Tính:
\(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
\(A=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
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
\(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+\frac{1}{19\cdot24}+.....+\frac{1}{44\cdot49}\right)1\frac{-3-5-7-...-49}{89}\)
\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+\frac{1}{19.24}+...+\frac{1}{44.49}\right)\frac{1-3-7-...-49}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{1+\left(-3\right)+\left(-7\right)+...+\left(-49\right)}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{1-\left(3+7+...+49\right)}{89}\)
\(=\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\left(49+3\right).24:2\right)}{89}\)
\(=\frac{9}{196}.\frac{-623}{89}\)
\(=\frac{9}{196}.\left(-7\right)\)
\(=\frac{-9}{28}\)
CHÚC BN HỌC TỐT!!!!
Thực hiện phép tính
a)\(\frac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}\)
b)\(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\)\(\cdot\frac{1-3-5-7-...-49}{89}\)
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
\(a,\frac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}=\frac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^3\cdot3^4\cdot3^6}=\frac{3^{10}\cdot2^3\cdot\left(3^2-2^3\right)}{2^3\cdot3^{10}}=3^2-2^3=1\)
\(b,\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
\(=\frac{1}{5}\left(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-\left(3+5+7+...+49\right)}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\cdot\frac{1-\left(3+49\right)\cdot24\div2}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{505}{89}\)
\(=\frac{1}{5}\cdot\frac{45}{196}\cdot\frac{505}{89}\)
Tính \(A=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
\(B=\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó
Giải giúp nhá mn!!!!!!
tìm kết quả của B:
B = \(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+....+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-...-49}{89}\)
tick cho giải nhanh!!!!!!!!!!!!
# love is the beautiful pain #
#)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 = \(-\frac{9}{28}\)
=>5 B = (1/4-1/49) X ( 1-(3+5+...+49))/89
=>5B= 45/196 X (-623)/89
=>5B=45/196 X (-7)
=> 5B = -45/28
=>B=-45/25:5
=>B=-9/28
vẬY B=-9/28
Tính:
a/\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)
b/\(\frac{2\cdot2306}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+230}}\)c/\(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+...+\frac{1}{44\cdot49}\right)\left(\frac{1-3-5-...-49}{89}\right)\)
\(b.\)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) \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2017}{2}+...+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2}+...+\frac{1}{2016}+\frac{1}{2017}\right)}\)
\(=\frac{1}{2017}\)
1 Tìm x,y
1) \(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
2 tìm giá trị nhỏ nhất hoac lớn nhat cua các biểu thức sau
A=\(\frac{x^2+5}{x^2+3}\)
3 chứng minh rằng: Với mọi số nguyên dương n thì :
\(3^{n+2}-2^{n+2}+3^n-2^n\)chia hết cho 10
4 tính
\(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+....+\frac{1}{44\cdot49}\right)\frac{1-3-5-7-...-49}{89}\)
A=\(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5 Tìm x biết
a) \(_{\left(x-7\right)^{x+1}-\left(x-7\right)^{x-11}=0}\)
b) \(\frac{1}{8}\cdot16^x=2^x\)
Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203