a)số bài trung binh chiếm số phần tổng số bài cả lớp là:

1-\(\dfrac{3}{8}\)-\(\dfrac{2}{5}\)=\(\dfrac{9}{40}\)(tổng số bài)

tổng số bai kiểm tra lớp 6A là:

9:\(\dfrac{9}{40}\)=40(bài)

b)tỉ số phần trăm số bài loại giỏi với số học sinh cả lớp là:

(40.\(\dfrac{3}{8}\)):40.100%=37,5%

6¹⁰ = (6⁵)² = (\(\overline{..6}\))² 

Vậy 6¹⁰ là một số chính phương (đpcm)

Cảm ơn cô

Câu 15;

a: \(A=\dfrac{10^8+2}{10^8-1}=\dfrac{10^8-1+3}{10^8-1}=1+\dfrac{3}{10^8-1}\)

\(B=\dfrac{10^8}{10^8-3}=\dfrac{10^8-3+3}{10^8-3}=1+\dfrac{3}{10^8-3}\)

Ta có: \(10^8-1>10^8-3\)

=>\(\dfrac{3}{10^8-1}< \dfrac{3}{10^8-3}\)

=>\(\dfrac{3}{10^8-1}+1< \dfrac{3}{10^8-3}+1\)

=>A<B

b: \(M=\dfrac{2^2}{1\cdot3}+\dfrac{2^2}{3\cdot5}+...+\dfrac{2^2}{197\cdot199}\)

\(=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{197\cdot199}\right)\)

\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{197}-\dfrac{1}{199}\right)\)

\(=2\left(1-\dfrac{1}{199}\right)=2\cdot\dfrac{198}{199}=\dfrac{396}{199}\)

em can cach giai

Số số hạng của S:

100 - 51 + 1 = 50 (số)

Ta có:

1/51 > 1/100

1/52 > 1/100

1/53 > 1/100

...

1/99 > 1/100

1/100 = 1/100

Cộng vế với vế, ta có:

S > 1/100 + 1/100 + 1/100 + ... + 1/100 (50 số 1/100)

= 50/100

= 1/2

Vậy S > 1/2

S = \(\dfrac{1}{51}\) + \(\dfrac{1}{52}\) + \(\dfrac{1}{53}\) +...+\(\dfrac{1}{98}\) + \(\dfrac{1}{100}\)

Tổng S có số phân số là: (100 - 51) : 1 + 1  = 50

Mặt khác ta có: \(\dfrac{1}{51}\) > \(\dfrac{1}{52}\) > \(\dfrac{1}{53}\)> ...> \(\dfrac{1}{100}\) 

     ⇒ \(\dfrac{1}{51}\) + \(\dfrac{1}{52}\) + \(\dfrac{1}{53}\) + ... + \(\dfrac{1}{100}\) > \(\dfrac{1}{100}\) + \(\dfrac{1}{100}\)+...+ \(\dfrac{1}{100}\)

         \(\dfrac{1}{51}\) + \(\dfrac{1}{52}\) + \(\dfrac{1}{53}\) + ... + \(\dfrac{1}{100}\) > \(\dfrac{1}{100}\) x 50

         \(\dfrac{1}{51}\) + \(\dfrac{1}{52}\) + \(\dfrac{1}{53}\) + ... + \(\dfrac{1}{100}\) > \(\dfrac{1}{2}\)

 Vậy S = \(\dfrac{1}{51}\) + \(\dfrac{1}{52}\) + \(\dfrac{1}{53}\) + ... + \(\dfrac{1}{100}\) > \(\dfrac{1}{2}\)

a) -24/x + 17/x = -7/x

Để -24/x + 7/x là số nguyên thì 7 ⋮ x

⇒ x ∈ Ư(7) = {-7; -1; 1; 7}

b) (x - 8)/(x + 1) + (x + 2)/(x + 1)

= (x - 8 + x + 2)/(x + 1)

= (2x + 6)/(x + 1)

= (2x + 2 + 4)/(x + 1)

= [2(x + 1) + 4)]/(x + 1)

= 2 + 4/(x + 1)

Để biểu thức đã cho là số nguyên thì 4 ⋮ (x + 1)

⇒ x + 1 ∈ Ư(4) = {-4; -2; -1; 1; 2; 4}

⇒ x ∈ {-5; -3; -2; 0; 1; 3}

a) \(3.\left(2x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=1\)

\(3.\left(2x-\dfrac{1}{2}\right)^3=1-\dfrac{1}{9}\)

\(3.\left(2x-\dfrac{1}{2}\right)^3=\dfrac{8}{9}\)

\(\left(2x-\dfrac{1}{2}\right)^3=\dfrac{8}{9}:3\)

\(\left(2x-\dfrac{1}{2}\right)^3=\dfrac{8}{27}\)

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

\(2x=\dfrac{2}{3}+\dfrac{1}{2}\)

\(2x=\dfrac{7}{6}\)

\(x=\dfrac{7}{6}:2\)

\(x=\dfrac{7}{12}\)

b) \(2.\left(x-\dfrac{1}{2}\right)^2+1\dfrac{1}{3}=2\dfrac{2}{9}\)

\(2\left(x-\dfrac{1}{2}\right)^2+\dfrac{4}{3}=\dfrac{20}{9}\)

\(2\left(x-\dfrac{1}{2}\right)^2=\dfrac{20}{9}-\dfrac{4}{3}\)

\(2\left(x-\dfrac{1}{2}\right)^2=\dfrac{8}{9}\)

\(\left(x-\dfrac{1}{2}\right)^2=\dfrac{8}{9}:2\)

\(\left(x-\dfrac{1}{2}\right)^2=\dfrac{4}{9}\)

\(x-\dfrac{1}{2}=-\dfrac{2}{3}\) hoặc \(x-\dfrac{1}{2}=\dfrac{2}{3}\)

*) \(x-\dfrac{1}{2}=-\dfrac{2}{3}\)

\(x=-\dfrac{2}{3}+\dfrac{1}{2}\)

\(x=-\dfrac{1}{6}\)

*) \(x-\dfrac{1}{2}=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}+\dfrac{1}{2}\)

\(x=\dfrac{7}{6}\)

Vậy \(x=-\dfrac{1}{6};x=\dfrac{7}{6}\)

c) \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{99}{101}\)

\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{99}{101}\)

\(2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{99}{101}\)

\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{99}{101}:2\)

\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{99}{202}\)

\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{99}{202}\)

\(\dfrac{1}{x+1}=\dfrac{1}{101}\)

\(x+1=101\)

\(x=101-1\)

\(x=100\)

a) -x/8 = -9/(2x)

x.2x = -9.(-8)

2x² = 72

x² = 72: 2

x² = 36

x = -6 hoặc x = 6

b) x/3 = 10/(x + 1)

x.(x + 1) = 3.10

x.(x + 1) = 30

x² + x - 30 = 0

x² - 5x + 6x - 30 = 0

(x² - 5x) + (6x - 30) = 0

x(x - 5) + 6(x - 5) = 0

(x - 5)(x + 6) = 0

x - 5 = 0 hoặc x + 6 = 0

*) x - 5 = 0

x = 0 + 5

x = 5

*) x + 6 = 0

x = 0 - 6

x = -6

c) 2 5/6 x - 1 2/3 + 2 3/4 = 1 1/3

17/6 x - 5/3 + 11/4 = 4/3

17/6 x = 4/3 + 5/3 - 11/4

17/6 x = 1/4

x = 1/4 : 17/6

x = 3/34

d) (2x - 1)/21 = 3/(2x + 1)

(2x - 1)(2x + 1) = 3.21

4x² + 2x - 2x - 1 = 63

4x² = 63 + 1

4x² = 64

x² = 64 : 4

x² = 16

x = -4 hoặc x = 4

Bài 6:

a:

ĐKXĐ: \(n\ne2\)

Để A>0 thì \(\dfrac{7}{n-2}>0\)

=>n-2>0

=>n>2

b:

ĐKXĐ: n<>1

Để B>0 thì \(\dfrac{n-1}{n-2}>0\)

=>\(\left[{}\begin{matrix}n-2>0\\n-1< 0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}n>2\\n< 1\end{matrix}\right.\)

Bài 5:

ĐKXĐ: n<>3

Để P là số nguyên thì \(n^2-2n+2⋮n-3\)

=>\(n^2-3n+n-3+5⋮n-3\)

=>\(5⋮n-3\)

=>\(n-3\in\left\{1;-1;5;-5\right\}\)

=>\(n\in\left\{4;2;8;-2\right\}\)

Sửa đề: \(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{50}>\dfrac{4}{5}\)

\(\dfrac{1}{11}>\dfrac{1}{50}\)

\(\dfrac{1}{12}>\dfrac{1}{50}\)

...

\(\dfrac{1}{50}=\dfrac{1}{50}\)

Do đó: \(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{50}>\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{40}{50}=\dfrac{4}{5}\)