\(\frac{2}{3}x-70\frac{10}{11};\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\right)=-5\)
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29 tháng 3 2017
\(\frac{3}{2}.x-70\frac{10}{11}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}=-5\right)\)
\(\frac{3}{2}.x-70\frac{10}{11}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\right)=-5\)
\(\frac{3}{2}.x-70\frac{10}{11}:\left[13\times\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\right]=-5\)
\(\frac{3}{2}.x-70\frac{10}{11}:\left[13\times\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\right]=-5\)
\(\frac{3}{2}.x-70\frac{10}{11}:\left[13\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\right]=-5\)
\(\frac{3}{2}.x-70\frac{10}{11}:\left[13\times\left(\frac{1}{3}-\frac{1}{11}\right)\right]=-5\)
\(\frac{3}{2}.x-\frac{45}{2}=-5\)
\(\frac{3}{2}.x=\frac{35}{2}\)
\(x=\frac{35}{3}\)
26 tháng 11 2018
Ta có : \(\frac{2}{3}x-70\frac{10}{11}(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{999999})=-5\)
\(\frac{2}{3}x-70\frac{10}{11}\cdot(\frac{13}{15}+\frac{13}{35}+\frac{13}{99})=-5\)
\(\frac{2}{3}x-70\frac{10}{11}\cdot(\frac{13}{3\cdot5}+\frac{13}{5\cdot7}+\frac{13}{9\cdot11})=-5\)
\(\frac{2}{3}x-70\frac{10}{11}\cdot\left[\frac{13}{2}\cdot(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{9\cdot11})\right]=-5\)
\(\frac{2}{3}x-70\frac{10}{11}\cdot\left[\frac{13}{2}\cdot(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11})\right]=-5\)
\(\frac{2}{3}x-70\frac{10}{11}\cdot[\frac{13}{2}\cdot(\frac{1}{3}-\frac{1}{11})]=-5\)
\(\frac{2}{3}x-70\frac{10}{11}\cdot\left[\frac{13}{2}\cdot\frac{8}{33}\right]=-5\)
\(\frac{2}{3}x-\frac{780}{11}\cdot\frac{52}{33}=-5\)
\(\frac{2}{3}x-\frac{13520}{121}=-5\)
Rồi còn khúc đó bạn tự làm
21 tháng 5 2017
d)
đặt A = 1 + 2 + 22 + ... + 280
2A = 2 + 22 + 23 + ... + 281
2A - A = ( 2 + 22 + 23 + ... + 281 ) - ( 1 + 2 + 22 + ... + 280 )
A = 281 - 1 > 281 - 2
e)
đặt \(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{899}{900}\)
\(A=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{900}\right)\)
\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\right)\)
\(A=29-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\right)\)
đặt \(B=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}\)
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{29.30}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{29}-\frac{1}{30}\)
\(=1-\frac{1}{30}=\frac{29}{30}< 1\)
\(\Rightarrow A< 29\)
30 tháng 9 2020
So sánh C và D biết
C=1+13+13^2+...+13^13/1+13+13^2+...+13^12
D=1+11+11^2+...+11^13/1+11+11^2+...+11^12
NK
0
KN
9 tháng 2 2020
\(ĐKXĐ:x\ne3;x\ne5;x\ne4;x\ne6\)
\(\frac{x}{x-3}-\frac{x}{x-5}=\frac{x}{x-4}-\frac{x}{x-6}\)
\(\Rightarrow\frac{x}{x-3}-\frac{x}{x-5}-\frac{x}{x-4}+\frac{x}{x-6}=0\)
\(\Rightarrow x\left(\frac{1}{x-3}-\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(tm\right)\\\frac{1}{x-3}-\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}=0\left(1\right)\end{cases}}\)
\(\left(1\right)\Rightarrow\frac{1}{x-3}+\frac{1}{x-6}=\frac{1}{x-5}+\frac{1}{x-4}\)
\(\Rightarrow\frac{2x-9}{\left(x-3\right)\left(x-6\right)}=\frac{2x-9}{\left(x-5\right)\left(x-4\right)}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\left(tm\right)\\\left(x-3\right)\left(x-6\right)=\left(x-5\right)\left(x-4\right)\left(2\right)\end{cases}}\)
\(\left(2\right)\Leftrightarrow x^2-9x+18=x^2-9x+20\)
\(\Leftrightarrow0=2\left(L\right)\)
Vậy pt có 2 nghiệm \(\left\{0;\frac{9}{2}\right\}\)
1 tháng 4 2020
a) Đk: x \(\ne\)-2
Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^2+8}=\frac{5}{x^2-2x+4}\)
<=> \(\frac{2\left(x^2-2x+4\right)-\left(2x^2+16\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)
<=> 2x2 - 4x + 8 - 2x2 - 16 = 5x + 10
<=> -4x - 8 = 5x + 10
<=> -4x - 5x = 10 + 8
<=> -9x = 18
<=> x = -2 (ktm)
=> pt vô nghiệm
b) Đk: x \(\ne\)2; x \(\ne\)-3
Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)
<=> \(\frac{x+3}{\left(x-2\right)\left(x+3\right)}-\frac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{5}{\left(x-2\right)\left(x+3\right)}\)
<=> x + 3 - 6x + 12 = -5
<=> -5x = -5 - 15
<=> -5x = -20
<=> x = 4
vậy S = {4}
c) Đk: x \(\ne\)8; x \(\ne\)9; x \(\ne\)10; x \(\ne\)11
Ta có: \(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)
<=> \(\left(\frac{8}{x-8}+1\right)+\left(\frac{11}{x-11}+1\right)=\left(\frac{9}{x-9}+1\right)+\left(\frac{10}{x-10}+1\right)\)
<=> \(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)
<=> \(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)
<=> x = 0 (vì \(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\)
Vậy S = {0}