a)co

b)ko

c)co

a﴿ 13! = 13.12.11....5.4.3.2.1 => 13! chia hết cho 2 và 5

11! = 11.10.9....5.4.3.2.1 => 11! chia hết cho 2 và 5

=> A = 13! ‐ 11! chia hết cho 2 và 5

c﴿ A = 13! ‐ 11! = 13.12.11! ‐ 11! = ﴾13.12 ‐ 1﴿ .11! = 155.11! chia hết cho 155

Vậy A chia hết cho 155 

x = 4

\(7^{x+1}=7^5\)

\(7^{x}\cdot7=7^5\)

\(7^{x}=7^5:7\)

\(7^{x}=7^4\)

\(\Rightarrow x=4\)

Vậy x = 4

\(x\left(x-1\right)=157\)

=>\(x^2-x-157=0\)

\(\Delta=\left(-1\right)^2-4\cdot1\cdot\left(-157\right)=629>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{1-\sqrt{629}}{2\cdot1}=\frac{1-\sqrt{629}}{2}\\ x=\frac{1+\sqrt{629}}{2\cdot1}=\frac{1+\sqrt{629}}{2}\end{array}\right.\)

bài này là cô lớp sáu giao cho em ý em còn chưa học cái đó nữa

bạn phải ghi rõ có dấu ra mới hiểu

Gọi \(a b = 10 a + b\), \(b a = 10 b + a\).
Khi đó:

\(a b - b a = \left(\right. 10 a + b \left.\right) - \left(\right. 10 b + a \left.\right) = 9 \left(\right. a - b \left.\right)\)

Vì \(9 \left(\right. a - b \left.\right)\) luôn chia hết cho 9, nên \(a b - b a\) chia hết cho 9.

Chúc em học tốt!

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2019.2021}\)

= \(2.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2019.2021}\right)\)

= \(1.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2019.2021}\right)\)

= \(1.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2019}-\dfrac{1}{2021}\right)\)

= \(1.\left(1-\dfrac{1}{2021}\right)\)

= \(1.\dfrac{2020}{2021}\)

= \(\dfrac{2020}{2021}\)

làm cho mn câu b đi

a)\(\frac{25-7.25}{6.15}\)

\(=\frac{25.\left(1-7\right)}{6.15}\)

\(=\frac{25.\left(-6\right)}{6.15}\)

\(=\frac{5.\left(-1\right)}{1.3}\)

\(=\frac{-5}{3}\)

b)\(\frac{15.11-15}{25.\left(-10\right)}\)

\(=\frac{15.\left(11-1\right)}{25.\left(-10\right)}\)

\(=\frac{15.10}{25.\left(-10\right)}\)

\(=\frac{3.1}{5.\left(-1\right)}\)

\(=\frac{-3}{5}\)

Ok thanks bn

cho bn 3

Hi

bài 1 tắt quá

Câu 1:

Đặt: \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+....+\frac{1}{100^2}\)

\(=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+....+\frac{1}{100.100}\)

\(A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{99.100}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}\)

\(\Rightarrow A< \frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)

\(\Rightarrow A< \frac{1}{2}\)

Vậy:.............

Câu 2:

\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{98}+1\right)\left(\frac{1}{99}+1\right)\)

\(=\left(\frac{1}{2}+\frac{2}{2}\right)\left(\frac{1}{3}+\frac{3}{3}\right)\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{98}+\frac{98}{98}\right)\left(\frac{1}{99}+\frac{99}{99}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{99}{98}.\frac{100}{99}\)

\(=\frac{3.4.5....99.100}{2.3.4...98.99}\)

\(=\frac{100}{2}=50\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=1.\left(1-\frac{1}{100}\right)\)

\(=1.\frac{99}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Hình như là fan BTS ms đúng mà~

bn cs đùa ko đấy?

Bạn thử làm với 18/58 thử nào