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YẾN ƠI TUI BẠN BÀ NÈ
NHƯNG TUI HỔNG PHẢI LỚP 6 THÔNG CẢM NHA
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Bài 1:
A,
(100+121+144)÷(169+196)
=365:365=1
B
,1.2.3...7.8.(9-1-8)
=1.2.3...7.8.0=0
C,
3^2.2^4.2^32/11.2^13.2^22-2^26
=3^2.2^36/11.2^35-2^36
=3^2.2^36/2^35-(11-2)
=9.2/9=2
D,
=1152-374-1152+(-65)+374
=(1152-1152)+(-374+374)+(-65)
=0+0+(-65)=-65
E,
=13-(12-11-10+9)+(8-7-6+5)-(4-3-2+1)
=13-0+0-0=13
Bài 2:
A,(19x+50)÷14=25-16
(19x+50)÷14=9
19x+50=126
19x=76
x=4
B,
31x+(1+2+3+...+30)=1240
31x+465=1240
31x=775
×=25
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\(\frac{3^2.4^2.2^{32}}{11.2^{13}.4^{11}-16^9}=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}=\frac{9.2}{9}=2\)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{6^9.2^{10}+6^{10}.2^{10}}=\frac{2^{19}.3^9+3^9.5.2^{18}}{6^9.2^{10}.\left(1+6\right)}=\frac{2^{18}.3^9.\left(2+5\right)}{2^9.3^9.2^{10}.7}=\frac{2^{18}.7}{2^{19}.7}=\frac{1}{2}\)
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\(b)\)\(9!-8!-7!.8^2\)
\(=\)\(8!\left(9-1\right)-7!.8^2\)
\(=\)\(7!.8.8-7!.8^2\)
\(=\)\(7!.8^2-7!.8^2\)
\(=\)\(0\)
\(c)\)\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\)\(\frac{\left(2^2\right)^2.\left(2^{16}\right)^2.3^2}{2^{13}.\left(2^2\right)^{11}.11-\left(2^4\right)^9}\)
\(=\)\(\frac{2^4.2^{32}.3^2}{2^{13}.2^{22}.11-2^{36}}\)
\(=\)\(\frac{2^{36}.3^2}{2^{35}.11-2^{36}}\)
\(=\)\(\frac{2^{36}.3^2}{2^{35}\left(11-2\right)}\)
\(=\)\(\frac{2.3^2}{9}\)
\(=\)\(\frac{2.3^2}{3^2}\)
\(=\)\(2\)
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a) A = 2 - 5 + 8 - 11 + ... - 161 + 164
A = ( 2 + 8 + ... + 164 ) - ( 5 + 11 + ... + 161 )
Đặt A = B - C
Số số hạng của B là : ( 164 - 2 ) : 6 + 1 = 28 ( số )
Tổng B là : ( 164 + 2 ) x 28 : 2 = 2324
Số số hạng của C là : ( 161 - 5 ) : 6 + 1 = 27 ( số )
Tổng C là : ( 161 + 5 ) x 27 : 2 = 2241
=> A = 2324 - 2241 = 83
Vậy A = 83
Đáp án
20628145.1
o
hok tốt !( 3 × 4 × 216)2
( 3 × 4 × 65536 )2
( 12 × 65536 ) 2
7864322
=> 7864322 = 618475290624