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ĐKXĐ : \(y\ge\frac{4}{\sqrt{3}}\) hoặc \(y\le\frac{-4}{\sqrt{3}}\)
\(B=-\left|1-2x\right|-2\left|x-3\right|-\sqrt{3y^2-16}+2021\)
\(B=-\left(\left|1-2x\right|+\left|2x-6\right|\right)-\sqrt{3y^2-16}+2021\)
\(B\le-\left|1-2x+2x-6\right|-0+2021=2016\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(1-2x\right)\left(2x-6\right)\ge0\left(1\right)\\3y^2-16=0\left(2\right)\end{cases}}\)
\(\left(1\right)\)
TH1 : \(\hept{\begin{cases}1-2x\ge0\\2x-6\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le\frac{1}{2}\\x\ge3\end{cases}}}\) ( loại )
TH2 : \(\hept{\begin{cases}1-2x\le0\\2x-6\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{2}\\x\le3\end{cases}\Leftrightarrow}\frac{1}{2}\le x\le3}\)
\(\left(2\right)\)\(\Leftrightarrow\)\(y^2=\frac{16}{3}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}y=\sqrt{\frac{16}{3}}\\y=-\sqrt{\frac{16}{3}}\end{cases}\Leftrightarrow\orbr{\begin{cases}y=\frac{4}{\sqrt{3}}\\y=\frac{-4}{\sqrt{3}}\end{cases}}}\) ( nhận )
Vậy GTNN của \(B\) là \(2016\) khi \(\frac{1}{2}\le x\le3\) và \(y=\frac{4}{\sqrt{3}}\) hoặc \(y=\frac{-4}{\sqrt{3}}\)
-,-
\(D=\sqrt{\left(2x-1\right)^2+16}+\left|y^2+1\right|+2\)
Ta có:\(\left\{{}\begin{matrix}\sqrt{\left(2x-1\right)^2+16}\ge\sqrt{16}=4\\\left|y^2+1\right|\ge1\end{matrix}\right.\)
Nên:\(D\ge4+1+2=7\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=0\end{matrix}\right.\)
\(\sqrt{\left(2x-\sqrt{16}\right)^2}+\left(y^2.64\right)^2+lx+y+zl=0\)
\(\Rightarrow\sqrt{2x-4}+8y^4+lx+y+zl=0\)
\(\sqrt{2x-4};8y^4;lx+y+zl\ge0\)mà \(\sqrt{2x-4}+8y^4+lx+y+zl=0\)
\(\Rightarrow\sqrt{2x-4}=8y^4=lx+y+zl=0\)
=>2x-4=y4=lx+y+zl=0
=>x=2;y=0;z=-2
Vậy x=2;y=0;z=-2
\(\left[-\sqrt{2,25}+4\sqrt{\left(-2,15\right)^2}-\left(3\sqrt{\dfrac{7}{6}}\right)^2\right]\sqrt{1\dfrac{9}{16}}\)
\(=\left[-1,5+4\sqrt{2,15^2}-9\cdot\dfrac{7}{6}\right]\sqrt{\dfrac{25}{16}}\)
\(=\left[4\cdot\dfrac{43}{20}-10,5-1,5\right]\cdot\dfrac{5}{4}\)
\(=\left[\dfrac{43}{5}-12\right]\cdot\dfrac{5}{4}\)
\(=\dfrac{43}{5}\cdot\dfrac{5}{4}-12\cdot\dfrac{5}{4}\)
\(=\dfrac{43}{4}-15=\dfrac{-17}{4}\)
a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)
b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)
c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)
\(\frac{3}{4}-\left(\frac{1}{4}-x\right)=\frac{2}{3}\)
\(\frac{1}{4}-x=\frac{3}{4}-\frac{2}{3}\)
\(\frac{1}{4}-x=\frac{1}{12}\)
\(x=\frac{1}{4}-\frac{1}{12}\)
\(x=\frac{2}{3}\)
\(\frac{3}{4}-\left(\frac{1}{4}-x\right)=\frac{2}{3}\)
=> \(\frac{3}{4}-\frac{1}{4}+x=\frac{2}{3}\)
=> \(\frac{1}{2}+x=\frac{2}{3}\)
=> x = \(\frac{2}{3}-\frac{1}{2}\)
=> x = \(\frac{4-3}{6}\)
=> x = \(\frac{1}{6}\).
\(\sqrt{\frac{9}{16}}+\frac{\frac{3}{5}}{\left|2x-20\%\right|}=\frac{3}{7}\)
=> \(\frac{3}{4}+\frac{\frac{3}{5}}{\left|2x-\frac{1}{5}\right|}=\frac{3}{7}\)
=> \(\frac{\frac{3}{5}}{\left|2x-\frac{1}{5}\right|}=\frac{3}{7}-\frac{3}{4}\)
=> \(\frac{\frac{3}{5}}{\left|2x-\frac{1}{5}\right|}=\frac{-9}{28}\)
=> \(-9\left|2x-\frac{1}{5}\right|=28.\frac{3}{5}\)
=> \(-9\left|2x-\frac{1}{5}\right|=\frac{84}{5}\)
=> \(\left|2x-\frac{1}{5}\right|=\frac{\frac{84}{5}}{-9}\)
=> \(\left|2x-\frac{1}{5}\right|=\frac{-28}{15}\)
=> Không có x thoả mãn đk.
<=>2x-\(\sqrt{3}\)=4 hoặc 2x-\(\sqrt{3}\)=-4
<=>2x=4+\(\sqrt{3}\) hoặc 2x=\(\sqrt{3}-4\)
<=>x=\(\frac{4+\sqrt{3}}{2}\) hoặc x=\(\frac{\sqrt{3}-4}{2}\)