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\(xy^2+\frac{2}{3}y^2x-\frac{3}{4}xy^2=\left(xy^2-\frac{3}{4}xy^2\right)+\frac{2}{3}y^2x=\frac{1}{4}xy^2+\frac{2}{3}y^2x\)
1) x2 = \(\frac{3^2}{5^2^{ }}\)
x = \(\frac{3}{5}\)
x2 = 0.09
x2 = \(\frac{9}{100}\)
x2 = \(\frac{3^2}{10^2}\)
x = \(\frac{3}{10}\)
1. \(x^2=\frac{9}{25}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-3}{5}\\x=\frac{3}{5}\end{cases}}\)
Vậy \(x=\frac{-3}{5}\)hoặc \(x=\frac{3}{5}\)
2. \(x^2=0,09\)\(\Rightarrow x^2=\frac{9}{100}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-3}{10}\\x=\frac{3}{10}\end{cases}}\)
Vậy \(x=\frac{-3}{10}\)hoặc \(x=\frac{3}{10}\)
3. \(\sqrt{2}.x=2\)\(\Rightarrow\left(\sqrt{2}.x\right)^2=2^2\)
\(\Rightarrow2x^2=4\)\(\Rightarrow x^2=2\)\(\Rightarrow x=\pm\sqrt{2}\)
Vì \(\sqrt{2}>0\); \(2>0\)\(\Rightarrow\)Để \(\sqrt{2}.x=2\)thì \(x>0\)
\(\Rightarrow x=\sqrt{2}\)
Vậy \(x=\sqrt{2}\)
2 tấn = 2000kg
Vậy 2 tấn thì cho \(2000:100\times70=1400\left(\text{kg gạo}\right)\)
Bài 7:
a: Xét ΔABE và ΔMBE có
BA=BM
BE chung
EA=EM
Do đó: ΔABE=ΔMBE
a) Có \(\left|x-3y\right|^5\ge0\);\(\left|y+4\right|\ge0\)
\(\rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\)
mà \(\left|x-3y\right|^5+\left|y+4\right|=0\)
\(\rightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
b) Tương tự câu a, ta có:
\(\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)
c. Tương tự, ta có:
\(\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\\left|y+2\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=-2\end{matrix}\right.\)
a. \(\left|x-3y\right|^5\ge0,\left|y+4\right|\ge0\Rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\) \(\Rightarrow VT\ge VP\)
Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\) Vậy...
b. \(\left|x-y-5\right|\ge0,\left(y-3\right)^4\ge0\Rightarrow\left|x-y-5\right|+\left(y-3\right)^4\ge0\) \(\Rightarrow VT\ge VP\)
Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\) Vậy ...
c. \(\left|x+3y-1\right|\ge0,3\cdot\left|y+2\right|\ge0\Rightarrow\left|x+3y-1\right|+3\left|y+2\right|\ge0\) \(\Rightarrow VT\ge VP\) Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\3\left|y+2\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-\left(-2\right)\cdot3=7\\y=-2\end{matrix}\right.\) Vậy...
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tham khảo k cho mk
|X +1/2 | =0
\(\Rightarrow x+\frac{1}{2}=0\)
\(x=0-\frac{1}{2}\)
\(x=-\frac{1}{2}\)
\(\left|x+\frac{1}{2}\right|=0\)
\(\Rightarrow x+\frac{1}{2}=0\)
\(x=0-\frac{1}{2}\)
\(x=-\frac{1}{2}\)
Vậy.......................................