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f(x) = 4 => (x - 4) - 3(x+1) = 4
=> x - 4 - 3x + 3 = 4
=> x - 4 - 3x + 3 - 4 = 0
=> x - 3x - 5 = 0
=> x - 3x = 5
=> x ( 1 - 3 ) = 5
=> -2x = 5
=> x = 5/-2 = -2,5
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5(x - 4)2 - 9 = 36
5(x - 4)2 = 45
(x - 4)2 = 9
\(\Rightarrow\) x - 4 = 3 hoặc x - 4 = - 3
\(\Rightarrow\) x = 7 hoặc x = 1
\(5\left(x-4\right)^2-9=36\)
\(5\left(x-4\right)^2=36+9=45\)
\(\left(x-4\right)^2=45:5=9\)
\(\left(x-4\right)^2=3^3=\left(-3\right)^2\)
Trường hợp 1: \(x-4=3\Leftrightarrow x=3+4=7\)
Trường hợp 2: \(x-4=-3\Leftrightarrow x=-3+4=1\)
Vậy x = 7 hoặc x = 1
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Ta có:\(\frac{4}{x}=\frac{5}{y}\Rightarrow\frac{x}{4}=\frac{y}{5}\)
Đặt \(\frac{x}{4}=\frac{y}{5}=k\)
\(\Rightarrow x=4k;y=5k\)
\(x+y=36\Rightarrow4k+5k=36\Rightarrow9k=36\Rightarrow k=4\)
Khi đó x=16;y=20
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Bài 4.
\(\left|x-1\right|+\left|y-2\right|+\left(z-x\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-2=0\\z-x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=z=1\\y=2\end{cases}}\)
Bài 3.
\(\left|x-1\right|+\left|2x-2\right|+\left|4x-4\right|+\left|5x-5\right|=36\)
\(\Leftrightarrow\left|x-1\right|+2\left|x-1\right|+4\left|x-1\right|+5\left|x-1\right|=36\)
\(\Leftrightarrow12\left|x-1\right|=36\)
\(\Leftrightarrow\left|x-1\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}\)
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\(\frac{x}{4}\): \(\frac{36}{x}\)=0
\(\frac{x^2-144}{4x}\)=0
x=-12, x=12