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25 tháng 1 2018
\(a\text{) }\left|2x-5\right|+\left|3y+1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x-5\right|=0\\\left|3y+1\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(\left|3x-4\right|+\left|3y-5\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x-4\right|=0\\\left|3y-5\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y-5=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=4\\3y=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{5}{3}\end{matrix}\right.\)
c) \(\left|2x-5\right|+\left|xy-3y+2\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x-5\right|=0\\\left|xy-3y+2\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\xy-3y+2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=5\\xy-3y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\xy-3y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\dfrac{5}{2}y-3y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\left(\dfrac{5}{2}-3\right)y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\left(-\dfrac{1}{2}\right)y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\\left(-\dfrac{1}{2}\right)y=-2\end{matrix}\right.\)
DL
2
1 tháng 9 2019
a) 2x - 5 = 3 + 2x - 7x
=> 2x - 2x + 7x = 3 +5
=> 7x = 8
=> x = 8/7
b) \(\left(2x-1\right)^2=\left(2x-1\right)^5\)
=> \(\left(2x-1\right)^2-\left(2x-1\right)^5=0\)
=> \(\left(2x-1\right)^2\left[1-\left(2x-1\right)^3\right]=0\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^3=1\end{cases}}\)
=> \(\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)
TL
0
NH
5
3 tháng 12 2019
a) Ta có: \(\frac{2x}{-3}=\frac{-3y}{5}=\frac{z}{4}\) => \(\frac{x}{-\frac{3}{2}}=\frac{y}{-\frac{5}{3}}=\frac{z}{4}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{-\frac{3}{2}}=\frac{y}{-\frac{5}{3}}=\frac{z}{4}=\frac{x+z}{-\frac{3}{2}+4}=\frac{30}{\frac{5}{2}}=12\)
=> \(\hept{\begin{cases}\frac{x}{-\frac{3}{2}}=12\\\frac{y}{-\frac{5}{3}}=12\\\frac{z}{4}=12\end{cases}}\) => \(\hept{\begin{cases}x=-18\\y=-20\\z=48\end{cases}}\)
vậy ...
M
3 tháng 12 2019
Làm tiếp Edogawa Conan
\(b,\text{ }\frac{3x}{-3}=\frac{-3y}{5}=4z\)
Bài giải
\(\frac{3x}{-3}=\frac{-3y}{5}=4z=\frac{x}{-1}=\frac{y}{\frac{5}{-3}}=\frac{z}{\frac{1}{4}}=\frac{x-z}{-1-\frac{1}{4}}=\frac{22}{-\frac{5}{4}}=\frac{11}{10}\)
\(\Rightarrow\text{ }x=-1\cdot\frac{11}{10}=-\frac{11}{10}\)
\(y=\frac{11}{10}\cdot\frac{5}{-3}=\frac{11}{-6}\)
\(z=\frac{11}{10}\cdot\frac{1}{4}=\frac{11}{40}\)
a) \(\left(3x-5\right)\left(x+4\right)=3x^2+12x-5x-20=3x^2+7x-20\)
b) \(\left(2x-3y\right)\left(2x+3y\right)=\left(2x\right)^2-\left(3y\right)^2=4x^2-9y^2\)