(3,2x-10):\(\frac{4}{5}\)= -48,5
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\(\left(2,1x+\frac{5}{3}\right)^6=\left(-3,2x+\frac{1}{2}\right)^6\Rightarrow2,1x+\frac{5}{3}=-3,2x+\frac{1}{2}\)
\(\Rightarrow2,1x+3,2x=\frac{1}{2}-\frac{5}{3}\)
\(5,3x=\frac{-7}{6}\)
\(x=\frac{-7}{6}:\frac{53}{10}=\frac{-35}{159}\)
Vậy \(x=\frac{-35}{159}\)
1: Ta có: 7x+6(3-x)=27-20+73
\(\Leftrightarrow7x+18-6x=80\)
\(\Leftrightarrow x=80-18=62\)
Vậy: x=62
2: Ta có: \(6x-5\left(x-7\right)=\left(27-514\right)-486-73\)
\(\Leftrightarrow6x-5x+35=27-514-486-73\)
\(\Leftrightarrow x+35=-1046\)
\(\Leftrightarrow x=-1081\)
Vậy: x=-1081
Bài 1:
a,\(0,75+\frac{9}{17}-1\frac{4}{5}-\frac{26}{17}-2\frac{4}{5}\)
\(=\frac{3}{4}+\left(\frac{9}{17}-\frac{26}{17}\right)-\left(1\frac{4}{5}+2\frac{4}{5}\right)\)
\(=\frac{3}{4}-1-\frac{23}{5}\)
\(=\frac{15}{20}-\frac{20}{20}-\frac{92}{20}=\frac{-97}{20}\)
Bài 2:
a, \(\left(2x+\frac{3}{4}\right)-\frac{10}{3}=\frac{-13}{3}\)
\(2x+\frac{3}{4}=\frac{-13}{3}+\frac{10}{3}\)
\(2x+\frac{3}{4}=-1\)
\(2x=-1-\frac{3}{4}\)
\(2x=\frac{-7}{4}\)
x = -7/8
b, 3,2x - 2,7x + 8,5 = 6
x(3,2 - 2,7) = -2,5
0,5x = -2,5
x = -5
b) \(\left(\frac{2}{3}x-1\right).\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-1=0\\\frac{3}{4}x+\frac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=1\\\frac{3}{4}x=-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1:\frac{2}{3}\\x=\left(-\frac{1}{2}\right):\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{2}{3}\right\}.\)
c) \(x:\frac{9}{14}=\frac{7}{3}:x\)
\(\Rightarrow\frac{x}{\frac{19}{4}}=\frac{\frac{7}{3}}{x}\)
\(\Rightarrow x.x=\frac{7}{3}.\frac{19}{4}\)
\(\Rightarrow x.x=\frac{133}{12}\)
\(\Rightarrow x^2=\frac{133}{12}\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\frac{133}{12}}\\x=-\sqrt{\frac{133}{12}}\end{matrix}\right.\)
Vậy \(x\in\left\{\sqrt{\frac{133}{12}};-\sqrt{\frac{133}{12}}\right\}.\)
d) \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)
\(\Rightarrow\left(3x-1\right)^{10}-\left(3x-1\right)^{20}=0\)
\(\Rightarrow\left(3x-1\right)^{10}.\left[1-\left(3x-1\right)^{10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^{10}=0\\1-\left(3x-1\right)^{10}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-1=0\\\left(3x-1\right)^{10}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x-1=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1:3\\3x-1=1\\3x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\3x=2\\3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=\frac{2}{3}\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{3};\frac{2}{3};0\right\}.\)
Chúc bạn học tốt!
A 3x-4x=-9-3
-x=-12
x=12
B 3.2x -5x +1=5+0.2x
3.2x-5x-0.2x=5-1
-2x=4
x=-2
C 1.5-x-2=-3x-0.3
-x+3x=-0.3-1.5+2
2x =0.2
x=0.1
E 2/3-1/2x-1=-x+1
-1/2x+x=1+1-2/3
1/2x=4/3
x=8/3
F 3t-4+13+2t+4-3t
=3t+2t-3t-4+13+4
=2t+13