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a/ ĐKXĐ: \(x\ge1\)
Khi \(x\ge1\) ta thấy \(\left\{{}\begin{matrix}VT>0\\VP=1-x\le0\end{matrix}\right.\) nên pt vô nghiệm
b/ \(x\ge1\)
\(\sqrt{\sqrt{x-1}\left(x-2\sqrt{x-1}\right)}+\sqrt{\sqrt{x-1}\left(x+3-4\sqrt{x-1}\right)}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-1\right)^2}+\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-2\right)^2}=\sqrt{x-1}\)
Đặt \(\sqrt{x-1}=a\ge0\) ta được:
\(\sqrt{a\left(a-1\right)^2}+\sqrt{a\left(a-2\right)^2}=a\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\Rightarrow x=1\\\sqrt{\left(a-1\right)^2}+\sqrt{\left(a-2\right)^2}=\sqrt{a}\left(1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left|a-1\right|+\left|a-2\right|=\sqrt{a}\)
- Với \(a\ge2\) ta được: \(2a-3=\sqrt{a}\Leftrightarrow2a-\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow a=\frac{9}{4}\Rightarrow\sqrt{x-1}=\frac{9}{4}\Rightarrow...\)
- Với \(0\le a\le1\) ta được:
\(1-a+2-a=\sqrt{a}\Leftrightarrow2a+\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-1}=1\Rightarrow...\)
- Với \(1< a< 2\Rightarrow a-1+2-a=\sqrt{a}\Leftrightarrow a=1\left(l\right)\)
c/ ĐKXĐ: \(x\ge\frac{49}{14}\)
\(\Leftrightarrow\sqrt{14x-49+14\sqrt{14x-49}+49}+\sqrt{14x-49-14\sqrt{14x-49}+49}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{14x-49}+7\right)^2}+\sqrt{\left(\sqrt{14x-49}-7\right)^2}=14\)
\(\Leftrightarrow\left|\sqrt{14x-49}+7\right|+\left|7-\sqrt{14x-49}\right|=14\)
Mà \(VT\ge\left|\sqrt{14x-49}+7+7-\sqrt{14x-49}\right|=14\)
Nên dấu "=" xảy ra khi và chỉ khi:
\(7-\sqrt{14x-49}\ge0\)
\(\Leftrightarrow14x-49\le49\Leftrightarrow x\le7\)
Vậy nghiệm của pt là \(\frac{49}{14}\le x\le7\)
a) \(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\)
\(\Leftrightarrow x+\dfrac{x+5}{x+3}=\dfrac{x+5}{x+3}\)
\(\Leftrightarrow x=0\)
b) \(2x+\dfrac{3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow x+x+\dfrac{3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow x+\dfrac{x\left(x-1\right)+3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow x+\dfrac{x^2-x+3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x}{x-1}-x\)
\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x-x\left(x-1\right)}{x-1}\)
\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x-x^2+x}{x-1}\)
\(\Leftrightarrow x^2-x+3=3x-x^2+x\) ( điều kiện \(x\ne1\) )
\(\Leftrightarrow2x^2-5x+3=0\)
\(\Delta=b^2-4ac\)
\(\Delta=1\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3}{2}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=1\left(loại\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{2}\)
c) \(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\)
\(\Leftrightarrow x^2-4x-2=\sqrt{\left(x-2\right)^2}\) ( điều kiện \(x>2\) )
\(\Leftrightarrow x^2-4x-2=x-2\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=5\end{matrix}\right.\)
Vậy \(x=5\)
d) \(\dfrac{2x^2-x-3}{\sqrt{2x-3}}=\sqrt{2x-3}\)
\(\Leftrightarrow2x^2-x-3=\sqrt{\left(2x-3\right)^2}\) ( điều kiện \(x>\dfrac{3}{2}\) )
\(\Leftrightarrow2x^2-x-3=2x-3\)
\(\Leftrightarrow2x^2-3x=0\)
\(\Leftrightarrow x\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)
Vậy phương trình vô nghiệm
ĐKXĐ: \(\hept{\begin{cases}x^2-5x+2\ge0\\2x-1>0\\x-2\ge0\end{cases}\Leftrightarrow x\ge2}\)
Phương trình
\(\Leftrightarrow\sqrt{x-2}\sqrt{2x-1}-x\sqrt{x-2}+3x-x^2-3\sqrt{2x-1}+x\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(\sqrt{2x-1}-x\right)\left(\sqrt{x-2}-3+x\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{2x-1}=x\\\sqrt{x-2}=3-x\end{cases}}\)
<=> 2x-1=x2 hoặc \(\hept{\begin{cases}3-x\ge0\\x-2=3-x^2\end{cases}}\)
<=> x2-2x+1=0 hoặc \(\hept{\begin{cases}x\le3\\x^2-7x+11=0\end{cases}}\)
<=> x=1 hoặc \(\hept{\begin{cases}x\le3\\x=\frac{7\pm\sqrt{3}}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{7-\sqrt{5}}{2}\end{cases}}\)
Đối chiếu điều kiện x>=2 => x=\(=\frac{7-\sqrt{5}}{2}\left(tm\right)\)
Vậy pt có nghiệm \(x=\frac{7-\sqrt{5}}{2}\)
\(TXĐ:D=R\)
\(pt\Leftrightarrow\sqrt{\left(2x-1\right)^2+1^2}+\sqrt{\left(\sqrt{3}x+1\right)^2+\left(x+1\right)^2}\)
\(+\sqrt{\left(\sqrt{3}x-1\right)^2+\left(x+1\right)^2}=3\sqrt{2}\left(1\right)\)
Chọn \(\hept{\begin{cases}\overrightarrow{u}=\left(1;1-2x\right)\\\overrightarrow{v}=\left(\sqrt{3}x+1;x+1\right)\\\overrightarrow{w}=\left(1-\sqrt{3}x;x+1\right)\end{cases}}\)\(\Rightarrow\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\left(3;3\right)\)
\(\Rightarrow\left|\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right|=3\sqrt{2}\)(2)
Ta có: \(\left|\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right|\le\left|\overrightarrow{u}\right|+\left|\overrightarrow{v}\right|+\left|\overrightarrow{w}\right|\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2+1^2}+\sqrt{\left(\sqrt{3}x+1\right)^2+\left(x+1\right)^2}\)
\(+\sqrt{\left(\sqrt{3}x-1\right)^2+\left(x+1\right)^2}\ge3\sqrt{2}\)
Dấu "=" xảy ra khi \(\overrightarrow{u};\overrightarrow{v};\overrightarrow{w}\)cùng hướng
Từ (1) và (2) suy ra \(\overrightarrow{u};\overrightarrow{v};\overrightarrow{w}\)cùng hướng
\(\Leftrightarrow\exists k,l>0\hept{\begin{cases}\overrightarrow{v}=k.\overrightarrow{u}\\\overrightarrow{v}=l.\overrightarrow{w}\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{3}x+1=k.1;x+1=k\left(1-2x\right)\\\sqrt{3}x+1=l\left(1-\sqrt{3}x\right);x+1=l\left(x+1\right)\end{cases}}\)
Vậy x = 0