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\(TXĐ:D=R\)
\(pt\Leftrightarrow\sqrt{\left(2x-1\right)^2+1^2}+\sqrt{\left(\sqrt{3}x+1\right)^2+\left(x+1\right)^2}\)
\(+\sqrt{\left(\sqrt{3}x-1\right)^2+\left(x+1\right)^2}=3\sqrt{2}\left(1\right)\)
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
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a.
\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1\le x\le3\)
b.
ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)
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1) \(\sqrt[]{3x+7}-5< 0\)
\(\Leftrightarrow\sqrt[]{3x+7}< 5\)
\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)
\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)
\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)
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\(\Leftrightarrow\left(\sqrt[3]{x+1}-1\right)+\left(\sqrt{2x+4}-2\right)< -x\sqrt{2}\)
=>\(\dfrac{x+1-1}{\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1}+\dfrac{2x+4-4}{\sqrt{2x+4}+2}+x\sqrt{2}< 0\)
=>x<0
=>-1<x<0
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\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x-3\ge0\\2x^2-3x+1\ge0\\x^2+2x-3\le2x^2-3x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge1\\x\le\dfrac{1}{2}\end{matrix}\right.\\x^2-5x+4\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x\le-3\\x\ge4\end{matrix}\right.\)
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\(\Leftrightarrow\frac{7x+4}{\sqrt{2\left(x-1\right)\left(x+1\right)}}+\frac{2\sqrt{2x+1}}{\sqrt{2\left(x+1\right)}}=3+\frac{3\sqrt{2x+1}}{\sqrt{x-1}}\)
\(\Leftrightarrow7x+4+2\sqrt{\left(2x+1\right)\left(x-1\right)}=3\sqrt{2\left(x-1\right)\left(x+1\right)}+3\sqrt{2\left(2x+1\right)\left(x+1\right)}\)
\(\Leftrightarrow\left(7x+4+\sqrt{8x^2-4x-4}\right)^2=\left(\sqrt{18x^2-18}+\sqrt{36^2+54x+18}\right)^2\)
\(\Leftrightarrow\left(7x+4\right)^2+8x^2-4x-4+2\left(7x+4\right)\sqrt{8x^2-4x-4}\)\(=18x^2-18+36x^2+54x+18+2\sqrt{\left(18x^2-18\right)\left(36x^2+54x+18\right)}\)
\(\Leftrightarrow3x^2-2x+12+4\left(7x+4\right)\sqrt{\left(x-1\right)\left(2x+1\right)}=36\left(x+1\right)\sqrt{\left(x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow3x^2-2x+12=4\left(2x+5\right)\sqrt{\left(x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\left(3x^2-2x+12\right)^2=16\left(2x+5\right)^2\left(x-1\right)\left(2x+1\right)\)
\(\Leftrightarrow119x^4+588x^3+1940x^2-672x-544=0\left(1\right)\)
Ta thấy x>1 => Vế trái (1) \(>119.1^4+588.1^3+1940.1^2-672.1-544=1431>0\)
=> pt vô nghiệm.
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Bạn xem lại đề
Dưới căn là \(\sqrt{2x^2+1}\) hay \(\sqrt{2x^2-1}\)
Nguyễn Việt Lâm Mình cũng đang thắc mắc, dường như đề bài thầy bình cho sai hay sao ấy