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a: ĐKXĐ: x>=3
Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)
=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)
=>\(\dfrac{3}{2}\sqrt{x-3}=3\)
=>\(\sqrt{x-3}=2\)
=>x-3=4
=>x=7(nhận)
b: ĐKXĐ: x>=0
\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)
=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)
=>\(7\sqrt{x}-5< =0\)
=>\(\sqrt{x}< =\dfrac{5}{7}\)
=>0<=x<=25/49
c: ĐKXĐ: x>=5
\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)
=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)
=>\(\dfrac{3}{2}\sqrt{x-5}=3\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$
$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$
$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$
Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$
$\Leftrightarrow \sqrt{x}=\frac{12}{5}$
$\Leftrightarrow x=5,76$ (thỏa mãn)
b. ĐKXĐ: $x^2\geq 5$
PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$
$\Leftrightarrow \sqrt{x^2-5}=0$
$\Leftrightarrow x=\pm \sqrt{5}$
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)
a) \(\sqrt{2}\cdot x-\sqrt{50}=0< =>\sqrt{2}\cdot x=\sqrt{50}\)
<=> x= 5
b) \(\sqrt{3}\cdot x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)
<=> \(\sqrt{3}\cdot\left(x+1\right)=\sqrt{3}\cdot\sqrt{4}+\sqrt{3}\cdot\sqrt{9}\)
<=> \(\sqrt{3}\cdot\left(x+1\right)=\sqrt{3}\cdot5< =>x+1=5\)
<=> x=4
c) \(\sqrt{3}\cdot x^2-\sqrt{12}=0\\ < =>x^2=\sqrt{4}=2;-2\\ < =>x=\sqrt{2};-\sqrt{2}\)
d) \(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\\ < =>x^2=\sqrt{100}=10;-10\\ < =>x=\sqrt{10};-\sqrt{10}\)
a, \(\sqrt{2}x-\sqrt{50}=0\Leftrightarrow\sqrt{2}x-5\sqrt{2}=0\Leftrightarrow\sqrt{2}\left(x-5\right)=0\Leftrightarrow x=5\)
b, \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3}\left(x+1\right)=5\sqrt{3}\Leftrightarrow x+1=5\Leftrightarrow x=4\)
c, \(\sqrt{3}x^2-\sqrt{12}=0\Leftrightarrow\sqrt{3}\left(x^2-2\right)=0\Leftrightarrow x^2-2=0\Leftrightarrow x=\pm\sqrt{2}\)
d, \(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\Leftrightarrow\dfrac{1}{\sqrt{5}}\left(x^2-10\right)=0\Leftrightarrow x^2-10=0\Leftrightarrow x=\pm\sqrt{10}\)
Trần Hữu Ngọc Minh xem tôi làm có đúng ko?
Giải:
a, \(\sqrt{2}.x-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}.x=\sqrt{50}\Leftrightarrow\sqrt{2}.x=\sqrt{25.2}\)
\(\Leftrightarrow\sqrt{2}.x=\sqrt{25}.\sqrt{2}\Leftrightarrow\sqrt{2}.x=5\sqrt{2}\)
\(\Leftrightarrow x=5\)
c, \(\sqrt{3}.x^2-\sqrt{12}=0\)
\(\Leftrightarrow\sqrt{3}.x^2=\sqrt{12}\)
\(\Leftrightarrow\sqrt{3}.x^2=\sqrt{4.3}\)
\(\Leftrightarrow\sqrt{3}.x^2=\sqrt{4}.\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}.x^2=2\sqrt{3}\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
d, \(\frac{x^2}{\sqrt{5}}-\sqrt{20}=0\)
\(\Leftrightarrow\frac{x^2}{\sqrt{5}}=\sqrt{20}\)
\(\Leftrightarrow x^2=\sqrt{5}.\sqrt{20}\)
\(\Leftrightarrow x^2=\sqrt{100}\)
\(\Leftrightarrow x=\pm10\)
a: ĐKXĐ: x>=-3/2
\(\sqrt{x^2+4}=\sqrt{2x+3}\)
=>\(x^2+4=2x+3\)
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1(nhận)
b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))
=>\(\sqrt{\left(x-3\right)^2}=2x-1\)
=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>x=4/3(nhận) hoặc x=-2(loại)
c:
Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)
ĐKXĐ: \(x>=-3\)
\(\sqrt{4x+12}=\sqrt{9x+27}-5\)
=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)
=>\(-\sqrt{x+3}=-5\)
=>x+3=25
=>x=22(nhận)
d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)
=>\(4x^2-6x+1=4x^2-20x+25\)
=>\(-6x+20x=25-1\)
=>\(14x=24\)
=>x=12/7(nhận)
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) \(\sqrt{2}x-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}x=\sqrt{50}\)
\(\Leftrightarrow x=\frac{\sqrt{50}}{\sqrt{2}}=\sqrt{\frac{50}{2}}=\sqrt{25}=5\)
b) \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)
\(\Leftrightarrow\sqrt{3}\left(x+1\right)=2\sqrt{3}+3\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}x=5\sqrt{3}\)
\(\Leftrightarrow x=5\)
c) \(\sqrt{3}x^2-\sqrt{12}=0\)
\(\Leftrightarrow\sqrt{3}\left(x^2-2\right)=0\)
\(\Leftrightarrow x^2-2=0\)
\(\Leftrightarrow x^2=2\Leftrightarrow\left[\begin{array}{nghiempt}x=\sqrt{2}\\x=-\sqrt{2}\end{array}\right.\)
d) \(\frac{x^2}{\sqrt{5}}-\sqrt{20}=0\)
\(\Leftrightarrow\)\(\frac{1}{\sqrt{5}}\left(x^2-10\right)=0\)
\(\Leftrightarrow x^2-10=0\)
\(\Leftrightarrow x^2=10\Leftrightarrow\left[\begin{array}{nghiempt}x=\sqrt{10}\\x=-\sqrt{10}\end{array}\right.\)