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\(\left(x+\dfrac{1}{3}\right)\times\dfrac{9}{14}\times\dfrac{7}{3}-\dfrac{1}{3}=1:\dfrac{9}{5}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{5}{9}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{5}{9}+\dfrac{1}{3}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{8}{9}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{8}{9}:\dfrac{3}{2}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{16}{27}\\ \Rightarrow x=\dfrac{16}{27}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{7}{27}\)
1)
1\(\frac{1}{2}\): 2\(\frac{1}{4}\)x 1\(\frac{3}{4}\)= \(\frac{3}{2}\): \(\frac{9}{4}\)x \(\frac{7}{4}\)= \(\frac{3x4x4}{2x9x7}\)= \(\frac{8}{21}\)
2)
Ta có : 1,23 < ... < ... < ... <1,24
= 1,230< ... < ... < ... <1,240
=> 3 số cần điền là : 1,231 ; 1,232 ; 1,234
\(\left(x+2\right)+\left(x+4\right)+...+\left(x+20\right)=160\)
\(x+2+x+4+...+x+20=160\)
\(\left(x+x+x+...+x\right)+\left(2+4+...+20\right)=160\)
\(x\cdot20+\frac{\left(20+2\right)\cdot\left(\left(20-2\right):2+1\right)}{2}=160\)
\(x\cdot20+110=160\)
\(x\cdot20=160-110\)
\(x\cdot20=50\)
\(x=50:20\)
\(x=2,5\)
Vậy \(x=2,5\)
[ x + 2 ] + [ x + 4 ] + [ x + 6 ] + ... + [ x + 20 ] = 160
( 20 + 2 ) x 10 : 2 + 10x = 160
22 x 10 : 2 + 10x = 160
220 : 2 + 10x = 160
110 + 10x = 160
10x = 160 - 110
10x = 50 ; x = 5
[ 1/2 + 1/4 + 1/8 + ... + 1.64 ] /x = 3/8
=> [ ( 1/2 + 1/4 ) + ( 1/8 + 1/16 ) + ( 1/32 + 1/64 ) ] / x = 3/8
[ 3/4 + 3/16 + 3/64 ] / x = 3/8
[ 48/64 + 12/64 + 3/64 ] / x = 3/8
=> 63/64x = 3/8
=> 64x = 21 x 8
64x = 168
x = 2,625
k chắc nha
\(x:0,1+x\times5=24\)
\(\Rightarrow x:\frac{1}{10}+x\times5=24\)
\(\Rightarrow x\times10+x\times5=24\)
\(\Rightarrow x\times\left(10+5\right)=24\)
\(\Rightarrow x\times15=24\)
\(\Rightarrow x=\frac{24}{15}=\frac{8}{5}\)
\(X:0,1+Xx5=24\)
\(\Rightarrow Xx10+Xx15=24\)
\(\Rightarrow Xx\left(10+15\right)=24\)
\(\Rightarrow Xx25=24\)
\(\Rightarrow X=\frac{24}{25}\)
Vậy \(X=\frac{24}{25}\)
Chúc bạn học tốt !!!
c,Dãy trên có số số hạng là:
(52-2):2+1=26 số hạng
Tổng của dãy trên là:
(52+2).26:2=702
Đáp số :.....
\(\frac{1}{2\cdot x}-2021-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{222}=\frac{6}{11}\)
\(\frac{1}{2\cdot x}-2021-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\right)=\frac{6}{11}\)
....
Cái dãy \(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\) nó không có quy luật, không tính được
Sửa đề\(\frac{1}{2x-2021}-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{220}=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{220}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{10}-\frac{1}{11}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(1-\frac{1}{11}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}.\frac{10}{11}=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{5}{11}=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}=1\)
=> 2x - 2021 = 1
=> 2x = 2022
=> x = 1011
Vậy x = 1011
\(\Leftrightarrow\left(x+1\right)\left[\left(x-1\right):3+1\right]:2=330\)
\(\Leftrightarrow\left(x+1\right)\left[\left(x-1\right):3+1\right]=660\)
\(\Leftrightarrow\left(x+1\right)\left[\left(x-1\right):3+1\right]=44.15\)
\(\Leftrightarrow x+1=44\Leftrightarrow x=43\)
Dấu \(.\)là dấu nhân
\(y+y.\frac{1}{4}:\frac{2}{7}+y:\frac{2}{9}=255\)
\(\Rightarrow y+y.\frac{1}{4}.\frac{7}{2}+y.\frac{9}{2}=255\)
\(\Rightarrow y+y.\frac{7}{8}+y.\frac{9}{2}=255\)
\(\Rightarrow y.\left(1+\frac{7}{8}+\frac{9}{2}\right)=255\)
\(\Rightarrow y.\left(\frac{8}{8}+\frac{7}{8}+\frac{36}{8}\right)=255\)
\(\Rightarrow y.\frac{51}{8}=255\)
\(\Rightarrow y=255:\frac{51}{8}\)
\(\Rightarrow y=255.\frac{8}{51}\)
\(\Rightarrow y=40\)
Vậy \(y=40\)
~ Ủng hộ nhé
A = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+..+9\right)}{1\times2+2\times3+3\times4+...+19\times20}\)
\(=\frac{\frac{1\times\left(1+1\right)}{2}+\frac{2\times\left(2+1\right)}{2}+\frac{3\times\left(3+1\right)}{2}...+\frac{9\times\left(9+1\right)}{2}}{1\times2+2\times3+3\times4+...+19\times20}\)
\(=\frac{\frac{1\times2}{2}+\frac{2\times3}{2}+\frac{3\times4}{2}+...+\frac{9\times10}{2}}{1\times2+2\times3+3\times4+...+9\times10}\)
\(=\frac{\frac{1}{2}\times\left(1\times2+2\times3+3\times4+...+9\times10\right)}{1\times2+2\times3+3\times4+...+9\times10}=\frac{\frac{1}{2}}{1}=\frac{1}{2}\)