarrow\left(x+4\right)\left(x+4\right)=2.20=40\)
\(x\left(x+4\right)+4\left(x+4\right)=40\)
\(x^2+4x+4x+16=40\)
\(x^2+8x=40-16=24\)
\(x\left(x+8\right)=24\)
\(x\in\left\{\varnothing\right\}\)
d) \(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=\left(x-1\right)\left(x+3\right)\)
\(x\left(x-2\right)+2\left(x-2\right)=x\left(x+3\right)-1\left(x+3\right)\)
\(x^2-2x+2x-4=x^2+3x-x-3\)
\(\)\(x^2-4=x^2+2x-3\)
\(\Leftrightarrow x^2-x^2-2x+3=4\)
\(-2x+3=4\)
\(-2x=1\)
\(x=-\dfrac{1}{2}\)
a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)
=>2/5x=8/5
=>x=4
b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)
=>1/3x=-6
=>x=-18
c: =>2|x-1/3|=0,24-4/5=-0,56<0
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
K chép lại đề, lm luôn nhé:
*\(\Rightarrow\) \(\left(\dfrac{7}{2}+2x\right)\cdot\dfrac{8}{3}=\dfrac{16}{3}\)
\(\Rightarrow\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)
\(\Rightarrow2x=2-\dfrac{7}{2}=-\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{3}{4}\)
* \(\Rightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{\dfrac{3}{4}-2}{2}=-\dfrac{5}{8}\)
=> K có gt x nào t/m đề
* Đề sai
* \(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
*\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)
\(\Rightarrow2x-1=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right)=-\dfrac{4}{63}\)
\(\Rightarrow2x=-\dfrac{4}{63}+1=\dfrac{59}{63}\)
\(\Rightarrow x=\dfrac{59}{63}:2=\dfrac{59}{126}\)
* \(\Rightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)
\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0\Rightarrow x=0\\2x=-\dfrac{6}{5}\Rightarrow x=-\dfrac{3}{5}\end{matrix}\right.\)
* \(\Rightarrow-5x-1-\dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(\Rightarrow-5x-\dfrac{1}{2}x-\dfrac{3}{2}x=-\dfrac{5}{6}+1-\dfrac{1}{3}\)
\(\Rightarrow-7x=-\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{1}{6}:7=-\dfrac{1}{42}\)
a)\(\left(3\dfrac{1}{2}+2x\right).2\dfrac{2}{3}=5\dfrac{1}{3}\)
\(\left(\dfrac{7}{2}+2x\right).\dfrac{8}{3}=\dfrac{16}{3}\)
\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)
\(2x=2-\dfrac{7}{2}=\dfrac{-3}{2}\Rightarrow x=\dfrac{-3}{4}\)
b)\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2=\dfrac{-1}{4}\)
\(\Rightarrow\left|2x-3\right|=\dfrac{-1}{8}\)
\(\Rightarrow x\in\varnothing\)
c) Đề sai,bạn có viết chữ x đâu,đó là phép tính mà.
d)\(\left(3x-1\right)\left(\dfrac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{-1}{2}x+5=0\Rightarrow x=10\)
e)\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=\dfrac{-21}{4}\)
\(2x-1=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{-4}{63}\)
\(\Rightarrow2x=\dfrac{59}{63}\Rightarrow x=\dfrac{59}{126}\)
g)\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(\left(2x+\dfrac{3}{5}\right)^2=0+\dfrac{9}{25}=\dfrac{9}{25}\)
\(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\)
\(th1:x=0\)
\(th2:x=\dfrac{-3}{5}\)
h)\(-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(-5x+-1-\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(\Leftrightarrow-5x+-1+\dfrac{5}{6}-\dfrac{1}{3}=2x\)
\(-5x+\dfrac{-1}{2}=2x\)
\(\dfrac{-1}{2}=2x+5x\)
\(\dfrac{-1}{2}=7x\Rightarrow x=\dfrac{-1}{14}\)
**\(\dfrac{2x-5}{6}=\dfrac{x-7}{5}\Leftrightarrow\dfrac{6\left(2x-5\right)}{30}=\dfrac{5\left(x-7\right)}{30}\)
\(\Leftrightarrow6\left(2x-5\right)=5\left(x-7\right)\Leftrightarrow12x-30=5x-35\)
\(\Leftrightarrow12x-5x=-35+30\Leftrightarrow7x=-5\Leftrightarrow x=-\dfrac{5}{7}\)
**\(\dfrac{4x+1}{3}=\dfrac{2x-3}{4}\Leftrightarrow\dfrac{4\left(x+1\right)}{12}=\dfrac{3\left(2x-3\right)}{12}\)
\(\Leftrightarrow4\left(x+1\right)=3\left(2x-3\right)\Leftrightarrow4x+4=6x-9\)
\(\Leftrightarrow4x-6x=-9-4\Leftrightarrow-2x=-13\Leftrightarrow x=\dfrac{13}{2}\)