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\(\left|x-\frac{1}{3}\right|+\left|x-y\right|=0\)
\(\Leftrightarrow\begin{cases}x-\frac{1}{3}=0\\x-y=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=\frac{1}{3}\\x=y\end{cases}\)\(\Leftrightarrow x=y=\frac{1}{3}\)
Ta có :
\(B=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot2}+\frac{1}{2\cdot15}+\frac{13}{15\cdot4}\)
\(=>\frac{B}{7}=\frac{5}{2\cdot7}+\frac{4}{7\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
\(=>\frac{B}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(=>\frac{B}{7}=\frac{1}{2}-\frac{1}{28}=\frac{14}{28}-\frac{1}{28}=\frac{13}{28}\)
\(=>B=\frac{13}{28}\cdot7=\frac{13}{4}\)
!Cám ơn các bạn nhiều...!
!
Theo đề bài ta có :
\(A=\frac{n+1}{n-1}=\frac{1}{2}\)
\(\Leftrightarrow2\left(n+1\right)=n-1\)
\(\Leftrightarrow2n+2=n-1\)
\(\Leftrightarrow2n-n=-1-2\)
\(\Rightarrow n=-3\)
Vậy với n = - 3 thì A = \(\frac{1}{2}\)
gọi ƯCLN của tử và mẫu =d
ta có;
2n+1chia hết cho d suy ra 6n+3chia hết cho d
3n +2 chia hết cho d suy ra 6n+4 chia hết cho d
suy ra {6n+4}-{6n+3} chia hết cho d
hay 1 chia hết cho d
suy ra d =1
vậy phân số trên là phân số tối giản
Giải:
Đặt ƯCLN (2n + 1; 3n +2) = d (d \(\in\) N*).
\(\Rightarrow\left\{{}\begin{matrix}2n+1⋮d\\3n+2⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6n+3⋮d\\6n+4⋮d\end{matrix}\right..}}\)\(\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d.\)
hay \(1⋮d\Rightarrow d=1.\)
Vậy \(\dfrac{2n+1}{3n+2}\) là phân số tối giản \(\forall\) n \(\in\) N*.
~ Học tốt!!! ~
\(\frac{1}{7}\)B=\(\frac{5}{2.7.1}+\frac{4}{1.7.11}+\frac{3}{11.2.7}+\frac{1}{2.7.15}+\frac{13}{15.4.7}\)
\(\frac{1}{7}\)B=\(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{28}\)
\(\frac{1}{7}B=\frac{13}{28}\)
B=\(\frac{13}{28}:\frac{1}{7}\)
B=\(\frac{13}{4}\)
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
Ta có : \(\dfrac{1}{2}xOy=\dfrac{1}{7}yOz\Rightarrow xOy=\dfrac{1}{7}yOz:\dfrac{1}{2}=\dfrac{2}{7}yOz\)
Ta lại có : góc xOy + góc yOz = 180 độ( hai góc kề bù )
\(\Rightarrow\) \(\dfrac{2}{7}yOz\) + góc yOz = 180 độ
\(\Rightarrow\)yOz(\(\dfrac{2}{7}+1\)) = 180 độ
\(\Rightarrow\)\(\dfrac{9}{7}yOz\)= 180 độ
\(\Rightarrow\)yOz = 180 : \(\dfrac{9}{7}\)=180 .\(\dfrac{7}{9}\)= 140 độ
Khi đó : xOy = 140 . \(\dfrac{2}{7}\)= 40 độ
\(\frac{3x-11}{2}-\frac{x-3}{3}=\frac{1}{6}\)
\(\frac{3\times\left(3x-11\right)}{3\times2}-\frac{2\times\left(x-3\right)}{2\times3}=\frac{1}{6}\)
\(\frac{9x-33}{6}-\frac{2x-6}{6}=\frac{1}{6}\)
\(\frac{\left(9x-33\right)-\left(2x-6\right)}{6}=\frac{1}{6}\)
\(9x-33-2x+6=1\)
\(\left(9x-2x\right)-\left(33-6\right)=1\)
\(7x-27=1\)
\(7x=1+27\)
\(7x=28\)
\(x=\frac{28}{7}\)
\(x=4\)
Chúc bạn học tốt
\(PT\Leftrightarrow\frac{3.\left(3x-11\right)-2.\left(x-3\right)}{6}=\frac{1}{6}\)
<=> 3.(3x - 11) - 2.(x - 3) = 1
<=> 9x - 33 - 2x + 6 = 1
<=> 7x = 28
<=> x = 4
\(\frac{2x+1}{3}=\frac{5}{2}\)
\(2x+1=\frac{5.3}{2}=\frac{15}{2}\)
2x= 15/2 - 1 = 13/2
x = 13/2 : 2
x = 13/4
b) 2x + 2x+1 + 2x+2 + 2x+3 = 480
2x.(1+ 2 +22 + 23) = 480
2x . 15 = 480
2x = 480 : 15 = 32
2x = 25 => x = 5
c) \(\left(\frac{3x}{7}+1\right):\left(-4\right)=-\frac{1}{28}\)
\(\frac{3x}{7}+1=\frac{-1}{28}.\left(-4\right)=\frac{1}{7}\)
\(\frac{3x}{7}=\frac{1}{7}-1=-\frac{6}{7}\)
< = > 3x= -6 => x = -2
=>B=\(\frac{12.\left(\frac{1}{13}+\frac{1}{1313}+\frac{1}{131}+\frac{1}{1313}\right)}{15.\left(\frac{1}{13}+\frac{1}{1313}+\frac{1}{131}+\frac{1}{1313}\right)}\)
=>B=\(\frac{12}{15}\)
=>B=\(\frac{4}{5}\)
B = \(\frac{12.\left(\frac{1}{13}+\frac{1}{1313}+\frac{1}{131}-\frac{1}{1313}\right)}{15.\left(\frac{1}{13}+\frac{1}{131}-\frac{1}{1313}+\frac{1}{1313}\right)}\)
=\(\frac{12.\left(\frac{1}{13}+\frac{1}{131}\right)}{15.\left(\frac{1}{13}+\frac{1}{131}\right)}\)
=\(\frac{12}{15}=\frac{4}{5}\)
Vậy B = 4/5.
Chào bạn, bạn hãy theo dõi bài giải của mình nhé!
\(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{946}+\frac{1}{990}\)
\(=\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{1892}+\frac{2}{1980}\)
\(=\frac{2}{5\cdot6}+\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+...+\frac{2}{43\cdot44}+\frac{2}{44\cdot45}\)
\(=2\left(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{43\cdot44}+\frac{1}{44\cdot45}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{43}-\frac{1}{44}+\frac{1}{44}-\frac{1}{45}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{45}\right)=2\left(\frac{9}{45}-\frac{1}{45}\right)=2\cdot\frac{8}{45}=\frac{16}{45}\)
Chúc bạn học tốt!