biết 1^2+2^2+3^2+...+8^2+9^2=285
tính hợp lí giá trị biểu thức: 2^2+4^2+6^2+...+16^2+18^2
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a) \(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)
\(=\frac{2^{15}.5^8+\left(-2\right)^5.10^9}{2^{16}.5^7+2.10^8}\)
\(=\frac{5-2^4.10}{2}\)
\(=5-8.10\)
\(=5-80\)
\(=-75\)
\(A=\left(\frac{21}{5}-\frac{3}{5}\right):\frac{16}{8}:\frac{9}{10}\)
\(A=\frac{18}{5}:\frac{16}{8}:\frac{9}{10}\)
\(A=\frac{18}{5}\times\frac{8}{16}\times\frac{10}{9}\)
\(A=\frac{9}{5}\times\frac{8}{8}\times\frac{10}{9}\)
\(A=\frac{72}{40}\times\frac{10}{9}\)
\(A=\frac{8}{4}\times\frac{1}{1}\)
\(A=\frac{8}{9}\)
\(B=\left(1\frac{1}{20}+\frac{36}{100}\right):\frac{84}{25}:\left(2+\frac{14}{16}\right)\)
\(B=\left(\frac{21}{20}+\frac{36}{100}\right):\frac{84}{25}:\left(\frac{32}{16}+\frac{14}{16}\right)\)
\(B=\left(\frac{105}{100}+\frac{36}{100}\right):\frac{84}{25}:\frac{46}{16}\)
\(B=\frac{141}{100}:\frac{84}{25}:\frac{46}{16}\)
\(B=\frac{141}{100}\times\frac{25}{84}\times\frac{16}{46}\)
\(B=\frac{141}{4}\times\frac{1}{84}\times\frac{16}{46}\)
\(B=\frac{141}{336}\times\frac{16}{46}\)
\(B=\frac{141}{21}\times\frac{1}{46}\)
\(B=\frac{141}{966}\)
\(A=\dfrac{5}{11}.\dfrac{5}{7}+\dfrac{5}{11}.\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}.1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\)
\(B=\dfrac{3}{13}.\dfrac{6}{11}+\dfrac{3}{13}.\dfrac{9}{11}-\dfrac{3}{13}.\dfrac{4}{11}=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}.1=\dfrac{3}{13}\)
\(C=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right).0=0\)
1.
\(B=20182018.2017-20172017.2018\)
\(B=2018.10001.2017-2017.10001.2018\)
\(B=0\)
1 B=20182018.2017-20172017.2018
B=2018.10001.2017-2017.10001.2018
B=0
2 C=12+22+32+...+1002
C=1(1+0)+2(1+1)+3(1+2)+...+100(1+99)
C=1+2+1.2+3+2.3+...+100+99.100
C=(1+2+3+...+100)+(1.2+2.3+...+99.100)
C=[(1+100).100:2]+[(99.100.101):3]
C=5050+333300
C=338350
A=100+(98-97)+(96-95)+...+(2-1)
=100+1+1+1+...+1
Từ 2 đến 98 có
(98-2):2+1=49 (số hạng)
=>A=100+1.49
=100+49
=149
Giải
Ta có:
22 + 42 + 62 + ... + 162 + 182
= 22 . (12 + 22 + 32 + ... + 82 + 92 )
= 22 . 285 (do 12 + 22 + 32 + ... + 82 + 92 = 285)
= 4 . 285
= 1140
Vậy giá trị của biểu thức là 1140