Tìm x , biết : 5x= 10 √ x
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| 5x - 10 | + 10 = 5x
<=> | 5x -10 | = 5x -10 . ĐK \(x\ge2\)
\(\Leftrightarrow\orbr{\begin{cases}5x-10=5x-10\\5x-10=10-5x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}0x=0\\10x=20\end{cases}\Leftrightarrow\orbr{\begin{cases}\text{vô số nghiệm }\\x=2\end{cases}}}\)
Vậy \(S=\left\{x\in R/x\ge2\right\}\)
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+TH1:\(x\le-7\)
Ta có: \(-x-1-x+2-x-7=5x-10\Rightarrow x=\dfrac{1}{2}\) (loại)
+TH2: \(-7\le x\le-1\)
Ta có: \(-x-1-x+2+x+7=5x-10\Rightarrow x=3\)(loại)
+TH3: \(-1\le x\le2\)
Ta có: \(x+1-x+2+x+7=5x-10\Rightarrow x=5\)(loại)
+TH4: \(x>2\)
Ta có: \(x+1+x-2+x+7=5x-10\Rightarrow x=8\)(tm)
Vậy x=8
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\(\left(4-2x\right)\left(5x-10\right)=0\\ \Leftrightarrow2\left(2-x\right)5\left(x-2\right)=0\\ \Leftrightarrow-10\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
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a) \(2.\left|5x-3\right|-2x=14\)
\(2\left|5x-3\right|=14+2x\)
\(\left|5x-3\right|=\frac{14+2x}{2}\)
\(\Rightarrow\orbr{\begin{cases}5x-3=\frac{-14-2x}{2}\\5x-3=\frac{14+2x}{2}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\left(5x-3\right).2=-14-2x\\\left(5x-3\right).2=14+2x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}10x-6+2x=-14\\10x-6-2x=14\end{cases}\Rightarrow\orbr{\begin{cases}12x=-14+6\\8x=14+6\end{cases}}}\Rightarrow\orbr{\begin{cases}12x=-8\\8x=20\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=2,5\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{-2}{3}\\x=2,5\end{cases}}\)
Những câu sau tương tự nhé.
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\(x^2-5x+10=0\)
\(\Leftrightarrow x^2-\frac{2.5x}{2}+\frac{25}{4}+\frac{15}{4}=0\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2+\frac{15}{4}=0\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=-\frac{15}{4}\)( Vô lí )
Vậy ko tìm đc x
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\(5x\left(3x^2y-2xy^2+1\right)-3xy\left(5x^2-3xy\right)+x^2y^2-10=0\)
\(\Leftrightarrow15x^3y-10x^2y^2+5x-15x^3y+9x^2y^2+x^2y^2-10=0\)
\(\Leftrightarrow5x=10\Leftrightarrow x=2\)
\(5x=10\sqrt{x}\)
\(\Leftrightarrow5x=10\cdot x^{\frac{1}{2}}\)
\(\Leftrightarrow x=2\cdot x^{\frac{1}{2}}\)
\(\Leftrightarrow x^{\frac{1}{2}}=2\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow\sqrt{x^2}=2^2\)
\(\Leftrightarrow x=4\)
Vậy x=4.
\(5x=10\sqrt{x}\)
\(\Rightarrow5\left(\sqrt{x}\right)^2=5.2\sqrt{x}\)
\(\Rightarrow\left(\sqrt{x}\right)^2=2\sqrt{x}\)
\(\Rightarrow\left(\sqrt{x}\right)^2-2\sqrt{x}=0\)
\(\Rightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\\sqrt{x}=2\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)