Giải phương trình: \(\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt[4]{4x-3}}=\frac{2}{x}\)
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LL
2
N
17 tháng 12 2019
\(DK:x\in\left(-\frac{1}{4};4\right)\)
PT\(\Leftrightarrow\frac{1}{4}\sqrt{4-x}+\frac{1}{\sqrt{4-x}}+2\sqrt{4x+1}+\frac{2}{\sqrt{4x+1}}+\frac{7}{4}\sqrt{4-x}-\sqrt{4x+1}=\frac{15}{2}\)
Ta co:
\(\frac{1}{4}\sqrt{4-x}+\frac{1}{\sqrt{4-x}}\ge^{ }1\left(1\right)\)
\(2\sqrt{4x+1}+\frac{2}{\sqrt{4x+1}}\ge4\left(2\right)\)
Dau '=' xay ra khi \(x=0\)
Xet
\(\frac{7}{4}\sqrt{4-x}-\sqrt{4x+1}=\frac{5}{2}\left(3\right)\)
\(\Leftrightarrow\frac{-\frac{7}{4}x}{\sqrt{4-x}+2}-\frac{4x}{\sqrt{4x+1}+1}=0\)
\(\Leftrightarrow x\left(\frac{7}{4\sqrt{4-x}+8}+\frac{4}{\sqrt{4x+1}+1}\right)=0\)
\(\Leftrightarrow x=0\left(n\right)\)
Tuc la \(\left(3\right)\)đúng khi \(x=0\) \(\left(4\right)\)
\(\left(1\right),\left(2\right),\left(4\right)\Rightarrow VT\ge\frac{15}{2}=VP\)
Khi \(x=0\)
7 tháng 1 2019
\(x^4+2x^3=4x+4\)
\(x^4+2x^3+x^2-x^2-4x-4=0\)
\(x^2\left(x^2+2x+1\right)-\left(x^2+4x+4\right)=0\)
\(\left[x\left(x+1\right)\right]^2-\left(x+2\right)^2=0\)
\(\left(x^2+x-x-2\right)\left(x^2+x+2\right)=0\)
\(\left(x^2-2\right)\left(x^2+x+2\right)=0\)
\(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x^2+x+2\right)=0\)
tự làm nốt nhé~
SN
7 tháng 1 2019
\(b,\frac{1}{x^2}+\sqrt{x+2}=\frac{1}{x}+\sqrt{2x+1}\)(1)
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x+2\ge0\\2x+1\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne0\\x\ge\frac{-1}{2}\end{cases}}\)
\(\left(1\right)\Leftrightarrow1+x^2\sqrt{x+2}=x+x^2\sqrt{2x+1}\)
\(\Leftrightarrow\left(1-x\right)+x^2\frac{1-x}{\sqrt{x+2}+\sqrt{2x+1}}=0\)
\(\Leftrightarrow\left(1-x\right)\left(1+\frac{x^2}{\sqrt{x+2}+\sqrt{2x+1}}\right)=0\)(2)
Vì\(\hept{\begin{cases}x\ne0\\x\ge\frac{-1}{2}\end{cases}}\Rightarrow1+\frac{x^2}{\sqrt{x+2}+\sqrt{2x+1}}>0\)
Nên từ (2) => Phương trình đã cho có nghiệm x = 1 (TMĐKXĐ)
8 tháng 10 2020
\(\Leftrightarrow\frac{7x+4}{\sqrt{2\left(x-1\right)\left(x+1\right)}}+\frac{2\sqrt{2x+1}}{\sqrt{2\left(x+1\right)}}=3+\frac{3\sqrt{2x+1}}{\sqrt{x-1}}\)
\(\Leftrightarrow7x+4+2\sqrt{\left(2x+1\right)\left(x-1\right)}=3\sqrt{2\left(x-1\right)\left(x+1\right)}+3\sqrt{2\left(2x+1\right)\left(x+1\right)}\)
\(\Leftrightarrow\left(7x+4+\sqrt{8x^2-4x-4}\right)^2=\left(\sqrt{18x^2-18}+\sqrt{36^2+54x+18}\right)^2\)
\(\Leftrightarrow\left(7x+4\right)^2+8x^2-4x-4+2\left(7x+4\right)\sqrt{8x^2-4x-4}\)\(=18x^2-18+36x^2+54x+18+2\sqrt{\left(18x^2-18\right)\left(36x^2+54x+18\right)}\)
\(\Leftrightarrow3x^2-2x+12+4\left(7x+4\right)\sqrt{\left(x-1\right)\left(2x+1\right)}=36\left(x+1\right)\sqrt{\left(x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow3x^2-2x+12=4\left(2x+5\right)\sqrt{\left(x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\left(3x^2-2x+12\right)^2=16\left(2x+5\right)^2\left(x-1\right)\left(2x+1\right)\)
\(\Leftrightarrow119x^4+588x^3+1940x^2-672x-544=0\left(1\right)\)
Ta thấy x>1 => Vế trái (1) \(>119.1^4+588.1^3+1940.1^2-672.1-544=1431>0\)
=> pt vô nghiệm.
\(Đkxđ:\hept{\begin{cases}2x-1>0\\4x-3>0\\x>0\end{cases}\Leftrightarrow x>\frac{3}{4}}\)
Phương trình tương đương với:
\(\left(\frac{x}{\sqrt{2x-1}}-1\right)+\left(\frac{x}{\sqrt[4]{4x-3}}-1\right)=0\)
\(\Leftrightarrow\frac{x-\sqrt{2x-1}}{\sqrt{2x-1}}+\frac{2-\sqrt[4]{4x-3}}{\sqrt[4]{4x-3}}=0\)
\(\Leftrightarrow\frac{x^2-2x+1}{\sqrt{2x-1}\left(x+\sqrt{2x-1}\right)}+\frac{x^2-\sqrt{4x-3}}{\sqrt[4]{4x-3}\left(x+\sqrt[4]{4x-3}\right)}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{\sqrt{2x-1}\left(x+\sqrt{2x-1}\right)}+\frac{x^4-4x+3}{\sqrt[4]{4x-3}\left(x+\sqrt[4]{4x-3}\right)\left(x^2+\sqrt{4x-3}\right)}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{\sqrt{2x-1}\left(x+\sqrt{2x-1}\right)}+\frac{\left(x-1\right)^2\left(x^2+2x+3\right)}{\sqrt[4]{4x-3}\left(x+\sqrt[4]{4x-3}\right)\left(x^2+\sqrt{4x-3}\right)}=0\)
\(\Leftrightarrow\left(x-1\right)^2\left[\frac{1}{\sqrt{2x-1}\left(x+\sqrt{2x-1}\right)}+\frac{\left(x+1\right)^2+2}{\sqrt[4]{4x-3}\left(x+\sqrt[4]{4x-3}\right)\left(x^2+\sqrt{4x-3}\right)}\right]=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy .............................