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a) ĐKXĐ: \(x\ne-5\)
\(\Leftrightarrow7x-7=6x+30\\ \Leftrightarrow x=37\)
b) \(\Leftrightarrow25x^2=144\\ \Leftrightarrow x^2=\dfrac{144}{25}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x^2}{6}=\dfrac{24}{25}\)
\(\Rightarrow x^2=\dfrac{24.6}{25}=\dfrac{144}{25}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(\dfrac{6}{2x+1}=\dfrac{2}{7}\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{21}\Rightarrow2x+1=21\)
\(\Rightarrow2x=21-1=20\Rightarrow x=\dfrac{20}{2}=10\)
Vậy x = 10
b. \(\dfrac{24}{7x-3}=\dfrac{-4}{25}\Rightarrow\dfrac{24}{7x-3}=\dfrac{24}{150}\Rightarrow7x-3=150\)
\(\Rightarrow7x=150+3=153\Rightarrow x=\dfrac{153}{7}\)
Vậy \(x=\dfrac{153}{7}\)
c. \(\dfrac{4}{x-6}=\dfrac{-12}{18}\Rightarrow-12\cdot\left(x-6\right)=4\cdot18=72\)
\(\Rightarrow x-6=\dfrac{72}{-12}=-6\Rightarrow x=-6+6=0\)
\(\dfrac{y}{24}=\dfrac{-12}{18}\Rightarrow y=\dfrac{-12\cdot24}{18}=-16\)
Vậy x = 0 ; y = -16
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a: \(\Leftrightarrow x^2=900\)
=>x=30 hoặc x=-30
b: \(\Leftrightarrow\dfrac{2}{3}:\left(-0.1x\right)=\dfrac{4}{3}:\dfrac{-2}{25}=-\dfrac{4}{3}\cdot\dfrac{25}{2}=-\dfrac{100}{6}=\dfrac{-50}{3}\)
=>0,1x=2/3:50/3=2/3x3/50=1/25
=>1/10x=1/25
hay x=1/25:1/10=10/25=2/5
d: \(\Leftrightarrow x^2=\dfrac{144}{25}\)
=>x=12/5 hoặc x=-12/5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{x-1}{x+5}=\dfrac{6}{7}\\ \Leftrightarrow\left(x-1\right).7=6\left(x+5\right)\\ \Rightarrow7x-7=6x+30\\ \Rightarrow7x-6x=7+30\\ \Rightarrow x=37\)
Vậy \(x=37\)
\(b,\dfrac{x^2}{6}=\dfrac{24}{25}\\ \Leftrightarrow x^2.25=24.6\\ \Rightarrow x^2.5^2=144\\ \Rightarrow\left(5x\right)^2=144\\ \Rightarrow\left(5x\right)^2=\left(\pm12\right)^2\\ \Rightarrow\left\{{}\begin{matrix}5x=12\\5x=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
Vậy \(x=\pm\dfrac{12}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
19) \(\sqrt{19-x}=19\)
\(\Rightarrow\sqrt{19-x}=\sqrt{19^2}\)
\(\Rightarrow19-x=19^2\)
\(\Rightarrow19-19^2=x\)
\(\Rightarrow x=19\left(1-19\right)=-19.18=-342\)
21) \(\sqrt{x-1}=\dfrac{1}{3}\)
\(\Rightarrow\sqrt{x-1}=\sqrt{\left(\dfrac{1}{3}\right)^2}\)
\(\Rightarrow x-1=\dfrac{1}{3^2}\)
\(x=\dfrac{1+9}{9}=\dfrac{10}{9}\)
24)\(\sqrt{2x+\dfrac{5}{4}}=\dfrac{3}{2}\)
\(\Rightarrow\sqrt{2x+\dfrac{5}{4}}=\sqrt{\left(\dfrac{3}{2}\right)^2}\)
\(\Rightarrow2x+\dfrac{5}{4}=\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)
\(\Rightarrow2x=\dfrac{9-5}{4}=1\)
\(\Rightarrow x=0,5\)
25) \(\sqrt{\dfrac{x}{3}-\dfrac{7}{6}}=\dfrac{1}{6}\)
\(\Rightarrow\sqrt{\dfrac{2x-7}{6}}=\sqrt{\left(\dfrac{1}{6}\right)^2}\)
\(\Rightarrow\dfrac{2x-7}{6}=\left(\dfrac{1}{6}\right)^2=\dfrac{1}{36}\)
\(\Rightarrow\dfrac{12x-42}{36}=\dfrac{1}{36}\)
\(\Rightarrow12x-42=1\)
\(\Rightarrow12x=43\)
\(\Rightarrow x=\dfrac{43}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
=> 2(2x+1) = 6.7
4x+2=42
4x=40
x=10
Vậy x=10
a)\(\dfrac{6}{2x+1}=\dfrac{2}{7}\\ =>6.7=2.\left(2x+1\right)\\ =>2x+1=\dfrac{6.7}{2}=\dfrac{42}{2}=21\\ =>2x=21-1=20\\ =>x=\dfrac{20}{2}=10\)
b) \(\dfrac{24}{7x-3}=-\dfrac{4}{25}\\ =>24.25=-4.\left(7x-3\right)\\ =>7x-3=\dfrac{24.25}{-4}=-150\\ =>7x=-150+3=-147\\ =>x=\dfrac{-147}{7}=-21\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=-\dfrac{12}{18}\\ =>x-6=\dfrac{4.18}{-12}=-6\\ =>x=-6+6=0\\ y=\dfrac{-12.24}{18}=-16\)
d) \(-\dfrac{1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\\ < =>-\dfrac{8}{40}\le-\dfrac{5x}{40}\le\dfrac{10}{40}\\ =>-8\le-5x\le10\\ Mà:-8< -5.1< -5.0< -5.\left(-1\right)< -5.\left(-2\right)=10\\ =>x\in\left\{-2;-1;0;1\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\\ < =>\dfrac{x+46}{20}=\dfrac{5x+2}{5}\\ =>5\left(x+46\right)=20\left(5x+2\right)\\ < =>5x+230=100x+40\\ < =>230-40=100x-5x\\ < =>190=95x\\ =>x=\dfrac{190}{95}=2\)
f) \(y\dfrac{5}{y}=\dfrac{56}{y}\\ < =>\dfrac{y^2+5}{y}=\dfrac{56}{y}\\ =>y\left(y^2+5\right)=56y\\ =>y^2+5=\dfrac{56y}{y}=56\\ =>y^2=56-5=51\\ =>y=\sqrt{51}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có : \(\dfrac{x^2}{6}=\dfrac{24}{25}\Leftrightarrow x^2=\dfrac{24.6}{25}=\dfrac{144}{25}\) \(\Rightarrow x=\pm\sqrt{\dfrac{144}{25}}=\pm\dfrac{12}{5}\)
vậy \(x=\pm\dfrac{12}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng t/c dtsbn:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{2-5+6}=\dfrac{24}{3}=8\\ \Rightarrow\left\{{}\begin{matrix}x=16\\y=40\\z=48\end{matrix}\right.\)
x²/6 = 24/25
x² = 24/25 . 6
x² = 144/25
x = -12/5 hoặc x = 12/5
\(\dfrac{x^2}{6}\)=\(\dfrac{24}{25}\)
\(x^2\).25=6.24
\(x^2\).25=144
\(x^2\) =\(\dfrac{144}{25}\)
\(x^2\) =\(\dfrac{12}{5}\)