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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 \(x\in\left\{0;\pm1;\pm2\right\}\)
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, \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow\left(10x+5\right)\left(x+1\right)=30\Leftrightarrow10x^2+10x+5x+5=30\)
\(\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(2x+5\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=1\end{cases}}\)
Ta có : \(\frac{x+1}{5}=\frac{x+2}{6}\)
\(\Rightarrow\left(x+1\right)6=5\left(x+2\right)\)
\(\Leftrightarrow6x+6=5x+10\)
\(\Leftrightarrow6x-5x=10-6\)
\(\Rightarrow x=4\)
\(\frac{x+1}{2}\)= \(\frac{8}{x+1}\)
x + 1 . x + 1 = 2 . 8
x . 2 = 16
x = 16 : 2
x = 8
a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)
\(\frac{1}{2}-x=\frac{57}{28}\)
\(x=-\frac{43}{28}\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow\left(2x-1\right)^2=5^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy ...
a) \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=\frac{-11}{4}\)
\(\Rightarrow\left(\frac{1}{2}-x\right)=\left(-\frac{5}{7}\right)+\frac{11}{4}\)
\(\Rightarrow\frac{1}{2}-x=\frac{57}{28}\)
\(\Rightarrow x=\frac{1}{2}-\frac{57}{28}\)
\(\Rightarrow x=-\frac{43}{28}\)
Vậy \(x=-\frac{43}{28}.\)
b) \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=20+5\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=5+1=6\\2x=\left(-5\right)+1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6:2\\x=\left(-4\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{3;-2\right\}.\)
d) \(\frac{x-6}{4}=\frac{4}{x-6}\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=4.4\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=16\)
\(\Rightarrow\left(x-6\right)^2=16\)
\(\Rightarrow x-6=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+6\\x=\left(-4\right)+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{10;2\right\}.\)
Chúc bạn học tốt!
Ta có : \(\frac{2x-1}{x+1}=\frac{2017}{2018}\)
=> 2018(2x - 1) = 2017(x + 1)
=> 4036x - 2018 = 2017x + 2017
=> 4036x - 2017x = 2017 + 2018
=> 2019x = 4035
=> x = \(\frac{4035}{2019}\)
\(a,\frac{2x-1}{x+1}=\frac{2017}{2018}\)
\(\Leftrightarrow2018.\left(2x-1\right)=2017.\left(x+1\right)\)
\(\Leftrightarrow4036x-2018=2017x+2017\) \(\Leftrightarrow4036x-2017x=2018+2017\)
\(\Leftrightarrow2019x=4035\Leftrightarrow x=\frac{4035}{2019}\)
\(b,\frac{x+2}{2x-5}=\frac{-x+3}{6-2x}\)( Điều kiện : \(x\ne3;x\ne2,5\))
\(\Leftrightarrow\left(x+2\right).\left(-2x+6\right)=\left(-x+3\right).\left(2x-5\right)\)
\(\Leftrightarrow-2x^2+6x-4x+12=-2x^2+5x+6x-15\)
\(\Leftrightarrow-2x^2+6x-4x+2x^2-5x-6x=-15-12\)
\(\Leftrightarrow-9x=-27\Leftrightarrow x=3\)( không thỏa mãn điều kiện )
\(\Rightarrow\)phương trình vô nghiệm .
\(\Rightarrow x\in\Phi\)