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bn nên ghi rõ r. Bn vt thế này mk ko hiểu bn vt cái j cả
\(\frac{x-1}{15}=\frac{-60}{1-x}\)
\(x-1=\frac{900}{1-x}\)
\(\left(x-1\right)\left(1-x\right)=-900\)
\(x-x^2-1+x=-900\)
\(2x-x^2-1=-900\)
\(2x-x^2-1+900=0\)
\(2x-x^2-899=0\)
\(\left(x-31\right)\left(x+29\right)=0\)
\(\orbr{\begin{cases}x-31=0\\x+29=0\end{cases}}\)
\(\orbr{\begin{cases}x=31\\x=-29\end{cases}}\)
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x/-15=-60/x
=> x.x = [-15]. [-60]
=> x^2 = 900
=> x= 30 hoặc -30
-2/x = -x/8
=> -x^2 = -16
=> x^2 = 16
=> x= -4 hoặc 4
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\(-\frac{x}{15}=\frac{-60}{x}\)
(=) \(\frac{x}{-15}=\frac{-60}{x}\)
(=) \(x^2=-15.-60\)
(=) \(x^2=900\)
=> \(\orbr{\begin{cases}x=30\\x=-30\end{cases}}\)
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a) Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{x}{2}=\frac{y}{3}=\frac{x+y}{2+3}=\frac{25}{5}=5\) ( do x + y = 25 )
\(\Rightarrow\hept{\begin{cases}x=5.2=10\\y=5.3=15\end{cases}}\)
b) Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{91}=\frac{y}{15}=\frac{z}{6}=\frac{-x+y-z}{-91+15-6}=\frac{60}{-82}=\frac{-30}{41}\)( do -x + y - z = 60 )
Từ đó tìm được x , y , z
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\(\left(x:23+45\right).67=8911\)
\(x:23+45=8911:67\)
\(x:23+45=133\)
\(x :23=133-45\)
\(x:23=88\)
\(x=88.23\)
\(x=2024\)
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\(\frac{x}{-15}=-\frac{60}{x}\)
\(\Rightarrow x^2=-15.\left(-60\right)\)
\(\Rightarrow x^2=900\)
\(\Rightarrow x=\sqrt{900}\)
\(x=30\)
\(\frac{x}{-15}=\frac{-60}{x}\)
\(\Rightarrow x\cdot x=\left(-60\right)\cdot\left(-15\right)\)
\(\Rightarrow x^2=900\)
=>x2=(-30)2 hoặc 302
=>x=30 hoặc -30
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Ta có: \(\frac{x}{-15}=\frac{-60}{x}\)
=>-15.(-60)=x.x
=>900=x2
=>x2=302=(-30)2
=>x=30 hoặc x=-30
Vậy x=30 hoặc x=-30
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\(\frac{X}{-15}\)=\(\frac{-60}{x}\)
=> \(^{x^2}\)=-15.(-60)
\(x^2\)=900
\(x^2\)=30.30
\(x^2\)=\(30^2\)
=> x=30
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x-1/-15 = -60/x-1
=> (x-1)2 = 900
=> (x-1)2 = 302
=> \(\hept{\begin{cases}x-1=-30\\x-1=30\end{cases}}\)
=>\(\hept{\begin{cases}x=-29\\x=31\end{cases}}\)
\(\frac{x}{-15}=\frac{-60}{x}\)
=>x.x=(-15).(-60)
=>x2=900=302=(-30)2
=>x=30,-30
Vậy x=30,-30
\(\frac{x}{-15}=\frac{-60}{x}\)
<=> (-15).(-60) = x.x
<=> 900 = x2
=> x2 = 302 = (-30)2
=> x \(\in\){30; -30}