Đặt \(A=4+4^2+4^3+...+4^{89}+4^{90}\)

Ta có: \(A=\left(4+4^2+4^3\right)+...+\left(4^{88}+4^{89}+4^{90}\right)\)

\(A=84+...+4^{87}.\left(4+4^2+4^3\right)\)

\(A=84+...+4^{87}.84\)

\(A=84.\left(1+...+4^{87}\right)\)

Vì \(84⋮21\) nên \(84.\left(1+...+4^{87}\right)⋮21\)

Vậy \(A⋮21\)

\(#\)  Hallowen vui vẻ 🎃

Bài 3:

a) \(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{4}{3}+\dfrac{11}{25}\)

\(=\left(\dfrac{5}{15}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)\)

\(=\left(\dfrac{5}{15}-\dfrac{20}{15}\right)+1\)

\(=-1+1\)

\(=0\)

b) \(12+8.\left(\dfrac{-1}{2}\right)^3+\left[\left(2018\right)^0:\dfrac{1}{2}\right]^2\)

\(=12+8.\dfrac{-1}{8}+\left[1.2\right]^2\)

\(=12+-1+2^2\)

\(=11+4\)

\(=15\)

c) \(\left(0,25\right)^{10}.4^{10}+\sqrt{5^2-3^2}\)

\(=\left(0,25.4\right)^{10}+\sqrt{25-9}\)

\(=1^{10}+\sqrt{16}\)

\(=1+4\)

\(=5\)

Bài 4:

a) \(x-\dfrac{7}{5}=0,2\)

\(=>x-\dfrac{7}{5}=\dfrac{1}{5}\)

\(=>x=\dfrac{1}{5}+\dfrac{7}{5}\)

\(=>x=\dfrac{8}{5}\)

b) \(\left(x-\dfrac{3}{5}\right):\dfrac{-1}{3}=0,4\)

\(=>\left(x-\dfrac{3}{5}\right):\dfrac{-1}{3}=\dfrac{2}{5}\)

\(=>x-\dfrac{3}{5}=\dfrac{2}{5}.\dfrac{-1}{3}\)

\(=>x-\dfrac{3}{5}=\dfrac{-2}{15}\)

\(=>x=\dfrac{-2}{15}+\dfrac{3}{5}=\dfrac{-2}{15}+\dfrac{9}{15}\)

\(=>x=\dfrac{7}{15}\)

=)) chăm đâu, lười quá nè ^^. Mà cảm ơn nhe 🥰

Tính:

\(5.2^3+7^{11}:7^9-1\)

\(=5.8+7^2-1\)

\(=40+49-1\)

\(=89-1\)

\(=88\)

Chọn khẳng định sai: Khẳng định " 24 là ước của 8 " sai. (Vì 8 không chia hết cho 24).

Liệt kê phần tử: \(A=\left\{15;17;19;21;23;25\right\}\)

Chữ số 8 trong dãy số 48 369 117 có giá trị là 8000000.

Cho biểu thức \(M=\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\), tìm giá trị nguyên của x để biểu thức M đạt giá trị nguyên.

1. They usually take a lot of beautiful photos.

2. Last week, they visited an orphanage in Ho Chi Minh City.

3. What does your brother do in his free time?

4. They cleaned the kitchen one week ago.

5. Do you do morning exercise every day?

6. I collected stamps when I was a child.

7. Having a healthy lifestyle is important.

8. Dickens wrote a lot of novels.

\(#WendyDang\)

a) \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{19.21}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{21}\right)\)

\(A=\dfrac{1}{2}.\left(\dfrac{21}{21}-\dfrac{1}{21}\right)\)

\(A=\dfrac{1}{2}.\dfrac{20}{21}\)

\(A=\dfrac{10}{21}\)

b) \(B=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(B=\dfrac{1}{99}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{96.97}+\dfrac{1}{97.98}+\dfrac{1}{98.99}\right)\)

\(B=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{96}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{99}-\left(1-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{99}-\left(\dfrac{99}{99}-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{99}-\dfrac{98}{99}\)

\(B=-\dfrac{97}{99}\)