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15 tháng 6 2018
\(\sqrt{3x^2-1}+\sqrt{x^2-x}-x\sqrt{x^2+1}=\frac{1}{2\sqrt{2}}\left(7x^2-x+4\right)\)
\(\Leftrightarrow2\sqrt{2}\left(\sqrt{3x^2-1}+\sqrt{x^2-x}-x\sqrt{x^2+1}\right)=7x^2-x+4\)
\(\Leftrightarrow\left[\left(3x^2-1\right)-2\sqrt{2}\sqrt{3x^2-1}+2\right]+\left[\left(x^2-x\right)-2\sqrt{2}\sqrt{x^2-x}+2\right]+\left[2x^2+2\sqrt{2}x\sqrt{x^2+1}+\left(x^2+1\right)\right]=0\)
\(\Leftrightarrow\left(\sqrt{3x^2-1}-\sqrt{2}\right)^2+\left(\sqrt{x^2-x}-\sqrt{2}\right)^2+\left(\sqrt{x^2+1}+\sqrt{2}x\right)^2=0\)
Làm nốt
HP
1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 6 2020
ĐKXĐ: \(-1\le x\le\frac{5}{3}\)
\(\Leftrightarrow6-2x+2\sqrt{-3x^2+2x+5}=3x^2-4x+4\)
\(\Leftrightarrow-3x^2+2x+5+2\sqrt{-3x^2+2x+5}-3=0\)
Đặt \(\sqrt{-3x^2+2x+5}=t\ge0\)
\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{-3x^2+2x+5}=1\)
\(\Leftrightarrow-3x^2+2x+4=0\)
\(\Leftrightarrow...\)
KR
0
1 tháng 12 2021
a,ĐKXĐ:\(x\ge2\)
\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
b,ĐKXĐ:\(x\in R\)
\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
c, ĐKXĐ:\(x\ge0\)
\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)
10 tháng 8 2020
\(P=\frac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{x+\sqrt{x}-2}\)
\(P=\frac{3x+3\sqrt{x}-3-x+1-x+4}{x+\sqrt{x}-2}\)
\(P=\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
NT
2
TN
6 tháng 7 2017
\(\sqrt{x^2+x+2}=\frac{3x^2+3x+2}{3x+1}\)
Đk:.... tự xác định :v
\(\Leftrightarrow\sqrt{x^2+x+2}-2=\frac{3x^2+3x+2}{3x+1}-2\)
\(\Leftrightarrow\frac{x^2+x+2-4}{\sqrt{x^2+x+2}+2}=\frac{3x^2-3x}{3x+1}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{3x\left(x-1\right)}{3x+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{3x}{3x+1}\right)=0\)
Dễ thấy: \(\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{3x}{3x+1}< 0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
LT
6
20 tháng 9 2018
ĐK: \(x\ge\frac{2}{5}\)
Ta có \(\sqrt{5x^3+3x^2+3x-2}+\frac{1}{2}=\frac{x^2}{2}+3x\)
<=> \(\sqrt{\left(5x-2\right)\left(x^2+x+1\right)}=\frac{x^2}{2}+3x-\frac{1}{2}\)
<=> \(2\sqrt{\left(5x-2\right)\left(x^2+x+1\right)}=x^2+6x-1\)
Đặt \(\sqrt{5x-2}=a\left(a\ge0\right),\sqrt{x^2+x+1}=b\left(b\ge0\right)\)
=> \(a^2+b^2=5x-2+x^2+x+1=x^2+6x+1\)
Ta có \(2ab=a^2+b^2\)
<=> \(\left(a-b\right)^2=0\) <=> a=b
Theo cách đặt ta có \(\sqrt{5x-2}=\sqrt{x^2+x+1}\)
=> \(5x-2=x^2+x+1\)
<=> \(\left(x-3\right)\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}x=3\left(TMĐK\right)\\x=1\left(TMĐK\right)\end{cases}}\)
Vậy
ĐKXĐ là x\(\ge\frac{2}{3}\)
\(\frac{x^2}{\sqrt{3x-2}}-\frac{3x-2}{\sqrt{3x-2}}=1-x\)
<=>x^2-3x+2=(1-x)\(\sqrt{3x-2}\)