Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
a, \(\sqrt{\left(2x+3\right)^2}=x+1\)
\(\Leftrightarrow\left|2x+3\right|=x+1\)
TH1: \(\left\{{}\begin{matrix}2x+3=x+1\\2x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x\ge-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.
Vậy phương trình vô nghiệm.
TH2: \(\left\{{}\begin{matrix}-2x-3=x+1\\2x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\x< -\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.
b,
a, \(\sqrt{\left(2x-1\right)^2}=x+1\)
\(\Leftrightarrow\left|2x-1\right|=x+1\)
TH1: \(\left\{{}\begin{matrix}2x-1=x+1\\2x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x\ge\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=2\)
TH2: \(\left\{{}\begin{matrix}-2x+1=x+1\\2x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=0\)
\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(1;1\right)\)
2.
ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)
\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)
\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)
\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\3-x+2\sqrt{x\left(3-2x\right)}=4\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)
\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)
\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
a: ĐKXĐ: \(\left[{}\begin{matrix}x>=2\\x< =-3\end{matrix}\right.\)
\(\sqrt{\left(x-2\right)\left(x+3\right)}=5\)
=>\(\sqrt{x^2+x-6}=5\)
=>\(x^2+x-6=25\)
=>\(x^2+x-31=0\)
=>\(\left[{}\begin{matrix}x=\dfrac{-1+5\sqrt{5}}{2}\left(nhận\right)\\x=\dfrac{-1-5\sqrt{5}}{2}\left(nhận\right)\end{matrix}\right.\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(2x+3\right)^2}=x-5\)
=>\(\left|2x+3\right|=x-5\)
=>\(\left\{{}\begin{matrix}x>=5\\\left(2x+3\right)^2=\left(x-5\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=5\\\left(2x+3-x+5\right)\left(2x+3+x-5\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=5\\\left(x+8\right)\left(3x-2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=5\\\left[{}\begin{matrix}x=-8\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
=>\(x\in\varnothing\)
c: ĐKXĐ: \(x\in R\)
\(\sqrt{x^2-6x+9}=x+7\)
=>\(\sqrt{\left(x-3\right)^2}=x+7\)
=>\(\left|x-3\right|=x+7\)
=>\(\left\{{}\begin{matrix}x+7>=0\\\left(x-3\right)^2=\left(x+7\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-7\\\left(x-3-x-7\right)\left(x-3+x+7\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-7\\-10\left(2x+4\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-7\\x+2=0\end{matrix}\right.\)
=>x=-2
d: ĐKXĐ: x>=3/2
\(\sqrt{2x-3}=x-1\)
=>\(\left\{{}\begin{matrix}2x-3=\left(x-1\right)^2\\x>=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-2x+1=2x-3\\x>=\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-4x+4=0\\x>=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\x>=\dfrac{3}{2}\end{matrix}\right.\)
=>x=2