tìm x thuộc z biết:
(2,8x-32):2/3=-90
help me, please
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a: \(=\dfrac{7}{5}\cdot\dfrac{15}{49}-\dfrac{12+10}{15}:\dfrac{11}{5}\)
\(=\dfrac{3}{7}-\dfrac{22}{15}\cdot\dfrac{5}{11}=\dfrac{3}{7}-\dfrac{2}{3}=\dfrac{9-14}{21}=\dfrac{-5}{21}\)
b: =>2,8x-32=-60
=>2,8x=-28
hay x=-10
A/\(\left(2,8x-32\right):\frac{2}{3}=-90\)
\(\left(\frac{28}{10}x-32\right)=\frac{-90}{1}.\frac{2}{3}\)
\(\left(\frac{14}{5}x-32\right)=\frac{-30}{1}.\frac{2}{1}\)
\(\left(\frac{14}{5}x-32\right)=-60\)
\(\frac{14}{5}x=-60+32\)
\(\frac{14}{5}x=-28\)
\(x=\frac{-28}{1}:\frac{14}{5}\)
\(x=\frac{-28}{1}.\frac{5}{14}\)
\(x=\frac{-2}{1}.\frac{5}{1}=-10\)
B/\(\left(4,5-2x\right).1\frac{4}{7}=\frac{11}{14}\)
\(\left(\frac{45}{10}-2x\right).\frac{11}{7}=\frac{11}{14}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}:\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{14}.\frac{7}{11}\)
\(\left(\frac{9}{2}-2x\right)=\frac{1}{2}.\frac{1}{1}=\frac{1}{2}\)
\(2x=\frac{9}{2}-\frac{1}{2}\)
\(2x=\frac{8}{2}\)
\(x=\frac{8}{2}:\frac{2}{1}=\frac{8}{2}.\frac{1}{2}\)
\(x=\frac{4}{2}.\frac{1}{1}=\frac{4}{2}=2\)
Em thích
a) | x + 12 - 41| = 11
\(\Rightarrow\orbr{\begin{cases}x+12-41=11\\x+12-41=-11\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=40\\x=18\end{cases}}\)
Vậy \(x\in\left\{40;18\right\}\)
À ... cái "Em thích" kia là đánh nhầm ạ :33
Xin các bạn đừng để ý :v
a ) (2,8x - 32 ) : 2/3 = -90
<=> 2,8x -32 = -60
<=> 2,8x = -28
<=> x = -10
b ) 4/5 + 5/7 :x =1/6
<=> 5/7:x = -19/30
<=> x = -150/133
a)\(\left(2,8x-32\right):\frac{2}{3}=-90\)
\(2,8x-32=-90\cdot\frac{2}{3}\)
\(2,8x-32=-60\)
\(2,8x=-60+32\)
\(2,8x=-28\)
\(x=-28:2,8\)
\(x=-10\)
Vậy x = -10
b) \(\frac{4}{5}+\frac{5}{7}:x=\frac{1}{6}\)
\(\frac{5}{7}:x=\frac{1}{6}-\frac{4}{5}\)
\(\frac{5}{7}:x=\frac{-19}{30}\)
\(x=\frac{5}{7}:\frac{-19}{30}\)
\(x=\frac{-150}{133}\)
Vậy x = -150/133
=))
\(xy+x+y=4\)
\(\Leftrightarrow xy+x+y+1=4+1\)
\(\Leftrightarrow x\left(y+1\right)+\left(y+1\right)=5\)
\(\Leftrightarrow\left(x+1\right)\left(y+1\right)=5\)
\(\Leftrightarrow x+1;y+1\inƯ\left(5\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1=1\\y+1=5\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=5\\y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-1\\y+1=-5\end{matrix}\right.\\\left\{{}\begin{matrix}x+1-5\\y+1=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-6\end{matrix}\right.\\\left\{{}\begin{matrix}x=-6\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
\(\frac{x+3}{x-2}=\frac{\left(x-2\right)+5}{x-2}=\frac{x-2}{x-2}+\frac{5}{x-2}=1+\frac{5}{x-2}\)
hàm số f(x) có giá trị ngyên \(\Leftrightarrow\) 5 \(⋮\)x-2
hay x-2 là các ước của 5
nên x-2\(\in\){-5;-1;1;5}
Vậy x\(\in\){-3;1;3;7}
Đó là đáp số cho bài toán của bạn
\(\left(2,8x-32\right):\frac{2}{3}=-90\)
\(2,8x-32=-90.\frac{2}{3}=\frac{-90.2}{3}=\frac{-30.2}{1}=-60\)
\(\frac{14}{5}x=-60+32=-28\)
\(x=-28:\frac{14}{5}=-28.\frac{5}{14}=\frac{-28.5}{14}=\frac{-2.5}{1}=-10\)
(2,8X-32):2/3=-90
2,8X -32=-90*2/3
2,8X-32=-60
2,8X=-60+32
2,8X=-28
X=-28/2,8
X=-10