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f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
a.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$
$\Leftrightarrow \sqrt{2x}=3$
$\Leftrightarrow 2x=9$
$\Leftrightarrow x=\frac{9}{2}$ (tm)
b.
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$
$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$
$\Leftrightarrow 3\sqrt{x+2}=15$
$\Leftrightarrow \sqrt{x+2}=5$
$\Leftrightarrow x+2=25$
$\Leftrightarrow x=23$ (tm)
c.
$\sqrt{(x-2)^2}=12$
$\Leftrightarrow |x-2|=12$
$\Leftrightarrow x-2=12$ hoặc $x-2=-12$
$\Leftrightarrow x=14$ hoặc $x=-10$
e.
PT $\Leftrightarrow |2x-1|-x=3$
Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$
$\Leftrightarrow x=4$ (tm)
Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$
$\Leftrightarrow x=\frac{-2}{3}$ (tm)
Sửa lại câu c) đặt \(\sqrt{x}+1=\)t \(\Rightarrow\left[2\left(t+\dfrac{1}{2}\right)\right]\left(t-3\right)\)=7⇒\(\left\{{}\begin{matrix}t=3\\t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\x=\dfrac{9}{4}\end{matrix}\right.\)
a) \(\left(\sqrt{4-3x}\right)^2=8^2\)\(\Leftrightarrow4-3x=64\Rightarrow x=-20\)
b) \(\sqrt{4x-8}+1=12\sqrt{\dfrac{x-2}{9}}\Leftrightarrow2\sqrt{x-2}+1\)\(=\left(12\sqrt{\left(x-2\right).\dfrac{1}{9}}\right)\)
\(\Leftrightarrow2t+1=12.\dfrac{1}{3}t\) (Đặt t = \(\sqrt{x-2}\))
\(\Rightarrow t=\dfrac{1}{2}\) \(\Rightarrow\sqrt{x-2}=\dfrac{1}{2}\)\(\Rightarrow x=\dfrac{9}{4}\)
c) pt\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+1=7\\\sqrt{x}-2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)
a) \(\Leftrightarrow\sqrt{3}\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\sqrt{3}-1\right)=0\Leftrightarrow x=1\)
b) \(\Leftrightarrow\sqrt{\left(x-3\right)^2}=7\)
\(\Leftrightarrow\left|x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)
c) \(\Leftrightarrow3\left|x-2\right|=45\)
\(\Leftrightarrow\left|x-2\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=15\\x-2=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-13\end{matrix}\right.\)
\(a,PT\Leftrightarrow\sqrt{3}\left(x-1\right)=1-x\\ \Leftrightarrow\sqrt{3}\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(\sqrt{3}+1\right)=0\\ \Leftrightarrow x=1\left(\sqrt{3}+1\ne0\right)\\ b,ĐK:x\in R\\ PT\Leftrightarrow\left|x-3\right|=7\Leftrightarrow\left[{}\begin{matrix}x-3=7\\3-x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\\ c,ĐK:x\in R\\ PT\Leftrightarrow3\left|x-2\right|=45\Leftrightarrow\left|x-2\right|=15\\ \Leftrightarrow\left[{}\begin{matrix}x-2=15\\2-x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-13\end{matrix}\right.\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290