M = \(\frac{9^2}{90}\)x \(\frac{10^2}{110}\) x \(\frac{11^2}{132}\) x ... x \(\frac{100^2}{10100}\)
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5 tháng 4 2017
ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)
ta gọi B là biểu thức thứ2
\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)
\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)
\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)
\(\Rightarrow x=1\)
mk nghĩ vậy bạn ạ, mk mong nó đúng
5 tháng 6 2015
Vê trái:
\(=\frac{2}{\left(x-1\right)\left(x+1\right)}+\frac{4}{\left(x-2\right)\left(x+2\right)}+...+\frac{20}{\left(x-10\right)\left(x+10\right)}\)
\(=\frac{\left(x+1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-10\right)}{\left(x+10\right)\left(x-10\right)}\)
\(=\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+2}+...+\frac{1}{x-10}-\frac{1}{x+10}\)
\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\)
Vế phải:
\(=\frac{\left(x+1\right)-\left(x-10\right)}{\left(x-10\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-9\right)}{\left(x-9\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-1\right)}{\left(x-1\right)\left(x+10\right)}\)
\(=\frac{1}{x-10}-\frac{1}{x+1}+\frac{1}{x-9}-\frac{1}{x+2}+...+\frac{1}{x-1}-\frac{1}{x+10}\)
\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\) = vế phải
=> đpcm
31 tháng 5 2017
c) x=-2 nha
d) =\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+......+\(\frac{1}{11.12}\)
=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+.....+\(\frac{1}{11}\)-\(\frac{1}{12}\)
=\(\frac{1}{5}\)-\(\frac{1}{12}\)= \(\frac{7}{60}\)
18 tháng 4 2018
mình biến đởi phần trong |......| rồi bạn thay vào nha
1/30 + 1/42 + 1/56 + 1/72 +1/ 90 + 1/110 + 1/132
=1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 +1/ 10.11
=1/5 -1/6 +1/6 - 1/7 +......+1/10 - 1/11
=1/5 - 1/11=11/55 - 5/55 =6/ 55
thay vào |....|=> |6/55 - x | = 2/3 => mở ra 2 trường hợp mà tính nha
chúc hok tốt
18 tháng 4 2018
=>(1/5.6+1/6.7+1/7.8+1/9.10+1/10.11+1/11.12)-x=2/3
=>(1/5-1/+1/6-1/7+...+1/11-1/12)-x=2/3
=>(1/5-1/12)-x=2/3
=>7/60-x=2/3
=>x=7/60-2/3
=>x=-11/20
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}
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 \(\left\{0;\frac{9}{2}\right\}\)
\(M=\frac{9^2}{9.10}.\frac{10^2}{10.11}.\frac{11^2}{11.12}.....\frac{100^2}{100.101}\)
\(=\frac{9}{10}.\frac{10}{11}.\frac{11}{12}.....\frac{100}{101}=\frac{9}{101}\)