K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
AH
Akai Haruma
Giáo viên
23 tháng 6 2018
Câu 1:
ĐK: \(x\in\mathbb{R}\)
\(\sqrt{x^2-2x+5}=x^2-2x-1=x^2-2x+5-6\)
Đặt \(\sqrt{x^2-2x+5}=t(t\geq 0)\). PT trở thành:
\(t=t^2-6\)
\(\Leftrightarrow t^2-t-6=0\Leftrightarrow (t-3)(t+2)=0\)
\(\Leftrightarrow \left[\begin{matrix} t=3\\ t=-2\end{matrix}\right.\). Vì $t\geq 0$ nên $t=3$
Do đó: \(\sqrt{x^2-2x+5}=3\Rightarrow x^2-2x+5=9\)
\(\Rightarrow x^2-2x-4=0\Rightarrow x=1\pm \sqrt{5}\)
Vậy........
AH
Akai Haruma
Giáo viên
23 tháng 6 2018
Câu 2:
ĐK: \(x\in\mathbb{R}\)
Ta có: \(x^2-4x-6=\sqrt{2x^2-8x+12}\)
\(\Rightarrow 2x^2-8x-12=2\sqrt{2x^2-8x+12}\)
\(\Leftrightarrow (2x^2-8x+12)-24-2\sqrt{2x^2-8x+12}=0\)
Đặt \(\sqrt{2x^2-8x+12}=t(t\geq 0)\). PT trở thành:
\(t^2-24-2t=0\)
\(\Leftrightarrow (t-6)(t+4)=0\Rightarrow \left[\begin{matrix} t=6\\ t=-4\end{matrix}\right.\)
Mà \(t\geq 0\Rightarrow t=6\)
Do đó: \(\sqrt{2x^2-8x+12}=6\Rightarrow 2x^2-8x+12=36\)
\(\Rightarrow x^2-4x-12=0\Rightarrow \left[\begin{matrix} x=6\\ x=-2\end{matrix}\right.\)
Vậy...........
24 tháng 8 2017
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) \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2< =>\sqrt{\dfrac{2x-3}{x-1}}=2< =>\dfrac{2x-3}{x-1}=4< =>2x-3=4x-4< =>-2x=-1=>x=\dfrac{1}{2}\)
NV
24 tháng 8 2017
bạn chưa có ĐKXĐ nên chưa xét kết quả có đúng vs Đk ko, có vài câu sai kết quả
6 tháng 10 2020
1) đk: \(x\ge1\)
Ta có: \(\sqrt{x-1}-\sqrt{2x\left(x-1\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}=\sqrt{2x\left(x-1\right)}\)
\(\Leftrightarrow x-1=2x^2-2x\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)
Vậy x = 1
2) đk: \(x\ge\frac{1}{2}\)
Ta có: \(\sqrt{5x^2}=2x-1\)
\(\Leftrightarrow5x^2=\left(2x-1\right)^2\)
\(\Leftrightarrow5x^2=4x^2-4x+1\)
\(\Leftrightarrow x^2+4x-1=0\)
\(\Leftrightarrow\left(x+2\right)^2-5=0\)
\(\Leftrightarrow\left(x+2-\sqrt{5}\right)\left(x+2+\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2+\sqrt{5}\left(ktm\right)\\x=-2-\sqrt{5}\left(ktm\right)\end{cases}}\)
=> PT vô nghiệm
3) đk: \(x\ge-1\)
Ta có: \(\sqrt{x+1}+\sqrt{9x+9}=4\)
\(\Leftrightarrow\sqrt{x+1}+3\sqrt{x+1}=4\)
\(\Leftrightarrow4\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=1\)
\(\Rightarrow x=0\)
6 tháng 10 2020
4) đk: \(x\ge2\)
Ta có: \(\sqrt{x-2}-\sqrt{x\left(x-2\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}=\sqrt{x\left(x-2\right)}\)
\(\Leftrightarrow x-2=x\left(x-2\right)\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)
Vậy x = 2
6) đk: \(x\ge-\frac{7}{5}\)
Ta có: \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
\(\Leftrightarrow\frac{2x-3}{x-1}=2\)
\(\Leftrightarrow2x-3=2x-2\)
\(\Leftrightarrow0x=1\) vô lý
=> PT vô nghiệm
AY
14 tháng 8 2018
a) \(\sqrt{x^2+2x+1}=9\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}+1\right)^2}=9\)
\(\Leftrightarrow\left|\sqrt{x}+1\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=9\\x+1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-10\end{matrix}\right.\)
b)\(\sqrt{1-4x+4x^2}=5\)
\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\)
\(\Leftrightarrow\left|1-2x\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
c)\(\sqrt{x^2-2x\sqrt{2}+2}=5\)
\(\Leftrightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=5\)
\(\Leftrightarrow\left|x-\sqrt{2}\right|=5\)
\(\left[{}\begin{matrix}x-\sqrt{2}=5\\x-\sqrt{2}=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5+\sqrt{2}\\x=-5+\sqrt{2}\end{matrix}\right.\)
Mình giải tới đây thôi
14 tháng 8 2017
I) xd mọi x
\(\sqrt{x^2-8x+16}+\sqrt{x^2-10x+25}=9\)
\(\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-5\right)^2}=9=>\left|x-4\right|+\left|x-5\right|=9\)
\(\left[{}\begin{matrix}x< 4\Rightarrow4-x+5-x=>x=0\left(n\right)\\4\le x< 5\Rightarrow x-4+5-x=9\left(vn\right)\\x\ge5\Rightarrow x-4+x-5=9\Rightarrow x=9\left(n\right)\\\end{matrix}\right.\)
kết luận
\(\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)
28 tháng 7 2020
d) \(\sqrt{x+1}+2=0\)( ko tìm đc )
e) \(9x^2=4\Leftrightarrow x^2=\frac{4}{9}\Leftrightarrow x=\pm\sqrt{\frac{4}{9}}\)
g) \(2x^2=\frac{9}{50}\Leftrightarrow x^2=\frac{9}{100}\Leftrightarrow x=\pm\sqrt{\frac{9}{100}}\)
28 tháng 7 2020
z) \(3-2x=1\Leftrightarrow2x=2\Leftrightarrow x=1\)
y) \(\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)=0\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=0\\1-\sqrt{x}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
1 tháng 7 2019
a) + \(VT=\sqrt{x^2+2x+10}+x^2+2x+1+7\)
\(=\sqrt{x^2+2x+1}+\left(x+1\right)^2+7>0\forall x\)
=> ptvn
d) ĐK : \(x^2+7x+7\ge0\)
Đặt \(t=\sqrt{x^2+7x+7}\ge0\) \(\Rightarrow t^2=x^2+7x+7\)
\(pt\Leftrightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}-2=0\)
\(\Leftrightarrow3t^2+2t-5=0\Leftrightarrow\left(3t+5\right)\left(t-1\right)=0\)
\(\Leftrightarrow t=1\) ( do \(3t+5>0\forall t\ge0\) )
\(\Leftrightarrow x^2+7x+1=0\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\) ( TM )
1 tháng 7 2019
f) ĐK : \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\ge0\\b=\sqrt{x+3}\ge0\end{matrix}\right.\) thì pt trở thành :
\(a+b-ab-1=0\)
\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)
\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)
S
5 tháng 7 2019
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=3\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\) \(+,x\ge2\Rightarrow\left\{{}\begin{matrix}x-2\ge0\\x-1\ge1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-2\right|=x-2\\\left|x-1\right|=x-1\end{matrix}\right.\Rightarrow\left|x-2\right|+\left|x-1\right|=x-2+x-1=3\Leftrightarrow2x-3=3\Leftrightarrow x=3\left(\text{t/m}\right)\) \(+,1\le x< 2\Rightarrow\left\{{}\begin{matrix}x-1\ge0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|x-2\right|=-\left(x-2\right)=2-x\end{matrix}\right.\Rightarrow\left|x-1\right|+\left|x-2\right|=x-1+2-x=1\left(l\right)\) \(+,x< 1\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=-\left(x-1\right)=1-x\\\left|x-2\right|=-\left(x-2\right)=2-x\end{matrix}\right.\Rightarrow\left|x-1\right|+\left|x-2\right|=1-x+2-x=3\Leftrightarrow3-2x=3\Leftrightarrow x=0\left(\text{t/m}\right)\) \(f,\left\{{}\begin{matrix}\sqrt{x^2-9}\ge0\\\sqrt{x^2-6x+9}\ge0\end{matrix}\right.mà:\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2-9}=0\\\sqrt{x^2-6x+9}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-9=0\\\sqrt{\left(x-3\right)^2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-9=0\\\left|x-3\right|=0\end{matrix}\right.\Leftrightarrow x=3\)\thay vào ta thấy thoa man => x=3
S
5 tháng 7 2019
\(ĐK:x\ge4\)\(\sqrt{x^2+x-20}=\sqrt{x^2+5x-4x-20}=\sqrt{x\left(x+5\right)-4\left(x+5\right)}=\sqrt{\left(x-4\right)\left(x+5\right)}=\sqrt{x-4}.\sqrt{x+5}=\sqrt{x-4}\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-4}=0\\\sqrt{x+5}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x+5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=-4\left(l\right)\end{matrix}\right.\Rightarrow x=4\) \(b,ĐK:x\le2;\sqrt{x+1}+\sqrt{2-x}=\sqrt{6}\Leftrightarrow x+1+2-x+2\sqrt{\left(x+1\right)\left(2-x\right)}=6\Leftrightarrow3+2\sqrt{\left(x+1\right)\left(2-x\right)}=6\Leftrightarrow2\sqrt{\left(x+1\right)\left(2-x\right)}=3\Leftrightarrow\sqrt{\left(x-1\right)\left(2-x\right)}=1,5\Leftrightarrow\left(x-1\right)\left(2-x\right)=\frac{9}{4}\Leftrightarrow\left(x-1\right)\left(x-2\right)=-\frac{9}{4}\Leftrightarrow x^2-3x+2=-\frac{9}{4}\Leftrightarrow x^2-3x+\frac{9}{4}=-2\Leftrightarrow\left(x-\frac{3}{2}\right)^2=-2\Rightarrow vonghiem\)
\(2x-5\sqrt{x}+2=0\)
\(\Leftrightarrow-5\sqrt{x}+2=0-2x\)
\(\Leftrightarrow-5\sqrt{x}+2=-2x\)
\(\Leftrightarrow-5\sqrt{x}=-2x-2\)
\(\Leftrightarrow\left(-5\sqrt{x}\right)^2=\left(-2x-2\right)^2\)
\(\Leftrightarrow25x=4x^2+8x+4\)
\(\Leftrightarrow4x^2+8x+4=25x\) (chuyển vế)
\(\Leftrightarrow4x^2+8x+4-25x=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
\(\Leftrightarrow x\left(4x-1\right)-\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=1\end{cases}}\)
Vậy nghiệm phương trình là: \(\left\{\frac{1}{4};1\right\}\)
ĐK:\(x\ge0\)
pt\(\Leftrightarrow\left(2x-4\sqrt{x}\right)-\left(\sqrt{x}-2\right)=0\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\Rightarrow\orbr{\begin{cases}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=2\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{4}\\x=4\end{cases}}}\left(TMĐK\right)\)
Vậy....