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Ta có: \(\frac{1-3-5-7-...-49}{89}=\frac{1-\left(3+5+7+...+49\right)}{89}=\frac{1-12.52}{89}=-\frac{623}{89}=-7\)
=> \(A=-7\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right)=-\frac{7}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{44.49}\right)\)
=>\(A=-\frac{7}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)=-\frac{7}{5}\left(\frac{1}{4}-\frac{1}{49}\right)=-\frac{7}{5}.\frac{45}{196}\)
=> \(A=-\frac{7}{5}.\frac{5.9}{28.7}=-\frac{9}{28}\)
Đáp số: A = -9/28
Ta có: 1−3−5−7−...−49 /89 =1−(3+5+7+...+49) /89 =1−12.52 /89 =−623 /89 =−7/5
=> A=−7(1/4.9 +1/9.14 +1/14.19 +...+1/44.49 )=−7/5 (5/4.9 +5/9.14 +5/14.19 +...+5/44.49 )
=> A=−75 .5.928.7 =−928
Đáp số: A = -9/28
bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó
b)=1/5.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/44-1/49).2-1-3-5-7-...-49/89
=1/5.(1/4-1/49).2-(1+3+5+7...+49)/89
=1/5.45/196.2-625/89
=9/196.-623/89
=9/196.-7
=9/28
h cho mình nha ! Chúc bạn học tốt
\(a,\frac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}=\frac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^3\cdot3^4\cdot3^6}=\frac{3^{10}\cdot2^3\cdot\left(3^2-2^3\right)}{2^3\cdot3^{10}}=3^2-2^3=1\)
\(b,\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
\(=\frac{1}{5}\left(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-\left(3+5+7+...+49\right)}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\cdot\frac{1-\left(3+49\right)\cdot24\div2}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{505}{89}\)
\(=\frac{1}{5}\cdot\frac{45}{196}\cdot\frac{505}{89}\)
#)Giải :
Đặt \(A=\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\)
\(\Rightarrow5A=\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{44.49}\)
\(\Rightarrow5A=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\)
\(\Rightarrow5A=\frac{1}{4}-\frac{1}{49}=\frac{45}{196}\)
\(\Rightarrow A=\frac{45}{196}\div5=\frac{9}{196}\)
Thay A vào B, ta được :
\(B=\frac{9}{196}.\frac{1-3-5-...-49}{89}\)
\(B=\frac{9}{196}.\frac{1-\left(3+5+7+...+49\right)}{89}\)
\(B=\frac{9}{196}.\frac{1-\left[\frac{\left(49+3\right).\left(\frac{49-3}{2}+1\right)}{2}\right]}{89}\)
\(B=\frac{9}{196}.\frac{-623}{89}=-\frac{9}{28}\)
B = \(\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right).\frac{1-3-5-...-49}{89}\)
= \(\frac{1}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{44.49}\right).\frac{1-\left(3+5+7+...+49\right)}{89}\)
= \(\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right).\frac{1-\left(24.52:2\right)}{89}\)
= \(\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{49}\right).\frac{1-624}{89}\)
= \(\frac{1}{5}.\frac{45}{196}.\left(-7\right)\)
= \(\frac{-9}{28}\)
Vậy B = \(-\frac{9}{28}\)
\(b.\)ghi lại đề nha bn
\(=\frac{2.2306}{1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{230.231}{2}}}\)
\(=\frac{2.2306}{1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{230.231}}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{230.231}\right)}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{230}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+2.\left(\frac{1}{2}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+1-\frac{2}{231}}\)
\(=\frac{2.2306}{2-\frac{2}{231}}\)
\(=\frac{2.2306}{2\left(1-\frac{1}{231}\right)}\)
\(=\frac{2306}{1-\frac{1}{231}}\)
mình nha bn thanks nhìu <3
a) \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2017}{2}+...+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2}+...+\frac{1}{2016}+\frac{1}{2017}\right)}\)
\(=\frac{1}{2017}\)
2.
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2a+2b+2c+2d}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)
\(\Rightarrow a=\frac{2b}{2}=b;b=\frac{2c}{2}=c;c=\frac{2d}{2}=d;d=\frac{2a}{2}=a\)
\(\Rightarrow a=b=c=d\)
Ta có : \(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}\)
\(=\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}\)
\(=\frac{4a}{2a}=2\)
3.
\(\left(x-1\right)\left(x-3\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-1< 0\\x-3>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-1>0\\x-3< 0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< 1\\x>3\end{cases}}\)( loại ) hoặc \(\hept{\begin{cases}x>1\\x< 3\end{cases}}\)
Vậy \(1< x< 3\)
Đặt \(A=\frac{1}{4\times9}+\frac{1}{9\times14}+\frac{1}{14\times19}+...+\frac{1}{44\times49}\)
Ta có : \(5\times A=\frac{5}{4\times9}+\frac{5}{9\times14}+\frac{5}{14\times19}+...+\frac{5}{44\times49}=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}=\frac{1}{4}-\frac{1}{49}\)
\(=\frac{49}{196}-\frac{4}{196}=\frac{45}{196}\)
\(\Rightarrow A=\frac{9}{196}\)
Đặt \(B=1-3-5-7-...-49=1-\left(3+5+...+49\right)\)
Đặt \(C=3+5+...+49\) ( khoảng cách là 2 )
Số số hạng là : \(\left(49-3\right):2+1=24\)
Tổng C là : \(\left(49+3\right)\times24:2=624\)
\(\Rightarrow B=1-264=-623\)
Vậy \(A=\frac{9}{196}\times\frac{-623}{89}=\frac{-9}{28}\)
Dòng cuối cùng mình không chắc là đúng nhé !
\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+\frac{1}{19.24}+...+\frac{1}{44.49}\right)\frac{1-3-7-...-49}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{1+\left(-3\right)+\left(-7\right)+...+\left(-49\right)}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{1-\left(3+7+...+49\right)}{89}\)
\(=\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\left(49+3\right).24:2\right)}{89}\)
\(=\frac{9}{196}.\frac{-623}{89}\)
\(=\frac{9}{196}.\left(-7\right)\)
\(=\frac{-9}{28}\)
CHÚC BN HỌC TỐT!!!!