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Lời giải:
a) ĐK: \(x>0; x\neq 25; x\neq 36\)
PT \(\Rightarrow (\sqrt{x}-2)(\sqrt{x}-6)=(\sqrt{x}-5)(\sqrt{x}-4)\)
\(\Leftrightarrow x-8\sqrt{x}+12=x-9\sqrt{x}+20\)
\(\Leftrightarrow \sqrt{x}=8\Rightarrow x=64\) (thỏa mãn)
Vậy.......
b)
ĐK: \(x\geq \frac{-1}{2}\)
PT \(\Leftrightarrow \sqrt{9(2x+1)}-\sqrt{4(2x+1)}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow 3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow \frac{4}{3}\sqrt{2x+1}=4\Leftrightarrow \sqrt{2x+1}=3\)
\(\Rightarrow x=\frac{3^2-1}{2}=4\) (thỏa mãn)
c)
ĐK: \(x\geq 2\)
PT \(\Leftrightarrow \sqrt{4(x-2)}-\frac{1}{2}\sqrt{x-2}+\sqrt{9(x-2)}=9\)
\(\Leftrightarrow 2\sqrt{x-2}-\frac{1}{2}\sqrt{x-2}+3\sqrt{x-2}=9\)
\(\Leftrightarrow \frac{9}{2}\sqrt{x-2}=9\Leftrightarrow \sqrt{x-2}=2\Rightarrow x=2^2+2=6\) (thỏa mãn)
ĐKXĐ:
a/ \(x-2020>0\Rightarrow x>2020\)
b/ \(x\ne0\)
c/ \(3x+5< 0\Rightarrow x< -\frac{5}{3}\)
d/ \(\frac{x-3}{1-x}\ge0\Rightarrow1< x\le3\)
Bài 2: ĐKXĐ tự tìm
a/ \(2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)
\(\Leftrightarrow13\sqrt{2x}=28\Rightarrow\sqrt{2x}=\frac{28}{13}\)
\(\Rightarrow x=\frac{392}{169}\)
b/ \(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\Rightarrow x=9\)
c/ \(3\sqrt{2x+1}>15\Rightarrow\sqrt{2x+1}>5\)
\(\Rightarrow2x+1>25\Rightarrow x>12\)
d/ \(\sqrt{x}+1>12\Rightarrow\sqrt{x}>11\Rightarrow x>121\)
a/ ĐK: \(x \ge -1\). Đặt \(\sqrt{x+1}=a \ge 0\)
PT: \(\Leftrightarrow6a-3a-2a=5\)
\(\Leftrightarrow a=5\)
\(\Leftrightarrow x+1=15\Leftrightarrow x=24\) (nhận)
b,c: Hai ý này đều làm theo cách bình phương hoặc đưa về phương trình chứa dấu giá trị tuyệt đối được nhé.
b) Cách 1: ĐKXĐ: Tự tìm
\(\sqrt{x^{2}-4x+4}=2\Leftrightarrow x^{2}-4x+4=4\Leftrightarrow x(x-4)=0\)
\(\Leftrightarrow x=0\) hoặc \(x=4\) cả 2 cái này đều TMĐK
Cách 2: \((\sqrt{x^2-4x+4}=2)\)
\(\Leftrightarrow \sqrt{(x-2)^2}=2\)
\(\Leftrightarrow \mid x-2\mid=2\)
Với \(x\geq 2\) thì :
\(x-2=2 \Leftrightarrow x=4\) (nhận)
Với \(x<2\) thì
\(-x-2=2\Leftrightarrow x=0\) (nhận)
Vậy \(S={0;4}\)
c) Cách 1: \(\sqrt{x^{2}-6x+9}=x-2\Leftrightarrow \left\{\begin{matrix}x\geq 2 \\ x^{2}-6x+9=x^{2}-4x+4 \end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix}x\geq 2 \\ x=\frac{5}{2} \end{matrix}\right.\)
Nghiệm TMĐK
Cách 2: \((\sqrt{x^2-6x+9}=x-2)\)
\(\Leftrightarrow \mid x-3\mid =x-2\)
Với \(x\geq 3\) thì
\(x-3=x-2\Leftrightarrow 0x=-1\) ( vô lý)
Với \(x<3\) thì
\(-x+3=x-2\Leftrightarrow -2x=-5 \Leftrightarrow x=\frac{5}{2}\)
Vậy \(S={\frac{5}{2}}\)
d) ĐKXĐ: Tự tìm
\(\sqrt{x^{2}+4}=\sqrt{2x+3}\Leftrightarrow x^{2}+4=2x+3\Leftrightarrow x^{2}-2x+1=0\Leftrightarrow (x-1)^{2}=0\)
\(\Leftrightarrow x=1\)
e) ĐKXĐ: \(x\geq \frac{3}{2}\)
\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\Leftrightarrow \frac{2x-3}{x-1}=4\Rightarrow 2x-3=4x-4\Leftrightarrow x=\frac{1}{2}\)
Nghiệm không TMĐK.
Phương trình vô nghiệm.
f) ĐKXĐ: \(x\geq \frac{-15}{2}\)
\(x+\sqrt{2x+15}=0\Leftrightarrow 2x+2\sqrt{2x+15}=0\Leftrightarrow 2x+15+2\sqrt{2x+15}+1-16=0\)
\(\Leftrightarrow (\sqrt{2x+15}+1)^{2}-4^{2}=0\Leftrightarrow (\sqrt{2x+15}+5)(\sqrt{2x+15}-3)=0\)
\(\Leftrightarrow \sqrt{2x+15}-3=0\Leftrightarrow \sqrt{2x+15}=3\Leftrightarrow 2x+15=9\Leftrightarrow x=-3\) (TMĐK)
a,
\(\Leftrightarrow\sqrt{1-x}=\frac{x-1}{\sqrt{6-x}+\sqrt{-5-2x}}\)
\(\Leftrightarrow-\sqrt{1-x}=\sqrt{6-x}+\sqrt{-5-2x}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{1-x}=\sqrt{6-x}-\sqrt{-5-2x}\\-\sqrt{1-x}=\sqrt{6-x}+\sqrt{-5-2x}\end{cases}}\)
b,tự nàm
c,
\(\Leftrightarrow64x^2-64x-64=64\sqrt{8x+1}\)
\(\Leftrightarrow\left(8x+1\right)^2=10\left(8x+1\right)+64\sqrt{8x+1}+55\)
đặt \(\sqrt{8x+1}=a\)
=>a4=10a2+64a+55
nhận thấy phương trình có dạng x4=ax2+bx+c
tìm số m sao cho b2-4(2m+a)(m2+c)=0
sau đó đưa về (x2+m)2=k2 với k là 1 số bất kì,sau đó giải ra
b)đk \(x\ge1\)
\(\sqrt{1+x^2+\frac{x^2}{\left(x+1\right)^2}}+\frac{x}{x+1}=\sqrt{\frac{\left(x+1\right)^2+x^2.\left(x+1\right)^2+x^2}{\left(x+1\right)^2}}+\frac{x}{x+1}\)
\(=\sqrt{\frac{x^4+2x^3+3x^2+2x+1}{\left(x+1\right)^2}}+\frac{x}{x+1}\)
\(=\sqrt{\frac{\left(x^2+x+1\right)^2}{\left(x+1\right)^2}}+\frac{x}{x+1}\)
\(=\frac{x^2+x+1}{x+1}+\frac{x}{x+1}=x+1\)
\(\Rightarrow\sqrt{1+2012^2+\frac{2012^2}{2013^2}}+\frac{2012}{2013}=2013\)
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2013\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2013\)
\(\Leftrightarrow x+\left|x-2\right|=2014\)
giai 2 pt
pt1 x+x-2=2014
x=1008
pt2 x+2-x=2014(vô lý)
a) \(\frac{3}{4}\sqrt{x}-\sqrt{9x}+5=\frac{1}{4}\sqrt{9x}\)
ĐK : x ≥ 0
⇔ \(\frac{3}{4}\sqrt{x}-\sqrt{3^2x}-\frac{1}{4}\sqrt{3^2x}=-5\)
⇔ \(\frac{3}{4}\sqrt{x}-3\sqrt{x}-\frac{1}{4}\cdot3\sqrt{x}=-5\)
⇔ \(-\frac{9}{4}\sqrt{x}-\frac{3}{4}\sqrt{x}=-5\)
⇔ \(-3\sqrt{x}=-5\)
⇔ \(\sqrt{x}=15\)
⇔ \(x=225\)( tm )
b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)
ĐK : x ≤ 3
⇔ \(\sqrt{3-x}-\sqrt{3^2\left(3-x\right)}+\frac{5}{4}\sqrt{4^2\left(3-x\right)}=6\)
⇔ \(\sqrt{3-x}-3\sqrt{3-x}+\frac{5}{4}\cdot4\sqrt{3-x}=6\)
⇔ \(-2\sqrt{3-x}+5\sqrt{3-x}=6\)
⇔ \(3\sqrt{3-x}=6\)
⇔ \(\sqrt{3-x}=2\)
⇔ \(3-x=4\)
⇔ \(x=-1\)( tm )
c) \(\sqrt{9x^2+12x+4}=4\)
⇔ \(\sqrt{\left(3x+2\right)^2}=4\)
⇔ \(\left|3x+2\right|=4\)
⇔ \(\orbr{\begin{cases}3x+2=4\\3x+2=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-2\end{cases}}\)
d) \(\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}\)
ĐK : x ≥ 1
⇔ \(\frac{1}{3}\sqrt{x-1}+2\sqrt{2^2\left(x-1\right)}-12\sqrt{\left(\frac{1}{5}\right)^2\cdot\left(x-1\right)}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+2\cdot2\sqrt{x-1}-12\cdot\frac{1}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+4\sqrt{x-1}-\frac{12}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{29}{15}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\sqrt{x-1}=1\)
⇔ \(x-1=1\)
⇔ \(x=2\)( tm )
Lời giải:
a) ĐK: $x\geq 2$
PT $\Leftrightarrow \sqrt{(x-2)(x+2)}-3\sqrt{x-2}=0$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x+2}-3)=0$
\(\Rightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x+2}-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=7\end{matrix}\right.\) (thỏa mãn)
Vậy..........
b) ĐK: $x\geq 0$
PT $\Leftrightarrow (\sqrt{x}-3)^2=0$
$\Leftrightarrow \sqrt{x}-3=0$
$\Leftrightarrow x=9$ (thỏa mãn)
c) ĐK: $x\geq 3$
PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{x-3}-\frac{1}{2}\sqrt{4(x-3)}=7$
$\Leftrightarrow 3\sqrt{x-3}+\sqrt{x-3}-\sqrt{x-3}=7$
$\Leftrightarrow 3\sqrt{x-3}=7$
$\Leftrightarrow x-3=(\frac{7}{3})^2$
$\Rightarrow x=\frac{76}{9}$
d)
ĐK: $x\geq \frac{-1}{2}$
PT $\Leftrightarrow 3\sqrt{4(2x+1)}-\frac{1}{3}\sqrt{9(2x+1)}-\frac{1}{2}\sqrt{25(2x+1)}+\sqrt{\frac{1}{4}(2x+1)}=6$
$\Leftrightarrow 6\sqrt{2x+1}-\sqrt{2x+1}-\frac{5}{2}\sqrt{2x+1}+\frac{1}{2}\sqrt{2x+1}=6$
$\Leftrightarrow 3\sqrt{2x+1}=6$
$\Leftrightarrow \sqrt{2x+1}=2$
$\Rightarrow x=\frac{3}{2}$ (thỏa mãn)
b)\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
\(pt\Leftrightarrow\frac{4}{x}+\sqrt{x-\frac{1}{x}}-\sqrt{\frac{3}{2}}=x+\sqrt{2x-\frac{5}{x}}-\sqrt{\frac{3}{2}}\)
\(\Leftrightarrow\left(\frac{4}{x}-x\right)+\frac{x-\frac{1}{x}-\frac{3}{2}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}=\frac{2x-\frac{5}{x}-\frac{3}{2}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\)
\(\Leftrightarrow\frac{-\left(x-2\right)\left(x+2\right)}{x}+\frac{\frac{\left(x-2\right)\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(x-2\right)\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{-\left(x+2\right)}{x}+\frac{\frac{\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\right)=0\)
Pt trong ngoặc VN suy ra x=2
a)\(x^2+3\sqrt{x^2-1}=\sqrt{x^4-x^2+1}\)
\(\Leftrightarrow x^2+3\sqrt{x^2-1}-1=\sqrt{x^4-x^2+1}-1\)
\(\Leftrightarrow\frac{x^2\left(3\sqrt{x^2-1}+1\right)}{3\sqrt{x^2-1}+1}+\frac{9\left(x^2-1\right)-1}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2+1-1}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{9x^2-10+3x^2\sqrt{x^2-1}+x^2}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{\sqrt{x^2-1}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}=\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{\sqrt{\left(x-1\right)\left(x+1\right)}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(\frac{\frac{1}{\sqrt{x^2-1}}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2}{\sqrt{x^4-x^2+1}+1}\right)=0\)
pt trong căn vô nghiệm
suy ra x=1; x=-1