A=\(\frac{[\left(\frac{2}{1000}-\frac{3}{2002}\right)\frac{1001}{17}+\frac{33}{44}]}{[\left(\frac{7}{1008}+\frac{11}{2016}\right)\frac{1008}{25}+\frac{1009}{2016}]}\)
Những câu hỏi liên quan
\(\left[\left(\frac{2}{1001}-\frac{3}{2002}\right).\frac{1001}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1008}+\frac{11}{2016}\right).\frac{1008}{25}+\frac{1009}{2016}\right]\)
giải hộ vs :((( tỉ mỉ nha !
Lm đúng mk tick cho
\(\left[\left(\frac{2}{1001}-\frac{3}{2002}\right).\frac{1001}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1008}+\frac{11}{2016}\right).\frac{1008}{25}+\frac{1009}{2016}\right]\)
\(=\left[\left(\frac{4}{2002}-\frac{3}{2002}\right).\frac{1001}{17}+\frac{33}{34}\right]:\left[\left(\frac{14}{2016}+\frac{11}{2016}\right).\frac{1008}{25}+\frac{1009}{2016}\right]\)
\(=\left(\frac{1}{2002}.\frac{1001}{17}+\frac{33}{34}\right):\left(\frac{25}{2016}.\frac{1008}{25}+\frac{1009}{2016}\right)\)
\(=\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{1009}{2016}\right)\)
\(=1:\frac{2017}{2016}\)
\(=\frac{2016}{2017}\)
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Câu này bạn có thể áp dụng phương pháp phân phối để rút gọn các phân số cho dễ nhé. Còn bạn cũng có thể áp dụng cách quy đồng cũng ko sai. Sorry bn nhé mình đang bận nên ko trình bày rõ đc ạ
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\(\left[\left(\frac{2}{1001}-\frac{3}{2002}\right).\frac{1001}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1008}+\frac{11}{2016}\right).\frac{1008}{25}+\frac{1009}{2016}\right]\)
\(=\left[\frac{1001}{17}.\frac{2}{1001}-\frac{3}{2002}.\frac{1001}{17}+\frac{33}{34}\right]:\left[\left(\frac{1008}{25}.\frac{7}{1008}+\frac{1008}{25}.\frac{11}{2016}+\frac{1009}{2016}\right)\right]\)
\(=\left[\frac{2}{17}-\frac{3}{34}+\frac{33}{34}\right]:\left[\frac{7}{25}+\frac{11}{50}+\frac{1009}{2016}\right]\)
\(=\left[\frac{2}{17}+\frac{15}{17}\right]:\left[\frac{14}{50}+\frac{11}{50}+\frac{1009}{2016}\right]\)
\(=1:\left[\frac{1}{2}+\frac{1009}{2016}\right]\)
\(=1:\left[\frac{1008}{2016}+\frac{1009}{2016}\right]\)
\(=1:\frac{2017}{2016}\)
\(=\frac{2016}{2017}\)
Học tốt nha!!!
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Xem thêm câu trả lời
\(A=\frac{[(\frac{2}{1001}-\frac{3}{2002})\frac{1001}{17}+\frac{33}{34}]}{[(\frac{7}{1008}+\frac{11}{2016})\frac{1008}{25}+\frac{1009}{2016}]}\)
cho \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{\left(a+b\right)},x^2+y^2=2\)
CMR: \(\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{1008}}=\frac{2}{\left(a+b\right)^{1008}}\)
Tính \(\left(S-P\right)^{2015}+\left(S+P\right)^{2016}\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2003}-\frac{1}{2004}+\frac{1}{2005}\)
\(P=\frac{1}{1008}+\frac{1}{1009}+.........+\frac{1}{2014}+\frac{1}{2015}\)
Giúp mình nhanh nhé!
Cho a,b,x,y là các số thực thỏa mãn: \(x^2+y^2=1\) và \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\). Chứng minh rằng:
\(\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{1008}}=\frac{2}{\left(a+b\right)^{1008}}\)
Ta có: \(\hept{\begin{cases}x^2+y^2=1\\\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\end{cases}}\)
\(\Leftrightarrow b\left(a+b\right)x^4+a\left(a+b\right)y^4=ab\left(x^4+2x^2y^2+y^4\right)\)
\(\Leftrightarrow b^2x^4+a^2y^4-2abx^2y^2=0\)
\(\Leftrightarrow\left(bx^2-ay^2\right)^2=0\)
\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\frac{x^{2016}}{a^{1008}}=\frac{y^{2016}}{b^{1008}}=\frac{1}{\left(a+b\right)^{1008}}\)
\(\Rightarrow\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{21008}}=\frac{2}{\left(a+b\right)^{1008}}\)
Em vào câu hỏi tương tự tham khảo:
Ta có: \(x^2+y^2=1\Leftrightarrow x^4+2x^2y^2+y^4=1\)
Khi đó: \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
<=> \(\left(a+b\right)\left(\frac{x^4}{a}+\frac{y^4}{b}\right)=x^4+2x^2y^2+y^4\)
<=> \(\frac{b}{a}x^4+\frac{a}{b}y^4=2x^2y^2\)
<=> \(\frac{x^4}{a^2}+\frac{y^4}{b^2}-\frac{2x^2y^2}{ab}=0\)
<=> \(\left(\frac{x^2}{a}-\frac{y^2}{b}\right)^2=0\)
<=> \(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)( dãy tỉ số bằng nhau)
Khi đó: \(\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{1008}}=2\frac{x^{2016}}{a^{1008}}=\frac{2}{\left(a+b\right)^{1008}}\)
A frac{frac{3}{4}-frac{3}{11}+frac{3}{13}}{frac{5}{4}-frac{5}{11}+frac{5}{13}}+frac{frac{1}{2}-frac{1}{3}+frac{1}{4}}{frac{5}{4}-frac{5}{6}+frac{5}{8}}B frac{2^{12}.3^5-4^6.9^2}{left(2^2.3right)^6+8^4.3^5}-frac{5^{10}.7^3-25^5.49}{left(125.7right)^3+5^9.14^3}C frac{left(a^{2016}+b^{2016}right)^{2017}}{left(c^{2016}+d^{2016}right)^{2017}} frac{left(a^{2017}-b^{2017}right)^{2016}}{left(c^{2017}-d^{2017}right)^{2016}}
Đọc tiếp
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
C = \(\frac{\left(a^{2016}+b^{2016}\right)^{2017}}{\left(c^{2016}+d^{2016}\right)^{2017}}\)= \(\frac{\left(a^{2017}-b^{2017}\right)^{2016}}{\left(c^{2017}-d^{2017}\right)^{2016}}\)
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
cs ng làm đung r
đag định lm
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+........+\frac{2}{x\cdot\left[x+1\right]}=\frac{1008}{1009}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1008}{1009}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1008}{1009}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{1008}{1009}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1008}{1009}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1008}{1009}\)
\(\Leftrightarrow1-\frac{2}{x+1}=\frac{1008}{1009}\)
\(\Leftrightarrow\frac{-2}{x-1}=\frac{1008}{1009}-1\)
\(\Leftrightarrow\frac{-2}{x+1}=\frac{-1}{1009}\)
\(\Leftrightarrow-1.\left(x+1\right)=-2.1009\)
\(\Leftrightarrow-x-1=-2018\)
\(\Leftrightarrow-x=-2018+1=-2017\)
\(\Leftrightarrow x=2017\)
Vậy x=2017
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Cho a,b,c,d\(\in\)N* ,a2+c2=1 và \(\frac{a^4}{b}+\frac{c^4}{d}=\frac{1}{b+d}\)CMR:
\(\frac{a^{2016}}{b^{1008}}+\frac{c^{2016}}{d^{1008}}=\frac{2}{\left(b+d\right)^{1008}}\)
a)Tìm số tự nhiên n mà \(\frac{5}{n-1}\)là số nguyên
b) Tính M = \((1-\frac{1000}{2016})×\left(1-\frac{1001}{2016}\right)×\left(1-\frac{1002}{2016}\right)×\dots×\left(1-\frac{2017}{2016}\right)\)
a) để 5/n-1 là số nguyên thì 5 chia hết cho n-1
=> n-1 thuộc Ư(5)=( 1, -1, 5, -5)
ta có
n-1=1=>n=2
n-1=-1=>n=0
n-1=5=>n=6
n-1=-5=>n=-4
mà n là số tự nhiên => n thuộc 2,0,6
máy mik bị lỗi bàn phím nên phải gõ ngoặc khác thay thế TvT, sorry nghen
b) M=(1-1000/2016) *...*(1-2016/2016)*(1-2017/2016)
=>M=(1-1000/2016)*.....*0*(1-2017/2016)
=>M=0