Tìm x:
\(\left(1^2+2^2+...+49^2\right)\left(2-x\right)=-1\frac{1}{5}\)
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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
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
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
Ta có \(1^2+2^2+3^2+...+49^2=\frac{49\left(49+1\right)\left(2.49+1\right)}{6}=40425\)
\(\Rightarrow40425\left(2-x\right)=-1\frac{1}{5}\Rightarrow2-x=-\frac{2}{67375}\Rightarrow x=\frac{134752}{67375}\)