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\(xy-3x-y=6\)
\(=>xy+3x-y-3=6-3\)
\(=>x\left(y+3\right)-\left(y+3\right)=3\)
\(=>\left(y+3\right)\left(x-1\right)=3\)
| y+3 | -1 | 3 | 1 | -3 | |
| x-1 | -3 | 1 | 3 | -1 |
| y+3 | -1 | 3 | -3 | 1 |
| y | -4 | -1 | -7 | -3 |
| x-1 | -3 | 1 | 3 | -1 |
| x | -2 | 2 | 4 | 0 |
a: \(=\dfrac{3}{2}\left(-21-\dfrac{1}{3}+1+\dfrac{1}{3}\right)=\dfrac{3}{2}\cdot\left(-20\right)=-30\)
b: \(=\dfrac{2018}{2019}\left(13-13-\dfrac{2018}{2019}-\dfrac{1}{2019}\right)=-\dfrac{2018}{2019}\)
Ta có: \(\dfrac{x+1}{2017}+\dfrac{x+1}{2018}=\dfrac{x+1}{2019}+\dfrac{x+1}{2020}\)
\(\Rightarrow\left(\dfrac{x+1}{2017}+\dfrac{x+1}{2018}\right)-\left(\dfrac{x+1}{2019}+\dfrac{x+1}{2020}\right)=0\)
\(\Rightarrow\dfrac{x+1}{2017}+\dfrac{x+1}{2018}-\dfrac{x+1}{2019}-\dfrac{x+1}{2020}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2020}\right)=0\)
Vì \(\dfrac{1}{2017}>\dfrac{1}{2018}>\dfrac{1}{2019}>\dfrac{1}{2020}>0\) nên
\(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2020}>0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
mik chỉ làm được một bài thôi cậu chọn đi bài nào nói với mik , mik làm cho
Bài 1:
a) \(\left|x-\dfrac{2}{3}\right|+\left|y+x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-\dfrac{2}{3}\right|=0\\\left|y+x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{2}{3}=0\\y+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{2}{3}\end{matrix}\right.\)
b) \(\left(x-2y\right)^2+\left|x+\dfrac{1}{6}\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2y\right)^2=0\\\left|x+\dfrac{1}{6}\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\x+\dfrac{1}{6}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=x\\x=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=-\dfrac{1}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{12}\\x=\dfrac{-1}{6}\end{matrix}\right.\)
a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)
=1-2/4=1/2
b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)
c: x-y=0 nên x=y
\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)
=2019