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a:Ta có: \(\sqrt{2x+9}=\sqrt{5-4x}\)
\(\Leftrightarrow2x+9=5-4x\)
\(\Leftrightarrow6x=-4\)
hay \(x=-\dfrac{2}{3}\left(nhận\right)\)
b: Ta có: \(\sqrt{2x-1}=\sqrt{x-1}\)
\(\Leftrightarrow2x-1=x-1\)
hay x=0(loại)
c: Ta có: \(\sqrt{x^2+3x+1}=\sqrt{x+1}\)
\(\Leftrightarrow x^2+3x=x\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
a. \(\sqrt{2x+9}=\sqrt{5-4x}\)
<=> 2x + 9 = 5 - 4x
<=> 2x + 4x = 5 - 9
<=> 6x = -4
<=> x = \(\dfrac{-4}{6}=\dfrac{-2}{3}\)
a) \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\) (x≥0) Đặt \(\sqrt{2x}\) = a ( a>0 )
Khi đó pt :
<=> 7+a =3 + \(\sqrt{5}\)
<=> 4+a = \(\sqrt{5}\)
<=> (4+a)\(^2\) = 5
<=> 16 + 8a + a\(^2\) = 5
<=>a\(^2\) + 8a+ 11 = 0
<=> a = -4 + \(\sqrt{5}\) (Loại) và a = -4-\(\sqrt{5}\)(Loại)
Vậy Pt vô nghiệm.
b) \(\sqrt{3x^2-4x}\) = 2x-3
<=> 3x\(^2\)- 4x = 4x\(^2\)-12x + 9
<=> x\(^2\)-8x+9 = 0
<=> x=1 , x=9
Vậy S={1;9}
c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}\) = 2
<=> \(\dfrac{\left(\sqrt{7-x}\right)^3+\left(\sqrt{x-5}\right)^3}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\dfrac{\left(\sqrt{7-x}+\sqrt{x-5}\right)\left(7-x-\sqrt{\left(7-x\right)\left(x-5\right)}+x-5\right)}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\sqrt{\left(7-x\right)\left(x-5\right)}=0\)
<=> x=7,x=5
Vậy x=5 hoặc x=7
a.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x-2\geq 0\\ x^2-2x+4=(2x-2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x^2-6x=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x(x-2)=0\end{matrix}\right.\Leftrightarrow x=2\)
b. ĐK: $x\geq 1$
PT $\Leftrightarrow \sqrt{(x-1)+2\sqrt{x-1}+1}=2$
$\Leftrightarrow \sqrt{(\sqrt{x-1}+1)^2}=2$
$\Leftrightarrow |\sqrt{x-1}+1|=2$
$\Leftrightarrow \sqrt{x-1}+1=2$
$\Leftrightarrow \sqrt{x-1}=1$
$\Leftrightarrow x=2$ (tm)
c.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 2x^2-2x+1=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x+1=4x^2-4x+1\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x=2x(x-1)=0\end{matrix}\right.\Leftrightarrow x=1\) (tm)
d.
ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow \sqrt{(x-4)+4\sqrt{x-4}+4}=2$
$\Leftrightarrow \sqrt{(\sqrt{x-4}+2)^2}=2$
$\Leftrightarrow |\sqrt{x-4}+2|=2$
$\Leftrightarrow \sqrt{x-4}+2=2$
$\Leftrightarrow \sqrt{x-4}=0$
$\Leftrightarrow x=4$ (tm)
a. ĐKXĐ: \(-1\le x\le1\)
Đặt \(\sqrt{1+x}+\sqrt{1-x}=t>0\)
\(\Rightarrow t^2=2+2\sqrt{1-t^2}\)
Pt trở thành:
\(t.t^2=8\Leftrightarrow t^3=8\Leftrightarrow t=2\)
\(\Rightarrow\sqrt{1+x}+\sqrt{1-x}=2\)
\(\Leftrightarrow2+2\sqrt{1-x^2}=2\)
\(\Leftrightarrow1-x^2=0\Rightarrow x=\pm1\)
b.
ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\)
Pt trở thành:
\(t=t^2-4-16\Leftrightarrow...\)
a: ĐKXĐ: x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x>=1/2
\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)
=>\(\sqrt{2x-1}-2\sqrt{2x-1}+5=0\)
=>\(5-\sqrt{2x-1}=0\)
=>\(\sqrt{2x-1}=5\)
=>2x-1=25
=>2x=26
=>x=13(nhận)
c: \(\sqrt{x^2-10x+25}=2\)
=>\(\sqrt{\left(x-5\right)^2}=2\)
=>\(\left|x-5\right|=2\)
=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
d: \(\sqrt{x^2-14x+49}-5=0\)
=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)
=>\(\sqrt{\left(x-7\right)^2}=5\)
=>|x-7|=5
=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)
\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)
\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
a: \(x^2\cdot2\sqrt{3}+x+1=\sqrt{3}\cdot\left(x+1\right)\)
=>\(x^2\cdot2\sqrt{3}+x\left(1-\sqrt{3}\right)+1-\sqrt{3}=0\)
\(\text{Δ}=\left(1-\sqrt{3}\right)^2-4\cdot2\sqrt{3}\left(1-\sqrt{3}\right)\)
\(=4-2\sqrt{3}-8\sqrt{3}\left(1-\sqrt{3}\right)\)
\(=4-2\sqrt{3}-8\sqrt{3}+24=28-10\sqrt{3}=\left(5-\sqrt{3}\right)^2>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x_1=\dfrac{-\left(1-\sqrt{3}\right)-\left(5-\sqrt{3}\right)}{2\cdot2\sqrt{3}}=\dfrac{-1+\sqrt{3}-5+\sqrt{3}}{4\sqrt{3}}=\dfrac{1-\sqrt{3}}{2}\\x_2=\dfrac{-\left(1-\sqrt{3}\right)+5-\sqrt{3}}{2\cdot2\sqrt{3}}=\dfrac{4}{4\sqrt{3}}=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
b: \(5x^2-3x+1=2x+31\)
=>\(5x^2-3x+1-2x-31=0\)
=>\(5x^2-5x-30=0\)
=>\(x^2-x-6=0\)
=>(x-3)(x+2)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
c: \(x^2+2\sqrt{2}x+4=3\left(x+\sqrt{2}\right)\)
=>\(x^2+2\sqrt{2}x+4-3x-3\sqrt{2}=0\)
=>\(x^2+x\left(2\sqrt{2}-3\right)+4-3\sqrt{2}=0\)
\(\text{Δ}=\left(2\sqrt{2}-3\right)^2-4\left(4-3\sqrt{2}\right)\)
\(=17-12\sqrt{2}-16+12\sqrt{2}=1\)>0
Do đó, phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x_1=\dfrac{-\left(2\sqrt{2}-3\right)-1}{2}=\dfrac{-2\sqrt{2}+3-1}{2}=-\sqrt{2}+1\\x_2=\dfrac{-\left(2\sqrt{2}-3\right)+1}{2}=\dfrac{-2\sqrt{2}+4}{2}=-\sqrt{2}+2\end{matrix}\right.\)
a, ĐKXĐ : \(x\ge\dfrac{1}{2}\)
PT <=> 2x - 1 = 5
<=> x = 3 ( TM )
Vậy ...
b, ĐKXĐ : \(x\ge5\)
PT <=> x - 5 = 9
<=> x = 14 ( TM )
Vậy ...
c, PT <=> \(\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy ...
d, PT<=> \(\left|x-3\right|=3-x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=x-3\\x-3=3-x\end{matrix}\right.\)
Vậy phương trình có vô số nghiệm với mọi x \(x\le3\)
e, ĐKXĐ : \(-\dfrac{5}{2}\le x\le1\)
PT <=> 2x + 5 = 1 - x
<=> 3x = -4
<=> \(x=-\dfrac{4}{3}\left(TM\right)\)
Vậy ...
f ĐKXĐ : \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)
PT <=> \(x^2-x=3-x\)
\(\Leftrightarrow x=\pm\sqrt{3}\) ( TM )
Vậy ...
a) \(\sqrt{2x-1}=\sqrt{5}\) (x \(\ge\dfrac{1}{2}\))
<=> 2x - 1 = 5
<=> x = 3 (tmđk)
Vậy S = \(\left\{3\right\}\)
b) \(\sqrt{x-5}=3\) (x\(\ge5\))
<=> x - 5 = 9
<=> x = 4 (ko tmđk)
Vậy x \(\in\varnothing\)
c) \(\sqrt{4x^2+4x+1}=6\) (x \(\in R\))
<=> \(\sqrt{\left(2x+1\right)^2}=6\)
<=> |2x + 1| = 6
<=> \(\left[{}\begin{matrix}\text{2x + 1=6}\\\text{2x + 1}=-6\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)(tmđk)
Vậy S = \(\left\{\dfrac{5}{2};\dfrac{-7}{2}\right\}\)