\(0,2-\left|4,2-2x\right|=0\)

\(\Leftrightarrow\left|4,2-2x\right|=0,2\)

\(\Leftrightarrow\orbr{\begin{cases}4,2-2x=0,2\\4,2-2x=-0,2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=2,2\end{cases}}\)

Vậy \(x\in\left\{2;2,2\right\}\)

\(0,2-\left|4,2-2x\right|=0\)

\(\Rightarrow\left|\frac{21}{5}-2x\right|=\frac{1}{5}\)

\(\Rightarrow\orbr{\begin{cases}\frac{21}{5}-2x=\frac{1}{5}\\\frac{21}{5}-2x=-\frac{1}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=4\\2x=\frac{22}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{11}{5}\end{cases}}\)

\(3\left(x+2\right)^2+\left(2x-3\right)^2-7\left(x-4\right)\left(x+4\right)=64\)

\(\Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-12x+9\right)-7\left(x^2-16\right)=64\)

\(\Leftrightarrow3x^2+12x+12+4x^2-12x+9-7x^2+112=64\)

\(\Leftrightarrow12+9+112=64\)(vô lí)

Vậy pt vô nghiệm

 TL:

\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-6x+9-7x^2+112=64\)

\(\Leftrightarrow6x+133=64\)  

\(\Leftrightarrow6x=-69\) 

\(\Leftrightarrow x=\frac{-23}{2}\) 

Vậy....

= \(\frac{11}{2}+\frac{11}{2}-\frac{29}{7}\)
= \(11-\frac{29}{7}\)
= \(\frac{48}{7}\)

\(5\frac{1}{2}+3\frac{2}{3}-4\frac{1}{7}=\frac{11}{2}+\frac{11}{3}-\frac{29}{7}=\frac{55}{6}-\frac{29}{7}=\frac{211}{42}\)

a)Ta có:\(17^{50}>17^{49}=\left(17^7\right)^7>30^7\)

b)Ta có:\(81^{75}=\left(81^3\right)^{25}=\left(3^{12}\right)^{25}=\left(3^3\right)^{100}=27^{100}< 30^{100}\)

Sai đề rồi bạn nhé

Vậy đề đúng là gì bạn

\(\frac{4}{7}+\frac{14}{4}=\frac{16}{28}+\frac{98}{28}=\frac{16+98}{28}=\frac{114}{28}=\frac{57}{14}\)

Chúc bạn học tốt !~!

\(\frac{4}{7}+\frac{14}{4}=\frac{16}{28}+\frac{98}{28}=\frac{114}{28}=\frac{57}{14}\)