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\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
a) \(\sqrt{25x+75}+3\sqrt{x-2}=2+4\sqrt{x+3}+\sqrt{9x-18}\) (ĐKXĐ : \(x\ge2\) )
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}-4\sqrt{x+3}-3\sqrt{x-2}=2\)
\(\Leftrightarrow\sqrt{x+3}=2\)
\(\Leftrightarrow x+3=4\)
\(\Leftrightarrow x=1\) ( Thỏa mãn ĐKXĐ )
c) \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\) (ĐKXĐ : \(x\ge-5\) )
\(\Leftrightarrow2\sqrt{x+5}+\sqrt{x+5}-\sqrt{x+5}=4\)
\(\Leftrightarrow2\sqrt{x+5}=4\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\) ( Thỏa mãn ĐKXĐ )
Vậy.......
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.
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.
a: \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
=>4x-4=2x-3
=>2x=1
hay x=1/2
b: \(\Leftrightarrow\sqrt{\dfrac{2x-3}{x-1}}=2\)
=>(2x-3)=4x-4
=>4x-4=2x-3
=>2x=1
hay x=1/2(nhận)
c: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=-3/2 hoặc x=7/2
e: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
=>căn (x-5)=2
=>x-5=4
hay x=9
a.
\(DK:49-28x-4x^2\ge0\)
PT\(\Leftrightarrow\sqrt{49-28x-4x^2}=5\)
\(\Leftrightarrow49-28x-4x^2=25\)
\(\Leftrightarrow4x^2+28x-24=0\)
\(\Leftrightarrow x^2+7x-6=0\)
Ta co:
\(\Delta=7^2-4.1.\left(-6\right)=73>0\)
\(\Rightarrow\hept{\begin{cases}x_1=\frac{-7+\sqrt{73}}{2}\left(n\right)\\x_2=\frac{-7-\sqrt{73}}{2}\left(n\right)\end{cases}}\)
Vay nghiem cua PT la \(\hept{\begin{cases}x_1=\frac{-7+\sqrt{73}}{2}\\x_2=\frac{-7-\sqrt{73}}{2}\end{cases}}\)
k) ĐK: $x^2\geq 5$
PT $\Leftrightarrow 2\sqrt{x^2-5}-\frac{1}{3}\sqrt{x^2-5}+\frac{3}{4}\sqrt{x^2-5}-\frac{5}{12}\sqrt{x^2-5}=4$
$\Leftrightarrow 2\sqrt{x^2-5}=4$
$\Leftrightarrow \sqrt{x^2-5}=2$
$\Rightarrow x^2-5=4$
$\Leftrightarrow x^2=9\Rightarrow x=\pm 3$ (đều thỏa mãn)
l) ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 2\sqrt{x+1}+3\sqrt{x+1}-\sqrt{x+1}=4$
$\Leftrightarrow 4\sqrt{x+1}=4$
$\Leftrightarrow \sqrt{x+1}=1$
$\Rightarrow x+1=1$
$\Rightarrow x=0$
m)
ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 4\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}+3\sqrt{x+1}$
$\Leftrightarrow 6\sqrt{x+1}=16+2\sqrt{x+1}$
$\Leftrightarrow 4\sqrt{x+1}=16$
$\Leftrightarrow \sqrt{x+1}=4$
$\Rightarrow x=15$ (thỏa mãn)
h)
ĐKXĐ: $x\geq -5$
PT $\Leftrightarrow \sqrt{x+5}=6$
$\Rightarrow x+5=36\Rightarrow x=31$ (thỏa mãn)
i) ĐKXĐ: $x\geq 5$
PT \(\Leftrightarrow \sqrt{x-5}+4\sqrt{x-5}-\sqrt{x-5}=12\)
\(\Leftrightarrow 4\sqrt{x-5}=12\Leftrightarrow \sqrt{x-5}=3\Rightarrow x-5=9\Rightarrow x=14\) (thỏa mãn)
j)
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 3\sqrt{2x}+\sqrt{2x}-6\sqrt{2x}+4=0$
$\Leftrightarrow -2\sqrt{2x}+4=0$
$\Leftrightarrow \sqrt{2x}=2$
$\Rightarrow x=2$ (thỏa mãn)
a) \(\sqrt{2x-3}=7\) ( ĐKXĐ : \(x\ge\dfrac{3}{2}\) )
\(\Leftrightarrow2x-3=49\)
\(\Leftrightarrow2x=52\)
\(\Leftrightarrow x=26\) ( thỏa mãn điều kiện xác định )
Vậy phương trình có nghiệm x = 26 .
b) \(\sqrt{x^2-4x+4}=\sqrt{6-2\sqrt{5}}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(\Leftrightarrow|x-2|-\sqrt{5}+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2-\sqrt{5}+1=0\\2-x-\sqrt{5}+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1-\sqrt{5}\\x=\sqrt{5}-3\end{matrix}\right.\)
Vậy ...............
P/s : Mình không chắc bài này có đúng không nữa .
\(a,đk:x\ge5\\ \Leftrightarrow\sqrt{x-5}+\sqrt{4\left(x-5\right)}-\dfrac{1}{5}\sqrt{9\left(x-5\right)}=3\\ \Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{1}{5}.3\sqrt{x-5}=3\\ \Leftrightarrow\dfrac{12}{5}\sqrt{x-5}=3\\ \Rightarrow\sqrt{x-5}=\dfrac{5}{4}\\ \Leftrightarrow\left(\sqrt{x-5}\right)^2=\left(\dfrac{5}{4}\right)^2\\ \Leftrightarrow x-5=\dfrac{25}{16}\\ \Rightarrow x=\dfrac{25}{16}+5\\ \Rightarrow x=\dfrac{105}{16}\left(t|m\right)\)
\(b,đk:x\ge1\\ \Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}=-2\\ \Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\\ \Leftrightarrow-2\sqrt{x-1}=-2\\ \Leftrightarrow\sqrt{x-1}=1\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\left(t|m\right)\)