1 và 1/9 x 1 và 1/10 x 1 và 1/11 x...x 1 và 1/2011 giúp mk vs mk đang cần gấp
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5 tháng 7 2021
Ta có ; \(1\dfrac{1}{10}.1\dfrac{1}{11}.....1\dfrac{1}{2011}.1\dfrac{1}{2012}\)
\(=\dfrac{11}{10}.\dfrac{12}{11}.....\dfrac{2012}{2011}.\dfrac{2013}{2012}\)
\(=\dfrac{2013}{10}=201,3\)
VV
5 tháng 7 2021
1 \(\dfrac{1}{10}\)x 1\(\dfrac{1}{11}\)x 1\(\dfrac{1}{12}\) x...x 1\(\dfrac{1}{2011}\)x 1\(\dfrac{1}{2012}\)
=\(\dfrac{11}{10}\)x \(\dfrac{12}{11}\)x \(\dfrac{13}{12}\)x \(\dfrac{2012}{2011}\)x \(\dfrac{2013}{2012}\)
VT
2 tháng 3 2023
\(\left(6:3,5-1\dfrac{1}{6}\times\dfrac{6}{7}\right):\left(4,2\times\dfrac{10}{11}+5\dfrac{2}{11}\right)\\ =\left(\dfrac{12}{7}-\dfrac{7}{6}\times\dfrac{6}{7}\right):\left(\dfrac{21}{5}\times\dfrac{10}{11}+\dfrac{57}{11}\right)\\ =\left(\dfrac{17}{7}-1\right):\left(\dfrac{42}{11}+\dfrac{57}{11}\right)\\ =\dfrac{10}{7}:9\\ =\dfrac{10}{63}\)
NK
2 tháng 3 2023
\(\left(6:\dfrac{3}{5}-1\dfrac{1}{6}\times\dfrac{6}{7}\right):\left(4,2\times\dfrac{10}{11}+5\dfrac{2}{11}\right)\)
\(=\left(6\times\dfrac{5}{3}-\dfrac{7}{6}\times\dfrac{6}{7}\right):\left(\dfrac{21}{5}\times\dfrac{10}{11}+\dfrac{57}{11}\right)\)
\(=\left(10-1\right):\left(\dfrac{42}{11}+\dfrac{57}{11}\right)=9:9=1\)
VT
2
TT
20 tháng 12 2020
Rồi sao? đề bài?
\(4(x+1)^2-(2x-1)^2-8(x-1)(x+1)=11\)
\(\Leftrightarrow4\left(x^2+2x+1\right)-\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)
\(\Leftrightarrow4x^2+8x+4-4x^2+4x-1-8x^2+8=11\)
\(\Leftrightarrow-8x^2+12x+11=11\)
\(\Leftrightarrow-4x\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
20 tháng 12 2020
Ta có:
\(4\left(x+1\right)^2-\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\\ \Leftrightarrow4x^2+8x+4-4x^2+4x-1-8x^2+8=11\\ \Leftrightarrow-8x^2+12x=0\\ \Leftrightarrow-4x\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
A
16 tháng 10 2020
a, Ta có :
\(N=x^2\left(y-1\right)-5x\left(1-y\right)=x^2\left(y-1\right)+5x\left(y-1\right)=x\left(x+5\right)\left(y-1\right)\)
Thay x = -20 ; y = 1001 ta được :
\(-20\left(-20+5\right)\left(1001-1\right)=-20.\left(-15\right).1000=300000\)
b, Ta có : \(x\left(x-y\right)^2-y\left(x-y\right)^2+xy^2-x^2y=\left(x-y\right)^3+xy\left(x-y\right)\)
\(=\left(x-y\right)^4\left(1+xy\right)\)
Thay x - y = 7 ; xy = 9 ta được :
\(7^4.\left(1+9\right)=2401.10=24010\)
16 tháng 10 2020
N = x2( y - 1 ) - 5x( 1 - y )
= x2( y - 1 ) + 5x( y - 1 )
= x( y - 1 )( x + 5 )
Tại x = -20 ; y = 1001 ta được :
N = -20( 1001 - 1 )( -20 + 5 )
= -20.1000.(-15)
= 1000.300
= 300 000
Q = x( x - y )2 - y( x - y )2 + xy2 - x2y
= x( x - y )2 - y( x - y )2 - xy( x - y )
= ( x - y )[ x( x - y ) - y( x - y ) - xy ]
= ( x - y )( x2 - xy - xy + y2 - xy )
= ( x - y )( x2 - 3xy + y2 )
= ( x - y )[ ( x2 - 2xy + y2 ) + 2xy - 3xy ]
= ( x - y )[ ( x - y )2 - xy ]
= 7[ 72 - 9 ]
= 7( 49 - 9 )
= 7.40 = 280
17 tháng 8 2021
\(a)\)
\(\frac{1}{x+1}-\frac{x-1}{x}=\frac{3x+1}{x\left(x+1\right)}\)
\(\Leftrightarrow x-x^2+1=3x+1\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b)\)
\(\frac{\left(x+2\right)^2}{2x-3}-\frac{1}{1}=\frac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x-3=x^2+10\)
\(\Leftrightarrow x^2+2x+1=x^2+10\)
\(\Leftrightarrow2x-9=0\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\frac{2}{9}\)
A
2
17 tháng 4 2019
\(x-\left[\frac{50x}{100}+\frac{25x}{200}\right]=11\frac{1}{4}\)
\(x-\left[\frac{50x\cdot2}{200}+\frac{25x}{200}\right]=11\frac{1}{4}\)
\(x-\left[\frac{100x}{200}+\frac{25x}{200}\right]=11\frac{1}{4}\)
\(x-\frac{125x}{200}=11\frac{1}{4}\)
\(\frac{200x}{200}-\frac{125x}{200}=11\frac{1}{4}\)
\(\frac{200x-125x}{200}=11\frac{1}{4}\)
\(\frac{75x}{200}=11\frac{1}{4}\)
\(\frac{3x}{8}=11\frac{1}{4}\)
\(3x=11\frac{1}{4}\cdot8\)
\(3x=90\)
\(x=30\)
Vậy x = 30
23 tháng 9 2021
\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
23 tháng 9 2021
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
31 tháng 3 2018
\(a,\frac{62}{7}:x=\frac{29}{9}:\frac{3}{56}\)
\(\frac{62}{7}:x=\frac{1624}{27}\)
\(x=\frac{62}{7}:\frac{1624}{27}=\frac{837}{5684}\)
\(b,\frac{1}{5}:x=\frac{1}{5}-\frac{1}{7}\)
\(\frac{1}{5}:x=\frac{2}{35}\)
\(x=\frac{1}{5}:\frac{2}{35}=\frac{7}{2}\)
\(c,\frac{2}{3}.x-\frac{4}{7}=\frac{1}{7}\)
\(\frac{2}{3}.x=\frac{1}{7}+\frac{4}{7}=\frac{5}{7}\)
\(x=\frac{5}{7}:\frac{2}{3}=\frac{15}{14}\)
\(d,\frac{2}{7}-\frac{8}{9}.x=\frac{2}{3}\)
\(\frac{8}{9}.x=\frac{2}{7}-\frac{2}{3}=-\frac{8}{21}\)
\(x=-\frac{8}{21}:\frac{8}{9}=-\frac{3}{7}\)
\(e,\frac{4}{7}+\frac{5}{9}:x=\frac{1}{5}\)
\(\frac{5}{9}:x=\frac{1}{5}-\frac{4}{7}=-\frac{13}{35}\)
\(x=\frac{5}{9}:-\frac{13}{35}=\frac{175}{117}\)
\(i,\frac{2}{5}-\frac{2}{5}.x=\frac{2}{5}\)
\(\frac{2}{5}.\left(1-x\right)=\frac{2}{5}\)
\(1-x=\frac{2}{5}:\frac{2}{5}=1\)
\(x=1-1=0\)
\(g,\frac{2}{3}+\frac{1}{3}:x=-1\)
\(\frac{1}{3}:x=-1-\frac{2}{3}=-\frac{5}{3}\)
\(x=\frac{1}{3}:-\frac{5}{3}=-\frac{1}{5}\)
học tốt nha
1\(\dfrac{1}{9}\) x 1\(\dfrac{1}{10}\) x 1\(\dfrac{1}{11}\) x .....x 1\(\dfrac{1}{2011}\)
= \(\dfrac{10}{9}\)x \(\dfrac{11}{10}\) x \(\dfrac{12}{11}\) x \(\dfrac{13}{12}\) x .....x\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{9}\)