1) Chứn tỏ rằng:
a) 0,(32)+0,(67)=1
b) 0,(33)x3=1
@
2) Tìm x:
0,(12):1,(6)=x:0,(3)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
9 tháng 10 2022
Câu 2:
a: 0,(32)+0,(67)
=32/99+67/99
=1
b: \(0.\left(33\right)\cdot3=\dfrac{1}{3}\cdot3=1\)
A
2
13 tháng 2 2022
e, \(\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2=0\Leftrightarrow\left(x-2\right)\left(x+2+x-2\right)=0\Leftrightarrow x=0;x=2\)
f, \(\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x-1\right)^2=0\Leftrightarrow x=1;x=-1\)
g, \(x^2\left(x-3\right)+4\left(3-x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\Leftrightarrow x=2;x=-2;x=3\)
h, \(\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\Leftrightarrow x=4;x=-\dfrac{2}{3}\)
25 tháng 10 2021
\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)
25 tháng 10 2021
a) \(\Rightarrow4\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
b) \(\Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Rightarrow x=-1\left(do.x^2+1\ge1>0\right)\)
c) \(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
d) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(\Rightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-1\end{matrix}\right.\)
1)
a)\(0,\left(32\right)+0,\left(67\right)\)
\(=0,\left(01\right).32+0,\left(01\right).67\)
\(=0,\left(01\right).\left(32+67\right)\)
\(=\frac{1}{99}.99\)
\(=1\left(đpcm\right)\)
b)\(0,\left(33\right).3\)
\(=0,\left(01\right).33.3\)
\(=\frac{1}{99}.33.3\)
\(=\frac{33}{99}.3\)
\(=\frac{99}{99}\)
\(=1\left(đpcm\right)\)
2)\(0,\left(12\right):1,\left(6\right)=x:0,\left(3\right)\)
\(\left[\left(0,01\right).12\right]:\left[1+0,\left(6\right)\right]=x:\left[0,\left(1\right).3\right]\)
\(\left(\frac{1}{99}.12\right):\left[1+0,\left(1\right).6\right]=x:\left(\frac{1}{9}.3\right)\)
\(\frac{4}{33}:\left[1+\frac{1}{9}.6\right]=x:\frac{1}{3}\)
\(\frac{4}{33}:\left[1+\frac{2}{3}\right]=x.3\)
\(3x=\frac{4}{33}:\frac{5}{3}\)
\(3x=\frac{4}{33}\cdot\frac{3}{5}\)
\(3x=\frac{4}{55}\)
\(x=\frac{4}{55}:3\)
\(x=\frac{4}{55}\cdot\frac{1}{3}\)
\(x=\frac{4}{165}\)