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Đề trước đó:
(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x
<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x
<=>x^2-11x-6=0
<=>x^2-2x. 11/2 + 121/4-145/4=0
<=>(x-11/2)^2=145/4
<=>|x-11/2|=căn(145)/2
<=>x=[11+-căn(145)]/2
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\(\left|3x-4\right|-\left|y+3\right|=0\)
\(\Rightarrow\left|3x-4\right|+\left|3-y\right|=0\)
\(\Rightarrow\hept{\begin{cases}3x-4=0\\3-y=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{4}{3}\\y=3\end{cases}}}\)
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a) \(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(\Rightarrow\left(\frac{3}{5}-\frac{2}{3}-1\right).x=\frac{-5}{21}:\frac{1}{7}=\frac{-5}{3}\)
\(\Rightarrow\frac{-16}{15}.x=\frac{-5}{3}\Rightarrow x=\frac{-5}{3}:\frac{-16}{15}=\frac{25}{16}\)
b) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(x-\frac{1}{4}\right)^2=\left(±\frac{1}{6}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{4}=\frac{1}{6}\\x-\frac{1}{4}=\frac{-1}{6}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{12}\\x=\frac{1}{12}\end{cases}}\)
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\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}-x=3\sqrt{3}\\\dfrac{2}{3}-x=-3\sqrt{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2-9\sqrt{3}}{3}\\x=\dfrac{2+9\sqrt{3}}{3}\end{matrix}\right.\)
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a) 2x^2 = 3x b) (x - 5)^2 = x - 5
=> 3x - 2x^2 = 0 =>(x - 5)^2 - (x - 5) = 0
=> x.(3 - 2x) = 0 => (x - 5).(x - 6) = 0
=> x = 0 hoặc 3 - 2x = 0 => x - 5 = 0 hoặc x - 6 = 0
=> x = 0 hoặc x = 3/2 => x = 5 hoặc x = 6
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\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{5}>\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{3}{5}< -\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x>1\\\dfrac{1}{2}x< \dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< \dfrac{2}{5}\end{matrix}\right.\)
(x-5)2=(1-3x)2
=> x-5 = 1- 3x
=> 4x = 6
=> x = \(\frac{3}{2}\)
( x - 5 )2 = ( 1 - 3x ) 2
x - 5 = 1 - 3x
x = 1 - 3x + 5
x = 6 - 3x
x + 3x = 6
( 3 + 1 )x = 6
4x=6
=> x = 6 : 4
=> x = 1,5