\(5xy\sqrt{\frac{x^2}{y^6}}=5\sqrt{\frac{x^4y^2}{y^6}}=5\sqrt{\frac{x^4}{y^4}}=5\left|\frac{x^2}{y^2}\right|=-5\)

\(5xy\sqrt{\frac{x^2}{y^6}}=5\sqrt{\frac{x^4y^2}{y^6}}=5\sqrt{\frac{x^4}{y^4}}=5\)

\(0.2x^3y^3\sqrt{\frac{16}{x^4y^8}}=0.2.4\sqrt{\frac{x^6y^6}{x^4y^8}}=0.8\frac{x}{y}\)

a, Ta có : \(\frac{y}{x}.\sqrt{\frac{x^2}{y^4}}=\frac{y}{x}.\frac{x}{y^2}=\frac{1}{y}\)

b , Ta có : \(5xy\sqrt{\frac{x^2}{y^6}}=5xy\frac{x}{y^3}=\frac{5x^2}{y^2}\)

c, Ta có : \(0,2x^3y^3\sqrt{\frac{16}{x^4y^8}}=0,2x^3y^3.\frac{4}{x^2y^4}=\frac{0,8x}{y}\)

Ta có: \(f\left(x\right)-g\left(x\right)=\left(0,7x^4+0,2x^2-5\right)-\left(0,3x^4+0,1x^2-8\right)\)

\(=0,7x^4+0,2x^2-5-0,3x^4-0,1x^2+8\)

\(=0,4x^4+0,1x^2+3\)

Vì \(\hept{\begin{cases}0,4x^4\ge0\\0,1x^2\ge0\end{cases}}\)nên \(0,4x^4+0,1x^2+3>0\)

\(\Rightarrow f\left(x\right)-g\left(x\right)>0\)hay \(f\left(x\right)>g\left(x\right)\forall x\)

Ta có :

\(\left|0,2x-3,1\right|+\left|0,2+3,1\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|0,2x-3,1\right|=0\\\left|0,2x+3,1\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}0,2x-3,1=0\\0,2x+3,1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}0,2x=3,1\\0,2x=-3,1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=15,5\\x=-15,5\end{matrix}\right.\) (vô lí)

Vậy ko tìm dc x thỏa mãn theo yêu cầu

Ta có: \(\left|0.2x-3.1\right|\)\(\ge\)0 vs mọi x

\(\left|0.2x+3.1\right|\)\(\ge\)0 vs mọi x

nên \(\left|0.2x-3.1\right|+\left|0.2x+3.1\right|=0\)

<=> \(\left\{{}\begin{matrix}\left|0.2x-3.1\right|=0\\\left|0.2x+3.1\right|=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}0.2x-3.1=0\\0.2x+3.1=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}0.2x=3.1\\0.2x=-3.1\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=15.5\\x=-15.5\end{matrix}\right.\) (vô lý, vì 1 giá trị x ko thể có 2 giá trị)

Vậy ko có giá trị x nào thỏa mãn \(\left|0.2x-3.1\right|+\left|0.2x+3.1\right|=0\)

Bài giải:(CHÚ Ý / LÀ PHẦN VÍ DỤ 1/2 LÀ 1 phần 2)

0,2x \(\frac{15}{36}\)-\(\orbr{\begin{cases}\\\end{cases}\frac{2}{5}}\)+\(\frac{2}{3}\)

1/5x15/36-16/15

1/12-16/15

-59/60

MIK KO CHẮC LÀ MIK LÀM ĐÚNG ĐÂU THÔNG CẢM HA :)

pạn ơi k mik đi mà

\(a,\frac{5}{3}-\frac{2}{3}.x=1\)

      \(\frac{5}{3}-1=\frac{2}{3}.x\)

               \(\frac{2}{3}=\frac{2}{3}.x\)

                  \(x=\frac{2}{3}:\frac{2}{3}\)

                 \(x=1\) 

\(b,\frac{4}{5}+\frac{5}{7}.x=\frac{1}{6}\)

               \(\frac{5}{7}.x=\frac{1}{6}-\frac{4}{5}\)

                \(\frac{5}{7}.x=-\frac{19}{30}\)

                     \(x=-\frac{19}{30}:\frac{5}{7}\)

                     \(x=-\frac{133}{150}\)