\(\sqrt{x+2}+\sqrt{5x+6}+2\sqrt{8x+9}=4x^2\)
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Những câu hỏi liên quan
16 tháng 6 2021
a) ĐK: \(x\ge3\)
PT \(\Leftrightarrow\sqrt{\left(x-3\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+1}-\sqrt{\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-3}-1\right)+\sqrt{x+1}\left(1-\sqrt{x-3}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+1}\right)\left(\sqrt{x-3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+1}\\\sqrt{x-3}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=x+1\\x-3=1\end{matrix}\right.\) \(\Leftrightarrow x=4\) (Thỏa mãn)
Vậy ...
25 tháng 11 2021
\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
a. ĐKXĐ: $x\geq 2$ hoặc $x=1$
PT $\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}$
$\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0$
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)
b.
PT $\Leftrightarrow \sqrt{(x-2)^2}=\sqrt{(2x-3)^2}$
$\Leftrightarrow |x-2|=|2x-3|$
\(\Leftrightarrow \left[\begin{matrix} x-2=2x-3\\ x-2=3-2x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
c. ĐKXĐ: $x=2$ hoặc $x\geq 3$
PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x-3}-1)=0$
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x-3}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=4\end{matrix}\right.\) (đều tm)
d.
PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$
$\Leftrightarrow |2x-1|=|x-3|$
\(\Leftrightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)
PA
25 tháng 7 2017
\(x^2-4x-6=\sqrt{2x^2-8x+12}\)
\(\Leftrightarrow\left(x^2+2x\right)-\left(6x+6+\sqrt{2x^2-8x+12}\right)=0\)
\(\Leftrightarrow x\left(x+2\right)-\dfrac{36x^2+72x+36-\left(2x^2-8x+12\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}=0\)
\(\Leftrightarrow x\left(x+2\right)-\dfrac{2\left(17x+6\right)\left(x+2\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}=0\)
\(\Leftrightarrow\left(x+2\right)\left[x-\dfrac{2\left(17x+6\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}\right]=0\)
Pt \(x-\dfrac{2\left(17x+6\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}\) vô nghiệm
=> x + 2 = 0
<=> x = - 2 (nhận)
PA
25 tháng 7 2017
\(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-2}-2\right|+\left|\sqrt{x-2}-3\right|=1\)
Ta có:
\(VT=\left|\sqrt{x-2}-2\right|+\left|3-\sqrt{x-2}\right|\ge\left|\sqrt{x-2}-2+3-\sqrt{x-2}\right|=1\)
Dấu "=" xảy ra khi \(\left(\sqrt{x-2}-2\right)\left(3-\sqrt{x-2}\right)\ge0\)
Bảng xét dấu:
Vậy \(6\le x\le11\)