sqrt(x-2+sqrt(2x-5)) + sqrt(x+2+sqrt(18x-45)) =7sqrt(2)
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AH
Akai Haruma
Giáo viên
1 tháng 10 2021
Lời giải:
a. ĐKXĐ: $x\geq 0$
$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$
$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$
$\Leftrightarrow 13\sqrt{2x}=28$
$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$
$\Leftrightarrow 2x=\frac{784}{169}$
$\Leftrightarrow x=\frac{392}{169}$
b. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x-5=4$
$\Leftrightarrow x=9$ (tm)
c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$
PT $\Leftrightarrow \frac{3x-2}{x+1}=9$
$\Rightarrow 3x-2=9(x+1)$
$\Leftrightarrow x=\frac{-11}{6}$ (tm)
Y
18 tháng 8 2023
\(a,3\sqrt{2x}+\sqrt{8x}-\sqrt{18x}=16\left(dk:x\ge0\right)\\ \Leftrightarrow3\sqrt{2x}+2\sqrt{2x}-3\sqrt{2x}=16\\ \Leftrightarrow\sqrt{2x}\left(3+2-3\right)=16\\ \Leftrightarrow2\sqrt{2x}=16\\ \Leftrightarrow\sqrt{2x}=8\\ \Leftrightarrow\left|2x\right|=64\\ \Leftrightarrow2x=64\\ \Leftrightarrow x=32\left(tm\right)\)
Vậy \(S=\left\{32\right\}\)
\(b,\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\left(dk:x\ge-5\right)\)
\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9\left(x+5\right)}=6\\ \Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}\left(2-3+4\right)=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=2\\ \Leftrightarrow\left|x+5\right|=4\\ \Leftrightarrow x+5=4\\ \Leftrightarrow x=-1\left(tm\right)\)
Vậy \(S=\left\{-1\right\}\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
BC
0
2 tháng 9 2021
a,ĐK: x≥4
Ta có: \(2\sqrt{x-4}-\dfrac{1}{3}\sqrt{9x-36}=4-\sqrt{x-4}\)
\(\Leftrightarrow2\sqrt{x-4}-\sqrt{x-4}=4-\sqrt{x-4}\)
\(\Leftrightarrow2\sqrt{x-4}=4\)
\(\Leftrightarrow\sqrt{x-4}=2\Leftrightarrow x-4=4\Leftrightarrow x=8\left(tm\right)\)
2 tháng 9 2021
b, ĐK: x≥2
Ta có: \(3\sqrt{x-2}-\sqrt{x^2-4}=0\)
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(3-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\3-\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)
28 tháng 1 2019
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
a. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$
$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$
$\Leftrightarrow -\sqrt{x-1}=-17$
$\Leftrightarrow \sqrt{x-1}=17$
$\Leftrightarrow x-1=289$
$\Leftrightarrow x=290$
b. ĐKXĐ: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$
$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$
$\Leftrihgtarrow \sqrt{2x-1}=2$
$\Leftrightarrow x=2,5$ (tm)
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm