Số gạo tẻ mẹ mua gấp số lần gạo nếp :
45 : 9 = 5 ( lần )
Đáp số : 5 lần
 

Bài 2 thiếu đề sửa.

\(A=3+3^2+3^3+3^4+3^5+...+3^{100}\)

\(\Rightarrow A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)

\(\Rightarrow A=\left(4.3\right)+\left(4.3^3\right)+...+\left(4.3^{99}\right)\)

\(\Rightarrow A=4.\left(3+3^3+...+3^{99}\right)\)

Vì tích có chứa thừa số 4 nên: \(A⋮4\)

Vậy A chia hết cho 4

Chứng minh chia hết cho 10 cũng tương tự nha bạn.

Bài 1:

\(A=\dfrac{2}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{89.93}\)

\(A=\dfrac{2}{1.5}+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{89}-\dfrac{1}{93}\right)\)

\(A=\dfrac{2}{5}+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{93}\right)\)

\(A=\dfrac{2}{5}+\dfrac{1}{4}.\dfrac{88}{465}\)

\(A=\dfrac{2}{5}+\dfrac{22}{465}=\dfrac{208}{465}\)

b) \(B=2+2^3+2^5+2^7+...+2^{99}\)

\(\Rightarrow2^2B=2^2.\left(1+2^3+2^5+2^7+...+2^{99}\right)\)

\(2^2B=2^3+2^5+2^7+2^9+...+2^{101}\)

\(2^2B-B=\left(2^3+2^5+2^7+2^9+...+2^{101}\right)-\left(2+2^3+2^5+2^7+...+2^{99}\right)\) \(3B=2^{101}-2\)

\(\Rightarrow B=\dfrac{2^{101}-2}{3}\)

Vậy \(B=\dfrac{2^{101}-1}{3}\)

~ Học tốt ~

a) \(A=2+2^2+2^3+2^4+...+2^{99}\)

\(\Rightarrow2A=2.\left(2+2^2+2^3+2^4+...+2^{99}\right)\)

\(2A=2^2+2^3+2^4+2^5+...+2^{100}\)

\(2A-A=\left(2^2+2^3+2^4+2^5+...+2^{100}\right)-\left(2+2^2+2^3+2^4+...2^{99}\right)\) \(\Rightarrow A=2^{100}-2\)

Vậy \(A=2^{100}-2\)

Giải:7)

\(7^2+24^2=49+576=625\)

\(9^2+40^2=81+1600=1681\)

8)

C1 :\(11^3:11^2=1331:121=11\)

C2 :\(11^3:11^2=11^{3-2}=11\)

C1 :\(16^2:4^2=256:16=16\)

C2 :\(16^2:4^2=\left(4^2\right)^2:4^2=4^4:4^2=16\)

9)

a) \(4^n=64\Rightarrow4^n=4^3\)

\(\Rightarrow n=3\)

b) \(2^n.16=128\Rightarrow2^n=8\)

\(\Rightarrow2^n=2^3\Rightarrow n=3\)

c) \(3^n:9=27\Rightarrow3^n=243\)

\(\Rightarrow3^n=3^5\Rightarrow n=5\)

~ Học tốt ~

\(\left|x+123\right|+\left|\left|y+1\right|-5\right|=0\)

Ta có: \(\left|x+123\right|\ge0\) với mọi x

\(\left|\left|y+1\right|-5\right|\ge0\) với mọi y

Nên \(\left|x+123\right|+\left|\left|y+1\right|-5\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}x+123=0\\\left|y+1\right|-5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-123\\\left|y+1\right|=5\end{matrix}\right.\)

\(\left|y+1\right|=5\Rightarrow y+1=\pm5\)

+ \(y+1=-5\Rightarrow y=-6\)

+ \(y+1=5\Rightarrow y=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=-123\\\left[{}\begin{matrix}y=-6\\y=4\end{matrix}\right.\end{matrix}\right.\)

~ Học tốt ~

Bài 1:

- 69 + 53 + 46 - 94 + (- 14) + 78

= - 69 + 99 - 100 + 78

= - 69 - 1 + 78 = -70 + 78

= 2

Bài 2:

a) \(\left|x+1\right|=0\)

\(\Rightarrow\) x + 1 = 0

\(\Rightarrow\) x = - 1

b) \(25-\left|x\right|=10\)

\(\Rightarrow\left|x\right|=15\)

\(\Rightarrow x=\pm15\)

c)\(\left|x-2\right|+7=12\)

\(\Rightarrow\left|x-2\right|=5\)

\(\Rightarrow x-2=\pm5\)

+ \(x-2=-5\Rightarrow x=-3\)

+ \(x-2=5\Rightarrow x=7\)

Vậy: \(\left\{{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

~ học tốt ~

Gọi phân số cần tìm là: \(\dfrac{a}{b}\)

Theo đề, ta có:

\(\dfrac{3a}{b}=\dfrac{a+b}{b+b}=\dfrac{a+b}{b^2}\)

\(\Rightarrow3a.2b=ab+b^2\)

\(\Rightarrow6ab=ab+b^2\)

\(\Rightarrow6=\left(ab+b^2\right):ab\)

\(\Rightarrow6=1+\dfrac{b}{a}\Rightarrow5=\dfrac{b}{a}\)

\(\Rightarrow...=\dfrac{2}{10}=\dfrac{1}{5}=\dfrac{a}{b}\)

Mà \(\dfrac{a}{b}\) tối giản nên \(\dfrac{a}{b}=\dfrac{1}{5}\)

Vậy phân số cần tìm là \(\dfrac{1}{5}\)

~ Học tốt ~