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2: =>2x-1/4=5/6-1/2x
=>5/2x=5/6+1/4=13/12
=>x=13/30
3: =>3x-5/6=2/3-1/2x
=>3,5x=2/3+5/6=4/6+5/6=9/6=3,2
hay x=32/35
a. \(\frac{2x+3}{15}=\frac{7}{5}\)
\(\Leftrightarrow5\left(2x+3\right)=15.7\)
\(\Leftrightarrow10x+15=105\)
\(\Leftrightarrow10x=90\)
\(\Leftrightarrow x=9\)
b. \(\frac{x-2}{9}=\frac{8}{3}\)
\(\Leftrightarrow3\left(x-2\right)=9.8\)
\(\Leftrightarrow3x-6=72\)
\(\Leftrightarrow3x=78\)
\(\Leftrightarrow x=26\)
c. \(\frac{-8}{x}=\frac{-x}{18}\)
\(\Leftrightarrow-x^2=-144\)
\(\Leftrightarrow x^2=12^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
Mấy câu kia tương tự
d, \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=6x-12\Leftrightarrow4x=-27\Leftrightarrow x=-\frac{27}{4}\)
e, \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x=132\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f, \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x=10\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x+2\right)\left(2x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{5}{2}\end{cases}}\)
g, \(\left(2x-1\right)\left(2x+1\right)=63\Leftrightarrow4x^2+2x-2x-1=63\Leftrightarrow4x^2-64=0\)
\(\Leftrightarrow x^2=16\Leftrightarrow x=\pm4\)
h, \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow\left(10x+5\right)\left(x+1\right)=30\Leftrightarrow10x^2+10x+5x+5=30\)
\(\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(2x+5\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=1\end{cases}}\)
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
Bài 2:
a, 1/3 + 1/2 : x = -4
=> 1/2 : x = -4 - 1/3
=> 1/2 : x = -13/3
=> x = 1/2 ; -13/3
=> x = -3/26
Vậy x = -3 / 26
Bài 2:
b, x2 - 4x = 0
=> x.(x - 4) =0
=> x=0 hoặc x - 4 = 0
x - 4= 0 => x=4
Vậy x=0 và x=4
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)
\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)
d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
=> \(x:\frac{1}{45}=\frac{1}{2}\)
=> \(x=\frac{1}{2}.\frac{1}{45}\)
=> \(x=\frac{1}{90}\)
Vậy \(x=\frac{1}{90}.\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)
Mình chỉ làm được thế thôi nhé, mong bạn thông cảm.
Chúc bạn học tốt!
a) \(4:\left(x-1\right)=\left(x-1\right):9\)
\(\frac{4}{x-1}=\frac{x-1}{9}\)
\(\left(x-1\right)^2=36\)
\(\left(x-1\right)^2=6^2\)
\(\Rightarrow x-1=6\)
\(\Rightarrow x=7\)
vậy \(x=7\)
c) \(3\frac{1}{2}:x\frac{1}{2}=5\frac{1}{3}:\frac{1}{2}.1\frac{1}{5}\)
\(\frac{7}{2}:\frac{1}{2}x=\frac{16}{3}:\frac{1}{2}.\frac{6}{5}\)
\(\frac{7}{2}:\frac{1}{2}x=\frac{64}{5}\)
\(\frac{1}{2}x=\frac{7}{2}:\frac{64}{5}\)
\(\frac{1}{2}x=\frac{35}{128}\)
\(x=\frac{35}{128}:\frac{1}{2}\)
\(x=\frac{35}{64}\)
d) \(\left|2x-3\right|=5\)
\(\Rightarrow\orbr{\begin{cases}2x-3=5\\2x-3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}2x=8\\2x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)
f) \(\left(2x-\frac{1}{2}\right)^2=\left(1-3x\right)^2\)
\(\Rightarrow2x-\frac{1}{2}=1-3x\)
\(\Rightarrow2x+3x=1+\frac{1}{2}\)
\(\Rightarrow5x=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{10}\)
\(\left(3-\frac{1}{2}:x\right)^2=14\)
\(\left(3-\frac{1}{2x}\right)^2=14\)
\(\frac{1}{4x^2}-2.\frac{1}{2x}.3+9=14\)
\(\frac{1}{4x^2}-\frac{3}{x}=5\)
\(\left(\frac{1}{4x}-3\right):x=5\)
a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) => \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)
Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)
b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)
=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=> \(\frac{27x}{4}=\frac{27}{40}\)
\(27x.40=27.4\)
\(1080.x=108\)
\(x=\frac{1}{10}\)
Vậy \(x=\frac{1}{10}\)
c) \(\left|x-1\right|+4=6\)
\(\left|x-1\right|=6-4\)
\(\left|x-1\right|=2\)
\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Vậy \(x=\left[3,-1\right]\)
d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)
e) \(\left(x^2-3\right)^2=16\)
\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)
\(x^2=7=>x=\sqrt{7}\)
Vậy \(x=\sqrt{7}\)
f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\)
\(\frac{2}{5}x=-\frac{4}{15}\)
\(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)
Vậy \(x=-\frac{2}{3}\)
g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)
\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)
Vậy \(x=-3\)
k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)
\(\frac{2}{5}x=\frac{4}{15}\)
\(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)
Vậy \(x=\frac{2}{15}\)
I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)
\(\frac{3}{5}x=\frac{5}{14}\)
\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}\)
\(\frac{2}{3}\)- \(\frac{1}{3}\). x - \(\frac{1}{3}\).\(\frac{3}{2}\)-\(\frac{1}{2}\).2x + \(\frac{1}{2}\)= 5
\(\frac{2}{3}\)- \(\frac{1}{3}\). x - \(\frac{1}{2}\)- x +\(\frac{1}{2}\) = 5
(\(\frac{2}{3}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)) - (\(\frac{1}{3}\). x + x) = 5
\(\frac{2}{3}\)- \(\frac{4}{3}\). x = 5
\(\frac{4}{3}\). x = \(\frac{2}{3}\)- 5
\(\frac{4}{3}\). x = \(\frac{-13}{3}\)
x = \(\frac{-13}{3}\): \(\frac{4}{3}\)
x = \(\frac{-13}{4}\)