Giải PT:
\(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
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NN
2
9 tháng 9 2021
a) \(\sqrt{\left(x-3\right)^2}=2\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\left(đk:x\ge-2\right)\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\Leftrightarrow\sqrt{x+2}=3\Leftrightarrow x+2=9\Leftrightarrow x=7\)
9 tháng 9 2021
a: \(\sqrt{\left(x-3\right)^2}=2\)
\(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
7 tháng 10 2021
c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=4\)
hay x=5
e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)
\(\Leftrightarrow\left|2x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
8 tháng 10 2021
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$
$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$
$\Leftrightarrow x\leq 2$
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 1=2\sqrt{x-2}$
$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$
$\Leftrightarrow \frac{1}{4}=x-2$
$\Leftrightarrow x=\frac{9}{4}$ (tm)
4 tháng 7 2021
a) Ta có: \(\sqrt{\left(x-3\right)^2}=2\)
\(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) ĐKXĐ: \(x\ge-2\)
Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+\dfrac{4}{5}\cdot5\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7(thỏa ĐK)
TG
4 tháng 7 2021
a) \(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy:.....
b) ĐKXĐ: x ≥ -2
\(\Leftrightarrow\sqrt{9}.\sqrt{x+2}-5.\sqrt{x+2}+\dfrac{4}{5}.\sqrt{25}.\sqrt{x+2}=6\)
<=> \(\sqrt{x+2}.\left(3-5+\dfrac{4}{5}.5\right)=6\)
\(\Leftrightarrow2.\sqrt{x+2}=6\)
\(\Leftrightarrow\sqrt{x+2}=3\)
<=> x + 2 = 9
<=> x = 7
19 tháng 8 2016
a/ \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐKXĐ : \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow2\sqrt{x-1}=2\Leftrightarrow x-1=1\Leftrightarrow x=2\)
b/ \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}+3=0\)
<=> 3 = 0 (vô lý)
=> pt vô nghiệm.
19 tháng 8 2016
c/ \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (ĐKXĐ : x>-5/7)
\(\Leftrightarrow9x-7=7x+5\Leftrightarrow2x=12\Leftrightarrow x=6\)
d/ \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\) (ĐKXĐ : \(x\ge\frac{3}{2}\))
\(\Leftrightarrow2x-3=4\left(x-1\Leftrightarrow\right)2x=1\Leftrightarrow x=\frac{1}{2}\) (loại)
Vậy pt vô nghiệm.
16 tháng 8 2023
\(\dfrac{1}{5}\sqrt[]{25x+50}-5\sqrt[]{x+2}+\sqrt[]{9x+18}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}\sqrt[]{25\left(x+2\right)}-5\sqrt[]{x+2}+\sqrt[]{9\left(x+2\right)}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}.5\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}\left(1-5+3\right)+9=0\)
\(\Leftrightarrow-\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}=9\)
\(\Leftrightarrow x+2=81\)
\(\Leftrightarrow x=79\)
1 tháng 11 2021
\(Đk:x\ge2\\ PT\Leftrightarrow\dfrac{10\sqrt{x-2}-\sqrt{x-2}+1}{2}=6\sqrt{x-2}\\ \Leftrightarrow9\sqrt{x-2}+1=12\sqrt{x-2}\\ \Leftrightarrow\sqrt{x-2}=\dfrac{1}{3}\Leftrightarrow x-2=\dfrac{1}{9}\\ \Leftrightarrow x=\dfrac{19}{9}\left(tm\right)\)
AT
4 tháng 7 2021
a) \(\sqrt{\left(x-3\right)^2}=2\Rightarrow\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Rightarrow\sqrt{9\left(x+2\right)}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25\left(x+2\right)}=6\)
\(\Rightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Rightarrow2\sqrt{x+2}=6\Rightarrow\sqrt{x+2}=3\Rightarrow x+2=9\Rightarrow x=7\)
\(Q=\dfrac{1}{x-2\sqrt{x}+3}\)
Ta có: \(x-2\sqrt{x}+3=x-2\sqrt{x}+1+2=\left(\sqrt{x}-1\right)^2+2\ge2\)
\(\Rightarrow\dfrac{1}{x-2\sqrt{x}+3}\le2\Rightarrow Q_{max}=2\) khi \(x=1\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\sqrt{1-8x+16x^2}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{1^2-2\cdot4x\cdot1+\left(4x\right)^2}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{\left(4x-1\right)^2}=\dfrac{1}{3}\)
\(\Leftrightarrow\left|4x-1\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=\dfrac{1}{3}\left(ĐK:x\ge\dfrac{1}{4}\right)\\4x-1=\dfrac{1}{3}\left(ĐK:x< \dfrac{1}{4}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=\dfrac{1}{6}\left(tm\right)\end{matrix}\right.\)
b) \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\) (ĐK: \(x\ge2\))
\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)
\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)
\(\Leftrightarrow6\sqrt{x-2}=18\)
\(\Leftrightarrow\sqrt{x-2}=3\)
\(\Leftrightarrow x-2=9\)
\(\Leftrightarrow x=9+2\)
\(\Leftrightarrow x=11\left(tm\right)\)
6 tháng 7 2021
a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)
\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)
\(\Leftrightarrow\sqrt{x+3}=3\)
\(\Leftrightarrow x+3=9\)
hay x=6
b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)
\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)
\(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
\(\Leftrightarrow\sqrt{9\left(x^2+2\right)}+2\sqrt{x^2+2}-\sqrt{25\left(x^2+2\right)}+3=0\)
\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}=-3\)
\(\Leftrightarrow0.\sqrt{x^2+2}=-3\)
\(\Leftrightarrow0=-3\)( vô lý)
Vậy pt vô nghiệm
Đặt \(\sqrt{x^2+2}=a\). Ta có:
\(\sqrt{9x^2+18}+2a-\sqrt{25x^2+50}+3=0\)
\(\Rightarrow\sqrt{9\left(x^2+2\right)}+2a-\sqrt{25\left(x^2+2\right)}+3=0\)
\(\Rightarrow3a+2a-5a+3=0\)
\(\Rightarrow0=-3\) (vô lí)
Vậy phương trình vô nghiệm