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1.
a)\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
b)\(\left(x-2\right)^2=1\Leftrightarrow\orbr{\begin{cases}x-2=\sqrt{1}\\x-2=-\sqrt{1}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{1}+2\\x=-\sqrt{1}+2\end{cases}}\)
Mấy câu kia tương tự,bạn tự làm nha :))
a. \(\frac{1}{2}\) - ( \(\frac{1}{3}\) + \(\frac{1}{4}\) ) < x < \(\frac{1}{48}\) - ( \(\frac{1}{16}\) - \(\frac{1}{6}\) )
\(\frac{1}{2}\) - \(\frac{7}{12}\) < x < \(\frac{1}{48}\) - \(\frac{-5}{48}\)
\(\frac{-1}{12}\) < x < \(\frac{1}{8}\)
Đề bài yêu cầu tìm x thuộc tập hợp gì bạn ơi. Bạn viết thiếu rồi .
M= 1 + 1/2.3 + 1/3.6 + 1/4.10 +.....+ 1/16.136
M=1 + 3/2 + 2 + 5/2 +...+ 17/2
M= 2/2 + 3/2 + 4/2 + 5/2 +...+ 17/2
M=\(\frac{2+3+4+5+...+17}{2}\)
M= 152/2=76
cho thêm điều kiện x,y thuộc Z nữa nhá
\(\frac{3}{x}+\frac{1}{3}=\frac{y}{3}\)
\(\frac{3}{x}=\frac{y-1}{3}\)
\(\Rightarrow x.\left(y-1\right)=9\)
Lập bảng ta có :
| x | 1 | 9 | -1 | -9 | 3 | -3 |
| y-1 | 9 | 1 | -9 | -1 | 3 | -3 |
| y | 10 | 2 | -8 | 0 | 4 | -2 |
Vậy ( x ; y ) = { ( 1 ; 10 ) ; ( 9 ; 2 ) ; ( -1 ; -8 ) ; ( -9 ; 0 ) ; ( 3 ; 4 ) ; ( -3 ; -2 ) }
mấy bài còn lại làm tương tự
\(c)\)
\(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=\left(7-\frac{1}{50}+x\right)\)
\(\Rightarrow2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\right)=\left(\frac{350}{50}-\frac{1}{50}+x\right)\)
\(\Rightarrow2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Rightarrow2x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)-x=\frac{349}{50}\)
\(\Rightarrow x-\left(1-\frac{1}{50}\right)=\frac{349}{50}\)
\(\Rightarrow x-\frac{49}{50}=\frac{349}{50}\)
\(\Rightarrow x=\frac{349}{50}+\frac{49}{50}\)
\(\Rightarrow x=\frac{199}{25}\)
Vậy \(x=\frac{199}{25}\)
~ Ủng hộ nhé
\(a)2.x-3=x+\frac{1}{2}\)
\(\Rightarrow2x-3-x=\frac{1}{2}\)
\(\Rightarrow x-3=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}+3\)
\(\Rightarrow x=\frac{1}{2}+\frac{6}{2}\)
\(\Rightarrow x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}\)
\(b)4.x-\left(2.x+1\right)=3-\frac{1}{3}+x\)
\(\Rightarrow4.x-2.x-1=\frac{9}{3}-\frac{1}{3}+x\)
\(\Rightarrow2.x-1=\frac{8}{3}+x\)
\(\Rightarrow2x-1-x=\frac{8}{3}\)
\(\Rightarrow x-1=\frac{8}{3}\)
\(\Rightarrow x=\frac{8}{3}+1\)
\(\Rightarrow x=\frac{8}{3}+\frac{3}{3}\)
\(\Rightarrow x=\frac{11}{3}\)
Vậy \(x=\frac{11}{3}\)
~ Ủng hộ nhé
Bài giải
a, \(3\frac{1}{3}\text{ : }2\frac{1}{2}-1< x< 7\frac{2}{3}\cdot\frac{3}{7}+\frac{5}{2}\)
\(\frac{10}{3}\text{ : }\frac{5}{2}-1< x< \frac{23}{3}\cdot\frac{3}{7}+\frac{5}{2}\)
\(\frac{4}{3}-1< x< \frac{23}{7}+\frac{5}{2}\)
\(\frac{1}{3}< x< \frac{81}{14}\)
\(\Rightarrow\text{ }0,\left(3\right)< x< 5,78...\)
\(\Rightarrow\text{ }x\in\left\{1\text{ ; }2\text{ ; }3\text{ ; }4\text{ ; }5\right\}\)
b, \(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)
\(\frac{1}{2}-\frac{7}{12}< x< \frac{1}{48}+\frac{5}{48}\)
\(-\frac{1}{12}< x< \frac{1}{8}\)
\(\Rightarrow\text{ }-0,08\left(3\right)< x< 0,125\)
\(\Rightarrow\text{ }x\in\varnothing\)
Ta có:
\(=\frac{6}{12}+\frac{6}{12}+\frac{20}{12}-\frac{9}{12}\)
\(=\frac{29}{12}\)
Hok tốt