b) 1.2.3...9-1.2.3...8-1.2.3... 0.7 .0.8 ^ 2
Giúp mình với
1.2.3...9-1.2.3...8-1.2.3...8^2
Giải thích giùm mình nha
1.2.3...9 - 1.2.3...8 - 1.2.3...82
= 1.2.3...8.(9 - 1) - 1.2.3...82
= 1.2.3...8.8 - 1.2.3...8.8
= 0
1.2.3...9 - 1.2.3...8 - 1.2.3...82
= 1.2.3...8.(9 - 1) - 1.2.3...82
= 1.2.3...8.8 - 1.2.3...8.8
= 0
\(1.2.3.....9-1.2.3.....8-1.2.3.....8^2\)
\(=1.2.3.....8.\left(9-1\right)-1.2.3.....8.8\)
\(=1.2.3.....8.8-1.2.3.....8.8\)
\(=0\)
1.2.3......9-1.2.3....8-1.2.3.....8.8 mng giúp với ạ
=1*2*3*...*8(9-1)-1*2*3*...*8*8
=0
b) 1.2.3...9-1.2.3...8-1.2.3...7.8^ 2
1.2.3...9 - 1.2.3...8 - 1.2.3....7.82
= 1.2.3.4.5.6.7.8.(9 - 1 - 8)
= 1.2.3.4.5.6.7.8.0
= 0
\(1.2.3.....9-1.2.3.....8-1.2.3...8.8\)
Giair nhanh mình tick!
1.2.3....9-1.2.3....8-1.2.3...8.8= 1.2.3....9-(12.3....8+1.2.3...8.8)=1.2.3...9-(1.2.3....8(1+8)=1.2.3....9-1.2.3....9=0
đang ký trang youtube này dùm nha https://www.youtube.com/channel/UCdMJRiuo_35tKETQtnAYOBQ
Câu 22. Cho P = 1.2.3...9 − 1.2.3...8 − 1.2.3...8 . Giá trị của biểu thức P là:A.0 B. 1 C. 2 D.8 ( ai biết làm bài này k chỉ với
1.2.3...9 - 1.2.3...8 - 1.2.3...7.82
\(1.2.3.....9-1.2.3.....8-1.2.3.....7.8^2\)
\(=1.2.3.....8\left(9-1-8\right)\)
\(=1.2.3.....8\cdot0\)
\(=0\)
1.2.3.....9−1.2.3.....8−1.2.3.....7.821.2.3.....9−1.2.3.....8−1.2.3.....7.82
=1.2.3.....8(9−1−8)=1.2.3.....8(9−1−8)
=1.2.3.....8⋅0=1.2.3.....8⋅0
=0
1.2.3...9-1.2.3...8-1.2.3...7.82
A= 1.2.3...9- 1.2.3...8-1.2.3...8.8
=1.2.3...8(9-1-8)
=1.2.3...8.0=0
Chúc bạn học tốt
\(A=1.2.3...9-1.2.3...8-1.2.3...8.8\)
\(=1.2.3...8.\left(9-1-8\right)\)
\(=1.2.3...8.0\)
\(=0\)
tính nhanh 1.2.3....9 - 1.2.3....8 - 1.2.3......8^2
nhưng bạn ơi mk cần cách tính làm sao cho nhanh
mk vẫn thắc mắc sao 1.2.3.....9- 1.2.3.4.8^2 lại bằng 1.2.3.4.5....9