a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\)                               ĐKXĐ : x #0, x#2, x#-2

<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)

<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)

=> 10 - 2x + 7x - 14 = 4x - 4 + x

<=>-2x + 7x - 4x + x  = -4 - 10 + 14

<=>x=-14

ko biết

a. 60%x + 0,4x + x : 3 = 2

0.6x + 0,4x + x : 3 = 2

x(0,6 + 0,4 : 3 ) = 2

\(x.\frac{1}{3}=2=>x=2:\frac{1}{3}=\frac{1}{6}\)

câu B tự làm nha .

\(\left(\frac{x}{-5}+1\frac{1}{2}\right):\frac{28}{75}-1,4.\frac{15}{49}=\left|-\frac{2}{3}\right|.\left(-\frac{3}{2}\right)^3\)

\(\left(\frac{x}{-5}+\frac{3}{2}\right).\frac{75}{28}-\frac{14}{10}.\frac{15}{49}=\frac{2}{3}.\frac{-27}{8}\)

\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}-\frac{3}{7}=\frac{-9}{4}\)

\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-9}{4}+\frac{3}{7}\)

\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-63}{28}+\frac{12}{28}\)

\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-51}{28}\)

\(\frac{-x}{5}+\frac{3}{2}=\frac{-51}{28}:\frac{75}{28}\)

\(\frac{-x}{5}+\frac{3}{2}=\frac{-51}{28}.\frac{28}{75}\)

\(\frac{-x}{5}+\frac{3}{2}=\frac{-17}{25}\)

\(\frac{-x}{5}=\frac{-17}{25}-\frac{3}{2}\)

\(\frac{-x}{5}=\frac{-34}{50}-\frac{75}{50}\)

\(\frac{-x}{5}=\frac{-109}{50}\)

\(\frac{-10x}{50}=\frac{-109}{50}\)

Hình như đề sai thì phải

ko biết nữa

mik chỉ biết là cô cho vậy thôi

ĐKXĐ: ...

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}+\frac{y}{x}=3\\\frac{x^2}{y^2}+\frac{y^2}{x^2}+2=49\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}+\frac{y}{x}=3\\\left(\frac{x}{y}+\frac{y}{x}\right)^2=49\end{matrix}\right.\)

Hệ vô nghiệm

\(\)đặt x+y=a , xy=b

=> \(\left\{{}\begin{matrix}a\left(1+\frac{1}{b}\right)=5\\\left(a^2-2b\right)\left(1+\frac{1}{b^2}\right)=49\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=\frac{5}{1+\frac{1}{b}}\\\left(\left(\frac{5}{1+\frac{1}{b}}\right)^2-2b\right)\left(1+\frac{1}{b^2}\right)=49\end{matrix}\right.\)

=> giải đc b²+7b+1=0 => b => a => x,y

số xấu vler , chả biết sếp nào nghĩ ra bài này nữa

ĐKXĐ: ...

\(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=5\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=49\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{x}+y+\frac{1}{y}=5\\\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2=53\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+\frac{1}{x}=a\\y+\frac{1}{y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=5\\a^2+b^2=53\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\ab=-14\end{matrix}\right.\) \(\Rightarrow a;b\) là nghiệm của \(t^2-5t-14=0\Rightarrow\left[{}\begin{matrix}t=7\\t=-2\end{matrix}\right.\)

\(\Rightarrow\left(a;b\right)=\left(-2;7\right);\left(7;-2\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x+\frac{1}{x}=-2\\y+\frac{1}{y}=7\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x+\frac{1}{x}=7\\y+\frac{1}{y}=-2\end{matrix}\right.\)

4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)

\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)

\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)

Tìm z thì dễ rồi