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\(\frac{3}{5}x+\frac{2}{5}x+\frac{x}{3}=2\)
\(x+\frac{x}{3}=2\)
\(\frac{4}{3}x=2\)
\(x=2:\frac{4}{3}=\frac{3}{2}\)
Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
1) \(S=2.2.2..2\left(2023.số.2\right)\)
\(\Rightarrow S=2^{2023}=\left(2^{20}\right)^{101}.2^3=\overline{....6}.8=\overline{.....8}\)
2) \(S=3.13.23...2023\)
Từ \(3;13;23;...2023\) có \(\left[\left(2023-3\right):10+1\right]=203\left(số.hạng\right)\)
\(\) \(\Rightarrow S\) có số tận cùng là \(1.3^3=27\left(3^{203}=\left(3^{20}\right)^{10}.3^3\right)\)
\(\Rightarrow S=\overline{.....7}\)
3) \(S=4.4.4...4\left(2023.số.4\right)\)
\(\Rightarrow S=4^{2023}=\overline{.....4}\)
4) \(S=7.17.27.....2017\)
Từ \(7;17;27;...2017\) có \(\left[\left(2017-7\right):10+1\right]=202\left(số.hạng\right)\)
\(\Rightarrow S\) có tận cùng là \(1.7^2=49\left(7^{202}=7^{4.50}.7^2\right)\)
\(\Rightarrow S=\overline{.....9}\)
\(2^{x+3}-2^x=56\)
\(2^x.2^3-2^x=56\)
\(2^x\left(8-1\right)=56\)
\(2^x=2^3\)
\(x=3\)
\(2^{x+3}-2^x=56\\ \Rightarrow2^x\left(2^3-1\right)=56\\ \Rightarrow2^x=56:7=8=2^3\\ \Rightarrow x=3\\ 3^{x+2}-3^x+3=651\\ \Rightarrow3^x\left(3^2-1\right)=651-3=648\\ \Rightarrow3^x=648:8=81=3^4\\ \Rightarrow x=4\)
Bài 2:
a: =>x=0 hoặc x+3=0
=>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
1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
\(60\%x+0,4.x+\frac{x}{3}=2\)
\(\frac{3}{5}.x+\frac{2}{5}.x+\frac{1}{3}.x=2\)
\(x.\left(\frac{3}{5}+\frac{2}{5}+\frac{1}{3}\right)=2\)
\(x.\frac{4}{3}=2\)
\(x=2:\frac{4}{3}\)
\(x=\frac{3}{2}=1,5\)
60% x + 0,4x + x : 3 = 2
3/5 . x + 2/5 . x + x : 3 = 2
(3/5 + 2/5) x + x : 3 = 2
x + x : 3 = 2
x : (1 + 3) = 2
x : 4 = 2
x = 8