Bài 1:

a) A = 2¹⁰+2¹¹+2¹² 

=2¹⁰*(1+2¹+2²)

=2¹⁰*(1+2+4)

=7*2¹⁰ chia hết 7

Đpcm

b)7*32=244

=32+64+128

=2⁵+2⁶+2⁷

Bài 2:

a)ko hiểu đề

b)nhân N với * x như dạng lp 6 âý

Đặt : \(\dfrac{1}{117}\) = x ; \(\dfrac{1}{119}\) = y .

A = ( 3 + x)( 4 + y) - (1 + 1 - x)(5 + 1 - y) - 5y

<=> A = 12 + 3y + 4x + xy - ( 2 - x)( 6 - y) - 5y

<=> A = 12 + 3y + 4x + xy - 12 + 2y + 6x - xy - 5y

<=> A = 10x

<=> A = \(\dfrac{10}{117}\).

Vậy A = \(\dfrac{10}{117}\)

Dịch k ra viết bằng ct toán đi

a: Đặt 117=a; 119=b

\(A=3\frac{1}{117}\cdot4\frac{1}{119}-1\frac{116}{117}\cdot5\frac{118}{119}-\frac{5}{119}\)

\(=3\frac{1}{a}\cdot4\frac{1}{b}-\left(1+\frac{a-1}{a}\right)\cdot\left(5+\frac{b-1}{b}\right)-\frac{5}{b}\)

\(=\frac{3a+1}{a}\cdot\frac{4b+1}{b}-\frac{2a-1}{a}\cdot\frac{6b-1}{b}-\frac{5}{b}\)

\(=\frac{\left(3a+1\right)\left(4b+1\right)-\left(2a-1\right)\left(6b-1\right)-5a}{ab}\)

\(=\frac{12ab+3a+4b+1-\left(12ab-2a-6b+1\right)-5a}{ab}\)

\(=\frac{12ab+3a+4b+1-12ab+2a+6b-1-5a}{ab}=\frac{10b}{ab}=\frac{10}{a}\)

\(=\frac{10}{117}\)

b: Đặt 105=a; 651=b

\(B=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)

\(=\left(2+\frac{1}{3a}\right)\cdot\frac{1}{b}-\frac{1}{a}\cdot\left(3+\frac{b-1}{b}\right)-\frac{4}{3a\cdot b}+\frac{4}{a}\)

\(=\frac{6a+1}{3a}\cdot\frac{1}{b}-\frac{1}{a}\cdot\frac{3b+b-1}{b}-\frac{4}{3ab}+\frac{4}{a}\)

\(=\frac{6a+1}{3ab}-\frac{4b-1}{ab}-\frac{4}{3ab}+\frac{4}{a}=\frac{6a+1-3\left(4b-1\right)-4+12b}{3ab}\)

\(=\frac{6a-3+12b-12b+3}{3ab}=\frac{6a}{3ab}=\frac{2}{b}=\frac{2}{651}\)