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KL
1
31 tháng 10 2016
\(\overline{4x7}+\overline{1y5}=407+105+10x+10y=512+10\left(x+y\right)=\left[513+9\left(x+y\right)\right]+\left(x+y-1\right)\)
Ta có 513 + 9(x+y) luôn chia hết cho 9
Vậy để số đã cho chia hết cho 9 thì x+y-1 chia hết cho 9
Vì \(0\le x,y\le9\) nên \(0\le x+y-1\le17\)
Bạn chọn các giá trị trong khoảng trên , rồi từ đó suy ra giá trị của y
BL
0
N
0
ML
3 tháng 8 2017
\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(\Leftrightarrow A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(\Leftrightarrow A=1-\frac{1}{46}\)
\(\Leftrightarrow A=\frac{45}{46}\)
7 tháng 5 2019
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{51}=\frac{17}{51}-\frac{1}{51}=\frac{16}{51}\)
\(B=5\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{100}-\frac{1}{103}\right)\)
\(\Rightarrow B=5\cdot\left(1-\frac{1}{103}\right)=5\cdot\frac{102}{103}=\frac{510}{103}\)
\(C=5\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}\right)\)
\(\Rightarrow C=5\cdot\left(1-\frac{1}{101}\right)=5\cdot\frac{100}{101}=\frac{500}{101}\)
7 tháng 5 2019
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(B=\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(B=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{5}{3}\left(1-\frac{1}{103}\right)\)
\(B=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
\(C=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(C=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(C=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(C=\frac{5}{2}\left(1-\frac{1}{101}\right)\)
\(C=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
19 tháng 6 2019
\(B=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{73.76}\)
\(\Leftrightarrow B=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{73}-\frac{1}{76}\)
\(\Leftrightarrow B=\frac{1}{4}+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{10}-\frac{1}{10}\right)+...+\left(\frac{1}{73}-\frac{1}{73}\right)-\frac{1}{76}\)
\(\Leftrightarrow B=\frac{1}{4}-\frac{1}{76}=\frac{9}{38}\)
~ Hok tốt ~
19 tháng 6 2019
\(\left[200-18:\left(372:3.x-1\right)\right]-28=166\)
\(\Leftrightarrow200-18:\left(124.x-1\right)=166+28\)
\(\Leftrightarrow200-18:\left(124.x-1\right)=194\)
\(\Leftrightarrow18:\left(372:3.x-1\right)=200-194\)
\(\Leftrightarrow18:\left(124.x-1\right)=6\)
\(\Leftrightarrow124.x-1=18:6\)
\(\Leftrightarrow124.x-1=3\)
\(\Leftrightarrow124.x=3+1\)
\(\Leftrightarrow124.x=4\)
\(\Leftrightarrow x=4:124\)
\(\Leftrightarrow x=\frac{1}{31}\)
~ Hok tốt ~
VN
27 tháng 6 2017
Q=\(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{7}+\frac{1}{7}-\frac{1}{19}+...+\frac{1}{252}-\frac{1}{509}\)
=\(\frac{1}{2}-\left(\frac{1}{9}+\frac{1}{9}\right)-\left(\frac{1}{7}+\frac{1}{7}\right)-...-\left(\frac{1}{252}+\frac{1}{252}\right)-\frac{1}{509}\)
=\(\frac{1}{2}-0+0+0+...+0-\frac{1}{509}\)
=\(\frac{1}{2}-\frac{1}{509}\)
=\(\frac{507}{1018}\)
MẤY CÂU KHÁC THÌ TƯƠNG TỰ, CHÚC BẠN MAY MẮN!!!:))
29 tháng 8 2023
\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)
\(\dfrac{8}{7}A=\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\)
\(\dfrac{8}{7}A=1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\)
\(\dfrac{8}{7}A=1-\dfrac{1}{89}=\dfrac{88}{89}\Rightarrow A=\dfrac{88}{89}:\dfrac{8}{7}=\dfrac{77}{89}\)
\(B=\dfrac{5^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{37.40}\)
\(B=\dfrac{25}{1.4}+\dfrac{9}{4.7}+\dfrac{9}{7.10}+...+\dfrac{9}{37.40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{37.40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{40}=\dfrac{277}{120}\Rightarrow B=\dfrac{277}{120}:\dfrac{1}{3}=\dfrac{277}{40}\)
29 tháng 8 2023
\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)
\(=7\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\right)\)
\(=7\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\right)\)
\(=7.\left(1-\dfrac{1}{89}\right)=7.\dfrac{88}{89}=\dfrac{616}{89}\)
giúp mik vs
7^4 * 7
= 7^(4+1)
= 7^5
= 16807