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Giải:
a) \(\dfrac{x}{-4}=\dfrac{-9}{x}\)
\(\Leftrightarrow x.x=-4.\left(-9\right)\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x=\pm6\)
Vậy ...
b) \(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\)
\(\Leftrightarrow\left(x-1\right)^2=900\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=31\\x=-29\end{matrix}\right.\)
Vậy ...
d) \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow5x-14=3x-4\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)
Vậy ...
Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x
=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x
=> 2x+3y-1 / 12 = 2x+3y-1 / 6x
=> 12 = 6x => x =2
1.
Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\)
\(\Rightarrow x^2-y^2=\left(5k\right)^2-\left(4k\right)^2=25k^2-16k^2=9k^2=4\)
\(\Rightarrow k^2=\dfrac{4}{9}\Rightarrow k=\pm\dfrac{2}{3}\)
\(\circledast k=\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{8}{3}\end{matrix}\right.\)
\(\circledast k=-\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=-\dfrac{8}{3}\end{matrix}\right.\)
2.
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\Rightarrow x=2\)
\(\Rightarrow y=\dfrac{\dfrac{2\cdot2+1}{5}\cdot7+2}{3}=3\)
3.
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-\left(z-3\right)}{4+9-4}=\dfrac{95-8+3}{9}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10\cdot4+2}{2}=21\\y=\dfrac{10\cdot9+6}{3}=32\\z=10\cdot4+3=43\end{matrix}\right.\)
\(\frac{x-1}{-15}=\frac{-60}{x-1}\)
\(\Leftrightarrow\left(x-1\right)^2=900\\ \Leftrightarrow\left(x-1\right)^2=\left(\pm30\right)^2\\ \Rightarrow x-1\in\left\{30;-30\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=31\\x=-29\end{matrix}\right.\)
Vậy...
* Đặt \(\dfrac{2x}{5}=\dfrac{-3y}{4}=k\Rightarrow2x=5k\Rightarrow x=\dfrac{5k}{2}\)
và\(-3y=4k\Rightarrow y=\dfrac{-4k}{3}\)
a) \(A=\dfrac{5x+3y}{6x-2y}\)
thay \(x=\dfrac{5k}{2}\)và \(y=\dfrac{-4k}{3}\), ta được
\(A=\dfrac{5.\dfrac{5k}{2}+3.\dfrac{-4k}{3}}{6.\dfrac{5k}{2}-2.\dfrac{-4k}{3}}=\dfrac{\dfrac{25k}{2}-4k}{15k+\dfrac{8k}{3}}=\dfrac{51}{106}\)
Bài B tương tự
Đặt:
\(\dfrac{2x}{5}=\dfrac{-3y}{4}=k\)
\(\Rightarrow\left\{{}\begin{matrix}2x=5k\Rightarrow x=2,5k\\-3y=4k\Rightarrow y=\dfrac{4}{-3}k\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{5x+3y}{6x-2y}\)
\(A=\dfrac{5.2,5k+3.\dfrac{4}{-3}k}{6.2,5k-2.\dfrac{4}{-3}k}\)
\(A=\dfrac{12,5k+-4k}{15k-\dfrac{8}{-3}k}\)
\(A=\dfrac{8,5k}{\dfrac{53}{3}k}\)
b Tương tự
Áp dụng tính chất dãy tỉ số bằng nhau ta có:\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}=\dfrac{2x+1+3y-2-2x-3y+1}{5+7-6x}=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\)
1) Tìm x, biết :
a) \(\dfrac{x-1}{3}=\dfrac{x+1}{5}\)
=> \(5\left(x-1\right)=3\left(x+1\right)\)
=> \(5x-5=3x+3\)
=> \(5x-5-3=3x\)
=> \(5x-8=3x\)
=> \(8=5x-3x\)
=> \(8=2x\)
=> x = 8 : 2
=> x = 4
a: =>x+1/2=5
=>x=9/2
b: =>(x-1)^2=900
=>x-1=30 hoặc x-1=-30
=>x=-29 hoặc x=31