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27 tháng 2 2021
\(\left(-x^2y\right)^3\cdot\dfrac{1}{2}\cdot x^2y^3\cdot\left(-2xy^2z\right)^2\\ =-x^6y^3\cdot\dfrac{1}{2}x^2y^3\cdot4x^2y^4z^2\\ =\left(-1\cdot\dfrac{1}{2}\cdot4\right)\cdot\left(x^6\cdot x^2\cdot x^2\right)\cdot\left(y^3\cdot y^3\cdot y^4\right)\cdot z^2\\ =-2x^{10}y^{10}z^2\)
13 tháng 1 2022
Bài 2:
a: =>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
XH
3
A
27 tháng 2 2021
\(\left(-x^3.z.y\right).\left(\dfrac{2}{3}.y.x^2\right)^2\)
\(=-x^3.z.y.\dfrac{4}{9}.y^2.x^4\)
\(=-\dfrac{4}{9}x^7.y^3.z\)
27 tháng 2 2021
`(-x^3zy)(2/3yx^2)^2`
`=-4/9x^3zy.y^2x^4`
`=-4/9x^{3+4}.y^{1+2}z`
`=-4/9x^7y^3z`
MP
1
22 tháng 8 2015
=> \(\frac{x-y}{1}=\frac{x+y}{7}=\frac{xy}{24}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{xy}{24}=\frac{x-y}{1}=\frac{x+y}{7}=\frac{\left(x-y\right)+\left(x+y\right)}{1+7}=\frac{\left(x-y\right)-\left(x+y\right)}{1-7}\)=> \(\frac{xy}{24}=\frac{x}{4}=\frac{y}{3}\)
\(\frac{xy}{24}=\frac{x}{4}\)=>\(\frac{x}{4}.\frac{y}{6}=\frac{x}{4}\)=> \(\frac{y}{6}=\frac{x}{4}:\frac{x}{4}=1\) ( do x khác 0) => y = 6
\(\frac{xy}{24}=\frac{y}{3}\Rightarrow\frac{x}{8}.\frac{y}{3}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{3}:\frac{y}{3}=1\) ( do y khác 0) => x = 8
Vậy...
5 tháng 1 2022
\(a,\left\{{}\begin{matrix}\left|x-3y\right|\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y=-12\\y=-4\end{matrix}\right.\)
\(b,Sửa:\left|x-y-5\right|+\left(y+3\right)^2=0\\ \left\{{}\begin{matrix}\left|x-y-5\right|\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-y-5=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+5=2\\y=-3\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}\left|x+y-1\right|\ge0\\\left(y-2\right)^4\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-y=-1\\y=2\end{matrix}\right.\)
\(d,\left\{{}\begin{matrix}\left|x+3y-1\right|\ge0\\3\left|y+2\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+3y-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-3y=7\\y=-2\end{matrix}\right.\)
\(e,Sửa:\left|2021-x\right|+\left|2y-2022\right|=0\\ \left\{{}\begin{matrix}\left|2021-x\right|\ge0\\\left|2y-2022\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2021-x=0\\2y-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\y=1011\end{matrix}\right.\)
YP
2
TH
17 tháng 11 2021
a,Ta có:
\(\dfrac{x}{y}=\dfrac{7}{4}=\dfrac{x}{7}=\dfrac{y}{4}\)
ÁP dụng tcdtsbn , ta có:
\(\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x+y}{7+4}=\dfrac{33}{11}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=21\\y=12\end{matrix}\right.\)
b,
\(\Rightarrow3.\left(x-1\right)=-24\)
\(\Rightarrow x-1=-8\)
\(\Rightarrow x=-7\)
17 tháng 11 2021
A)\(\dfrac{x}{y}=\dfrac{7}{4}\Rightarrow\dfrac{x}{7}=\dfrac{y}{4}\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x+y}{7+4}=\dfrac{33}{11}=3\)
\(\dfrac{x}{7}=3\Rightarrow x=21\\ \dfrac{y}{4}=3\Rightarrow y=12\)
B) \(3\left(x-1\right)+5=-19\\ \Rightarrow3\left(x-1\right)=-24\\ \Rightarrow x-1=-8\\ \Rightarrow x=-7\)
28 tháng 9 2021
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
1.
a.
x(y-1)-(y-1)=0
(x-1)(y-1)=0
x=1
y=1
b.x,y khác không
(x+y)/xy=1/2
2x+2y=xy
x(2-y)-2(2-y)=4
(x-2)(y-2)=4
x-2={-4,-2,1,4}=>x={-2,0,3,6}
y-2={-1,-2,4,1}=>y={3,0,6,3}