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ML
13 tháng 8 2016
1.
\(\text{ĐK: }x\ge\frac{1}{2}\)
\(pt\Leftrightarrow\left(x^2+1\right)\left(x-\sqrt{2x-1}\right)+\)\(\left(x-\sqrt[3]{2x^2-x}\right)=0\)
\(\Leftrightarrow\left(x^2+1\right).\frac{x^2-\left(2x-1\right)}{x+\sqrt{2x-1}}+\frac{x^3-\left(2x^2-x\right)}{x^2+Ax+A^2}=0\text{ }\left(A=\sqrt[3]{2x^2-x}\right)\)
\(\Leftrightarrow\left(x-1\right)^2\left[\frac{x^2+1}{x+\sqrt{2x-1}}+\frac{2x}{x^2+A^2+\left(x+A\right)^2}\right]=0\)
\(\Leftrightarrow x=1\text{ }\left(do\text{ }....................................................>0\right)\)
T
13 tháng 9 2019
ĐK: \(x\ge-7\)
PT \(\Leftrightarrow\left(\sqrt[3]{x-8}-\left(x-8\right)\right)+\left[\sqrt{x+7}-4\right]+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\frac{-\left(x-9\right)\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}+\frac{x-9}{\sqrt{x+7}+4}+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left[x^2+x+2+\frac{1}{\sqrt{x+7}+4}-\frac{\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}\right]=0\)
\(\Leftrightarrow x=9\)
P/s:em chả biết đánh giá cái ngoặc to thế nào nữa:((((
1 tháng 10 2015
D) ĐK x>= 1
đặt \(\sqrt{x-1}=a;\sqrt{x^3+x^2+x+1}=b\)
pt <=> \(a+b=1+ab\Rightarrow a+b-1-ab=0\)
<=> \(\left(a-1\right)\left(1-b\right)=0\)
12 tháng 10 2021
a, Với x >= 0 ; x khác 4
\(=\frac{x-3\sqrt{x}+2-\left(x+4\sqrt{x}+3\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-3\sqrt{x}-3-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{-7\sqrt{x}-6-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{-\sqrt{x}-6}{\sqrt{x}-2}\)
b, \(Q+1>0\Leftrightarrow\frac{-\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}-2}>0\Leftrightarrow\frac{-8}{\sqrt{x}-2}>0\)
\(\Rightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\Rightarrow0\le x< 4\)
c, \(\frac{-\left(\sqrt{x}+6\right)}{\sqrt{x}-2}=\frac{-\left(\sqrt{x}-2+8\right)}{\sqrt{x}-2}=-1-\frac{8}{\sqrt{x}-2}\)
\(\Rightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
| \(\sqrt{x}-2\) | -1 | 1 | -2 | 2 | -4 | 4 | -8 | 8 |
| x | 1 | 9 | 0 | 16 | loại | 36 | loại | 100 |
\(ĐKXĐ:-3\le x\le2\)
\(\sqrt{x+3}-\sqrt{7-x}=\sqrt{2-x}\)
\(x+3-7+x-2\sqrt{\left(x+3\right)\left(7-x\right)}=2-x\)
\(2x-4-2\sqrt{7x+21-x^2-3x}=2-x\)
\(2\sqrt{-x^2+4x+21}=6-3x\)
\(4\left(-x^2+4x+21\right)=36-36x+9x^2\)
\(-4x^2+16x+21=9x^2-36x+36\)
\(13x^2-52x+15=0\)
\(\sqrt{\Delta}=\sqrt{\left(-52\right)^2-4.13.15}=2\sqrt{481}\)
\(\orbr{\begin{cases}x=\frac{52+2\sqrt{481}}{26}=\frac{26+\sqrt{481}}{13}\left(KTM\right)\\x=\frac{52-2\sqrt{481}}{26}=\frac{26-\sqrt{481}}{13}\left(TM\right)\end{cases}}\)
ĐK : \(\hept{\begin{cases}\sqrt{x+3}\ge0\\\sqrt{7-x}\ge0\\\sqrt{2-x}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-3\\x\le7\\x\le2\end{cases}}\Leftrightarrow-3\le x\le2\)
\(\Leftrightarrow\left(\sqrt{x+3}-\sqrt{5}\right)-\left(\sqrt{7-x}-\sqrt{5}\right)-\sqrt{2-x}=0\)
\(\Leftrightarrow\frac{x+3-5}{\sqrt{x+3}+5}-\frac{7-x-5}{\sqrt{7-x}+5}-\sqrt{2-x}=0\)
\(\Leftrightarrow-\frac{2-x}{\sqrt{x+3}+5}-\frac{2-x}{\sqrt{7-x}+5}-\sqrt{2-x}=0\)
\(\Leftrightarrow-\sqrt{2-x}\left(\frac{\sqrt{2-x}}{\sqrt{x+3}+5}+\frac{\sqrt{2-x}}{\sqrt{7-x}+5}+1\right)=0\)(1)
Dễ thấy với \(-3\le x\le2\)thì \(\frac{\sqrt{2-x}}{\sqrt{x+3}+5}+\frac{\sqrt{2-x}}{\sqrt{7-x}+5}+1>0\)
nên (1) <=> \(-\sqrt{2-x}=0\Leftrightarrow x=2\left(tm\right)\)