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a) \(\sqrt{\left(x-3\right)^2}=3\Leftrightarrow\left|x-3\right|=3\) \(\Leftrightarrow\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(N\right)\\x=0\left(N\right)\end{matrix}\right.\)
b) \(\sqrt{4x^2-20x+25}+2x=5\Leftrightarrow\left|2x-5\right|+2x-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x-5+2x-5=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\5-2x+2x-5=0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x-10=0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\0x=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x=\dfrac{10}{4}\left(N\right)\end{matrix}\right.\\x\le\dfrac{5}{2}\end{matrix}\right.\) ** 10/4 = 5/2 rồi**
Kl: x \< 5/2
c) \(\sqrt{1-12x+36x^2}=5\Leftrightarrow\left|1-6x\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}1-6x=5\\1-6x=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\left(N\right)\\x=1\left(N\right)\end{matrix}\right.\)
Kl: x=-2/3, x=1
d) Đk: x >/ 1
\(\sqrt{x+2\sqrt{x-1}}=2\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}+1=2\left(1\right)\\\sqrt{x-1}+2=-2\left(VN\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\)(N)
Kl: x=2
e) Đk: x >/ 1
\(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}\ge1\\\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{x-1}-1=\sqrt{x-1}-1\) (luôn đúng)
kl: x >/ 1
f) \(\sqrt{x^2-\dfrac{1}{2}x+\dfrac{1}{16}}=\dfrac{1}{4}-x\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{4}\\\left|\dfrac{1}{4}-x\right|=\dfrac{1}{4}-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{4}\\\dfrac{1}{4}-x=\dfrac{1}{4}-x\end{matrix}\right.\)
(luôn đúng)
Kl: x \< 1/4
Lần sau xé nhỏ câu hỏi giùm con nha má, để nhiều thế này thất thu T_T!
a)\(\sqrt{3x+1}+2x=\sqrt{x-4}-5\left(ĐKXĐ:x\ge4\right)\)
\(\Leftrightarrow\left(\sqrt{3x+1}-\sqrt{x-4}\right)+\left(2x+5\right)=0\)
\(\Leftrightarrow\frac{3x+1-x+4}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)
\(\Leftrightarrow\frac{2x+5}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1\right)=0\)
a') (tiếp)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2,5\left(KTMĐKXĐ\right)\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\)
Xét phương trình \(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\)(1)
Với mọi \(x\ge4\), ta có:
\(\sqrt{3x+1}>0\); \(\sqrt{x-4}\ge0\)
\(\Rightarrow\sqrt{3x+1}+\sqrt{x-4}>0\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}>0\)
\(\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1>0\)
Do đó phương trình (1) vô nghiệm.
Vậy phương trình đã cho vô nghiệm.
a)\(\sqrt{4x+20}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)\(\sqrt{9x-45}\)=4 ; ĐKXĐ : x ≥_+ 5
⇔ \(\sqrt{2^2x+2^2.5}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)\(\sqrt{3^2x-3^2.5}\) =4
⇔ 2\(\sqrt{x+5}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)3\(\sqrt{x-5}\) =4 ⇔ 2\(\sqrt{x+5}\) +\(\sqrt{x-5}\) -\(\sqrt{x-5}\) =4⇔2\(\sqrt{x+5}\)=4(tm)
⇔\(\sqrt{x+5}\)=2⇔x+5=4 ⇔x=-1
Vậy x=-1
b) \(\sqrt{x^2-36}\) - \(\sqrt{x-6}\) =0 ; ĐKXĐ: x≥_+6
⇔ \(\sqrt{\left(x-6\right)\left(x+6\right)}\) - \(\sqrt{x-6}\) =0 ⇔ \(\sqrt{x-6}\).\(\sqrt{x+6}\) - \(\sqrt{x-6}\) =0
⇔ \(\sqrt{x-6}\)(\(\sqrt{x+6}\) -1 )=0 ⇔\([\) \(\begin{matrix}\sqrt{x-6}&=0\\\sqrt{x+6}-1&=0\end{matrix}\) ⇔ \([\) \(\begin{matrix}x-6&=0\\x+6-1&=0\end{matrix}\) ⇔\([\) \(\begin{matrix}x&=6\left(ktm\right)\\x&=-5\left(tm\right)\end{matrix}\)
Vậy x=-5
c) \(\sqrt{4-x^2}\) -x +2 =0 ; ĐKXĐ: -2≤x≤2
⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) -x+2 =0 ⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) -(x-2)=0
⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) =(x-2) ⇔ (2-x)(2+x)=(x-2)2 ⇔ 4-x2 = x2-4x+4 ⇔ -x2-x2+4x=4-4
⇔-2x2+4x=0 ⇔ -2x(x-2)=0 ⇔ \([\) \(\begin{matrix}-2x&=0\\x-2&=0\end{matrix}\) ⇔\([\) \(\begin{matrix}x&=0\left(tm\right)\\x&=2\left(tm\right)\end{matrix}\)
Vậy S=\(\left\{0;2\right\}\)
d) \(\sqrt{\left(2x-3\right)\left(x-1\right)}-\sqrt{x-1}=0\) ; ĐKXĐ: x≥\(\dfrac{3}{2}\);x ≥ 1
⇔\(\sqrt{2x-3}.\sqrt{x-1}-\sqrt{x-1}=0\) ⇔ \(\sqrt{x-1}.\left(\sqrt{2x-3}-1\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{2x-3}-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x-1=0\\2x-3-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Vậy s=\(\left\{1:2\right\}\)
a) \(\frac{1}{x-1+\sqrt{x^2-2x+3}}+\frac{1}{x-1-\sqrt{x^2-2x+3}}=1\)
ĐKXĐ : \(x\inℝ\)
\(\Leftrightarrow\frac{x-1-\sqrt{x^2-2x+3}}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}+\frac{x-1+\sqrt{x^2-2x+3}}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}=\frac{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}\)
\(\Rightarrow2x-2=\left[\left(x-1\right)+\left(\sqrt{x^2-2x+3}\right)\right]\left[\left(x-1\right)-\left(\sqrt{x^2-2x+3}\right)\right]\)
\(\Leftrightarrow2x-2=\left(x-1\right)^2-\left(\sqrt{x^2-2x+3}\right)^2\)
\(\Leftrightarrow2x-2=x^2-2x+1-\left(x^2-2x+3\right)\)
\(\Leftrightarrow2x-2=x^2-2x+1-x^2+2x-3\)
\(\Leftrightarrow2x-2=-2\)
\(\Leftrightarrow2x=0\)
\(\Leftrightarrow x=0\)
Vậy phương trình có nghiệm duy nhất x = 0
a) dat x-1=a
x=a+1
\(a+1+\sqrt{5+\sqrt{a}}=6\)
\(5-a=\sqrt{5+\sqrt{a}}\)
\(25-10a+a^2=5+\sqrt{a}\)
\(20-10a+a^2-\sqrt{a}=0\)
(a - \sqrt{5} - 5) (a + \sqrt{a} - 4) = 0
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
a) \(\sqrt{3}x-\sqrt{12}=0< =>\sqrt{3}x=\sqrt{12}=>x=2\)
Vay S = { 2 }
b) \(\sqrt{2}x+\sqrt{2}=\sqrt{8}+\sqrt{18}< =>\sqrt{2}x=\sqrt{8}+\sqrt{18}-\sqrt{2}< =>\sqrt{2}x=2\sqrt{2}+3\sqrt{2}-\sqrt{2}\) <=> \(\sqrt{2}x=4\sqrt{2}=>x=4\)
Vay S = { 4 }
c) \(\sqrt{5}x^2-\sqrt{20}=0< =>\sqrt{5}x^2=\sqrt{20}< =>x^2=2=>x=\sqrt{2}\)
Vay S = {\(\sqrt{2}\) }
d) \(\sqrt{x^2+6x+9}=3x+6< =>\sqrt{\left(x+3\right)^2}=3x+6< =>x+3=3x+6< =>-2x=\) \(3=>x=-\dfrac{3}{2}\)
Vay S = { - 3/2 }
e) \(\sqrt{x^2-4x+4}-2x+5=0< =>\sqrt{\left(x-2\right)^2}-2x+5=0< =>x-2-2x+5=0\) <=> \(-x+3=0< =>-x=-3=>x=3\)
Vay S = { 3 }
F) \(\sqrt{\dfrac{2x-3}{x-1}}=2\)
<=> \(\dfrac{2x-3}{x-1}=4< =>2x-3=4x-4< =>-2x=-1=>x=\dfrac{1}{2}\)
Vay S = { 1/2 }
g) \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2< =>\sqrt{\dfrac{2x-3}{x-1}}=2< =>\dfrac{2x-3}{x-1}=4< =>2x-3=4x-4< =>-2x=-1=>x=\dfrac{1}{2}\)
bạn chưa có ĐKXĐ nên chưa xét kết quả có đúng vs Đk ko, có vài câu sai kết quả
(a) Điều kiện: \(\left\{{}\begin{matrix}x+1\ge0\\x-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x>5\end{matrix}\right.\Rightarrow x>5\).
Phương trình tương đương: \(\sqrt{x+1}=2\sqrt{x-5}\)
\(\Leftrightarrow x+1=4\left(x-5\right)\Leftrightarrow x=7\left(TM\right)\).
Vậy: \(S=\left\{7\right\}.\)
(b) Phương trình tương đương: \(x^2-1=8\)
\(\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\).
Vậy: \(S=\left\{\pm3\right\}\)
a: ĐKXĐ: x+1>=0 và x-5>0
=>x>5
\(\dfrac{\sqrt{x+1}}{\sqrt{x-5}}=2\)
=>\(\sqrt{\dfrac{x+1}{x-5}}=2\)
=>\(\dfrac{x+1}{x-5}=4\)
=>4x-20=x+1
=>3x=21
=>x=7
b: ĐKXĐ: \(x\in R\)
\(\sqrt[3]{x^2-1}=2\)
=>x^2-1=8
=>x^2=9
=>x=3 hoặc x=-3