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x2 = 5/7.x
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\(\Leftrightarrow\frac{5}{7}+\left|\frac{1}{2}-x\right|=\frac{11}{4}\)
\(\Leftrightarrow\left|\frac{1}{2}-x\right|=\frac{11}{4}-\frac{5}{7}=\frac{77-20}{28}=\frac{57}{28}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}-x=\frac{57}{28}\\\frac{1}{2}-x=-\frac{57}{28}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}-\frac{57}{28}=\frac{14-57}{28}=\frac{-43}{28}\\x=\frac{1}{2}+\frac{57}{28}=\frac{14+57}{28}=\frac{71}{28}\end{cases}}\)
PT có 2 nghiệm là: -43/28 và 71/28
TH1 : \(x< \frac{1}{2}\), ta có:
\(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)
\(-\frac{5}{7}-\frac{1}{2}+x=-\frac{11}{4}\)
\(-\frac{17}{14}+x=-\frac{11}{4}\)
\(x=-\frac{11}{4}-\left(-\frac{17}{14}\right)\)
\(x=-\frac{43}{28}\)( thỏa mãn )
TH2 : \(x\ge\frac{1}{2}\); ta có:
\(-\frac{5}{7}-\left(x-\frac{1}{2}\right)=-\frac{11}{4}\)
\(-\frac{5}{7}-x+\frac{1}{2}=-\frac{11}{4}\)
\(-\frac{3}{14}-x=-\frac{11}{4}\)
\(x=-\frac{3}{14}-\left(-\frac{11}{4}\right)\)
\(x=\frac{71}{28}\)(thỏa mãn)
Vậy \(\orbr{\begin{cases}x=\frac{-43}{28}\\x=\frac{71}{28}\end{cases}}\)
a: Ta có: \(2x\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)
a. 6x2 - (2x + 5)(3x - 2) = 7
<=> 6x2 - 6x2 + 4x - 15x + 10 = 7
<=> -11x = -3
<=> \(x=\dfrac{3}{11}\)
b. (5 - x)(25 + 5x + x2) + x(x2 - 7) = 25
<=> 125 - x3 + x3 - 7x = 25
<=> -7x = 25 - 125
<=> -7x = -100
<=> \(x=\dfrac{100}{7}\)
c. (7 - 2x)2 + (3 + 2x)(3 - 2x) = 30
<=> 49 - 28x + 4x2 + 9 - 4x2 = 30
<=> 4x2 - 4x2 - 28x = 30 - 49 - 9
<=> -28x = -28
<=> x = 1
a: \(x^2-6x+5=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
b: \(x\left(x+5\right)+x\left(x+15\right)=0\)
\(\Leftrightarrow x\left(2x+20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-10\end{matrix}\right.\)
\(-2,5\times\left(x+\dfrac{1}{5}\right)=\dfrac{7}{5}\)
\(\Leftrightarrow x+\dfrac{1}{5}=\dfrac{7}{5}:\left(-2,5\right)\)
\(\Leftrightarrow x+\dfrac{1}{5}=\dfrac{7}{5}\times\left(-\dfrac{2}{5}\right)\)
\(\Leftrightarrow x+\dfrac{1}{5}=-\dfrac{14}{25}\)
\(\Leftrightarrow x=-\dfrac{19}{25}\)
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = (9 + 1) + (8 + 2) + (7 + 3) + (6 + 4) + 5
= 10 + 10 + 10 + 10 + 5 = 45
\(x^2=\frac{5}{7}x\)
\(\Rightarrow x^2-\frac{5}{7}x=0\)
\(\Rightarrow x\left(x-\frac{5}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-\frac{5}{7}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{7}\end{cases}}}\)