Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐK:(tự tìm)
Bình phương 2 vế
\(\Rightarrow2x+2\sqrt{x^2-14x+49}=14\)
\(\Leftrightarrow2x+2\sqrt{\left(x-7\right)^2}=14\)
\(\Leftrightarrow2x+2\left|x-7\right|=14\)
Xét \(x\ge7\)\(\Rightarrow2x+2x-14=14\)
\(\Leftrightarrow x=7\left(tm\right)\)
Xét x<7\(\Rightarrow2x-2x+14=14\)
\(\Leftrightarrow14=14\)(luôn đúng)
Thử lại,kết hợp với đk rồi kết luận
ĐK : \(x\ge\frac{7}{2}\)
Đặt \(\sqrt{14x-49}=a\) , ta có :
\(\sqrt{x+a}+\sqrt{x-a}=\sqrt{14}\)
\(\Leftrightarrow\left(\sqrt{x+a}+\sqrt{x-a}\right)^2=14\)
\(\Leftrightarrow x+a+x-a+2\sqrt{x^2-a^2}=14\)
\(\Leftrightarrow2x+2\sqrt{x^2-14x+49}=14\)
\(\Leftrightarrow2x+2\left|x-7\right|=14\)
TH 1 : \(x\ge7\) \(\Rightarrow4x-14=14\Leftrightarrow x=7\) ( t/m )
TH 2 : \(\frac{7}{2}\le x\le7\)
\(\Rightarrow2x+14-2x=14\)
\(\Leftrightarrow14=14\) ( t/m )
Vậy ...
a) ĐK : \(x\ge1\)
pt <=> \(\sqrt{3^2\left(x-1\right)}-\frac{1}{2}\sqrt{2^2\left(x-1\right)}=2\)
<=> \(\left|3\right|\sqrt{x-1}-\frac{1}{2}\cdot\left|2\right|\sqrt{x-1}=2\)
<=> \(3\sqrt{x-1}-1\sqrt{x-1}=2\)
<=> \(2\sqrt{x-1}=2\)
<=> \(\sqrt{x-1}=1\)
<=> \(x-1=1\)=> \(x=2\)( tm )
b) \(3x-\sqrt{49-14x+x^2}=15\)
<=> \(\sqrt{x^2-14x+49}=3x-15\)
<=> \(\sqrt{\left(x-7\right)^2}=3x-15\)
<=> \(\left|x-7\right|=3x-15\)(1)
Với x < 7
(1) <=> 7 - x = 3x - 15
<=> -x - 3x = -15 - 7
<=> -4x = -22
<=> x = 11/2 ( tm )
Với x ≥ 7
(1) <=> x - 7 = 3x - 15
<=> x - 3x = -15 + 7
<=> -2x = -8
<=> x = 4 ( ktm )
Vậy x = 11/2
a) \(ĐKXĐ:x\ge1\)
\(\sqrt{9x-9}-\frac{1}{2}\sqrt{4x-4}=2\)
\(\Leftrightarrow\sqrt{9.\left(x-1\right)}-\frac{1}{2}.\sqrt{4\left(x-1\right)}=2\)
\(\Leftrightarrow3\sqrt{x-1}-\frac{1}{2}.2\sqrt{x-1}=2\)
\(\Leftrightarrow3\sqrt{x-1}-\sqrt{x-1}=2\)
\(\Leftrightarrow2\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)\(\Leftrightarrow x=2\)( thỏa mãn ĐKXĐ )
Vậy phương trình có nghiệm là \(x=2\)
b) \(3x-\sqrt{49-14x+x^2}=15\)
\(\Leftrightarrow3x-\sqrt{\left(7-x\right)^2}=15\)
\(\Leftrightarrow3x-\left|7-x\right|=15\)
+) TH1: Nếu \(7-x< 0\)\(\Leftrightarrow x>7\)
thì \(3x-\left(x-7\right)=15\)
\(\Leftrightarrow3x-x+7=15\)\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\)( không thỏa mãn )
+) TH2: Nếu \(7-x\ge0\)\(\Leftrightarrow x\le7\)
thì \(3x-\left(7-x\right)=15\)
\(\Leftrightarrow3x-7+x=15\)
\(\Leftrightarrow4x=22\)\(\Leftrightarrow x=\frac{22}{4}\)( thỏa mãn ĐKXĐ )
Vậy nghiệm của phương trình là \(x=\frac{22}{4}\)
a/ \(\sqrt{x^2-14x+49}+4x-7=0\)
\(\Leftrightarrow\sqrt{\left(x-7\right)^2}=7-4x\)
\(\Leftrightarrow\left|x-7\right|=7-4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=7-4x\\x-7=4x-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\left(KTM\right)\\x=0\left(TM\right)\end{matrix}\right.\)
Vậy pt có 1 nghiệm x = 0
b/ đkxđ: x ≥2
\(\sqrt{x+2+4\sqrt{x-2}}=4\sqrt{x-2}-5\)
Đặt \(\sqrt{x-2}\) = t (t ≥ 0)
PT \(\Leftrightarrow\sqrt{t^2+4t+4}=4t-5\)
\(\Leftrightarrow\sqrt{\left(t+2\right)^2}=4t-5\)
\(\Leftrightarrow\left|t+2\right|=4t-5\)
Vì t ≥ 0 => t + 2 > 0
=> \(t+2=4t-5\)
\(\Leftrightarrow-3t=-7\Leftrightarrow t=\dfrac{7}{3}\left(TM\right)\)
\(\Rightarrow\sqrt{x-2}=\dfrac{7}{3}\Rightarrow x-2=\dfrac{49}{9}\)
\(\Leftrightarrow x=\dfrac{67}{9}\)(TM)
Vậy pt có nghiệm \(x=\dfrac{67}{9}\)
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. ĐKXĐ: \(x\ge\dfrac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a+b=\sqrt{3a^2-b^2}\)
\(\Leftrightarrow\left(a+b\right)^2=3a^2-b^2\)
\(\Leftrightarrow a^2-ab-b^2=0\Leftrightarrow\left(a-\dfrac{1+\sqrt{5}}{2}b\right)\left(a+\dfrac{\sqrt{5}-1}{2}b\right)=0\)
\(\Leftrightarrow a=\dfrac{1+\sqrt{5}}{2}b\Leftrightarrow\sqrt{x^2+2x}=\dfrac{1+\sqrt{5}}{2}\sqrt{2x-1}\)
\(\Leftrightarrow x^2+2x=\dfrac{3+\sqrt{5}}{2}\left(2x-1\right)\)
\(\Leftrightarrow x^2-\left(\sqrt{5}+1\right)x+\dfrac{3+\sqrt{5}}{2}=0\)
\(\Leftrightarrow\left(x-\dfrac{\sqrt{5}+1}{2}\right)^2=0\)
\(\Leftrightarrow x=\dfrac{\sqrt{5}+1}{2}\)
b. ĐKXĐ: \(x\ge5\)
\(\Leftrightarrow\sqrt{5x^2+14x+9}=\sqrt{x^2-x-20}+5\sqrt{x+1}\)
\(\Leftrightarrow5x^2+14x+9=x^2-x-20+25\left(x+1\right)+10\sqrt{\left(x+1\right)\left(x-5\right)\left(x+4\right)}\)
\(\Leftrightarrow2x^2-5x+2=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-4x-5}=a\ge0\\\sqrt{x+4}=b>0\end{matrix}\right.\)
\(\Rightarrow2a^2+3b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-4x-5}=\sqrt{x+4}\\2\sqrt{x^2-4x-5}=3\sqrt{x+4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x+4\\4\left(x^2-4x-5\right)=9\left(x+4\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
b)đk:\(x\ge\dfrac{1}{2}\)
Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)
\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)
=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\)
Dấu = xảy ra\(\Leftrightarrow x=1\)
Vậy....
c) đk: \(x\ge0\)
\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)
\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)
pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)
\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...
a)ĐKXĐ: x≥-1/3; x≤6
<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)
(vì x≥-1/3 nên3x+1≥0 )
ĐKXĐ: \(-\dfrac{1}{3}\le x\le6\)
\(\left(\sqrt{3x+1}-4\right)+\left(1-\sqrt{6-x}\right)+\left(3x^2-14x-5\right)=0\)
\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1\right)=0\)
\(\Leftrightarrow x-5=0\) (do \(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1>0;\forall x\))
\(\Rightarrow x=5\)
ĐKXĐ: \(\left\{{}\begin{matrix}3x+1>=0\\6-x>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{3}\\x< =6\end{matrix}\right.\)
\(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)
=>\(\sqrt{3x+1}-4+1-\sqrt{6-x}+3x^2-14x-5=0\)
=>\(\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{1-6+x}{1+\sqrt{6-x}}+3x^2-15x+x-5=0\)
=>\(\dfrac{3\cdot\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{6-x}+1}+\left(x-5\right)\left(3x+1\right)=0\)
=>\(\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{6-x}+1}+3x+1\right)=0\)
=>x-5=0
=>x=5(nhận)
a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)
\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)
\(\Leftrightarrow\sqrt{x+3}=3\)
\(\Leftrightarrow x+3=9\)
hay x=6
b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)
\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)
ĐKXĐ:...
Bình phương 2 vế ta được:
\(2x+2\sqrt{x^2-14x+49}=14\)
\(\Leftrightarrow x-7+\sqrt{\left(x-7\right)^2}=0\)
\(\Leftrightarrow x-7+\left|x-7\right|=0\)
- Với \(\frac{49}{14}\le x\le7\Rightarrow...\)
- Với \(x>7\Rightarrow...\)
Đơn giản nên bạn tự phá trị tuyệt đối và giải