ABCD là hình chữ nhật 

=> CD = AB = 4 (cm) 

=> AD = BC = 3 (cm) 

=> BD = AC = 5 (cm)

Số sách đủ để chia là:

\(3+3=6\left(quyển\right)\)

Tổ được chia 8  quyển nhiều hơn tổ được chia 7 quyển là:

\(8-7=1\left(quyển\right)\)

Số tổ được chia sách là:

\(6:1=6\left(tổ\right)\)

Số sách văn và toán là:

Ta có:

\(Q=\dfrac{1}{x^2-4x+11}=\dfrac{1}{\left(x^2-4x+4\right)+7}\\ =\dfrac{1}{\left(x-2\cdot x\cdot2+2^2\right)+7}=\dfrac{1}{\left(x-2\right)^2+7}\) 

\(\left(x-2\right)^2\ge0\forall x=>\left(x-2\right)^2+7\ge7\forall x\\ =>Q=\dfrac{1}{\left(x-2\right)^2+7}\le\dfrac{1}{7}\forall x\)

Dấu "=" xảy ra: `x-2=0<=>x=2` 

1) 12 ⋮ x => x ∈ Ư(12) = {1; -1; 2; -2; 3; -3; 4; -4; 6; -6; 12; -12} 

Mà: x > 2 

=> x ∈ {3; 4; 6; 12} 

2) 24 ⋮ x => x ∈ Ư(24) = {1; -1; 2; -2; 3; -3; 4; -4; 6; -6; 8; -8; 12; -12; 24; -24} 

Mà: x > 4 

=> x ∈ {6; 8; 12; 24} 

3) 36 ⋮ x => x ∈ Ư(36) = {1; -1; 2; -2; 3; -3; 4; -4; 6; -6; 9; -9; 12; -12; 18; -18; 36; -36}

Mà: x ≥ 3 

=> x ∈ {3; 4; 6; 9; 12; 18; 36} 

4) 40 ⋮ x => x ∈ Ư(40) = {1; -1; 2; -2; 4; -4; 5; -5; 8; -8; 10; -10; 20; -20; 40; -40} 

Mà: x < 10 và x là số tự nhiên

=> x ∈ {1; 2; 4; 8}

\(10x-x^2+2\\ =\left(-x^2+10x-25\right)+27\\ =-\left(x^2-10x+25\right)+27\\ =-\left(x-5\right)^2+27\)
Ta có: \(-\left(x-5\right)^2\le0\forall x=>-\left(x-5\right)^2+27\le27\forall x\)

Dấu "=" xảy ra: `x-5=0<=>x=5` 

          Bài 1:

A = 8.(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)

A = (3^{2 -} 1)(3² + 1)(3⁴+ 1)(3⁸ +1)(3¹⁶ + 1)

A = (3⁴ - 1)(3⁴ + 1)(3⁸+ 1)(3¹⁶ + 1)

A = (3⁸ - 1)(3⁸ + 1)(3¹⁶ + 1)

A = (3¹⁶ - 1)(3¹⁶ +1)

 A = (3¹⁶)² - 1²  

A = 3³² - 1

1: \(A=8\left(3^2+1\right)\left(3^4+1\right)\cdot...\cdot\left(3^{16}+1\right)\)

\(=\left(3^2-1\right)\left(3^2+1\right)\cdot\left(3^4+1\right)\left(3^8+1\right)\cdot\left(3^{16}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3^8-1\right)\cdot\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3^{16}-1\right)\left(3^{16}+1\right)=3^{32}-1\)

2: \(B=\left(1-3\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=-\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=-\left(3^4-1\right)\left(3^4+1\right)\cdot\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=-\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)=-\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(=-\left(3^{32}-1\right)=1-3^{32}\)

3: \(C=24\left(5^2+1\right)\left(5^4+1\right)\cdot...\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)

\(=\left(5^2-1\right)\left(5^2+1\right)\cdot\left(5^4+1\right)\left(5^{128}+1\right)+\left(5^{256}-1\right)\)

\(=\left(5^4-1\right)\cdot\left(5^4+1\right)\cdot...\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)

\(=\left(5^8-1\right)\left(5^8+1\right)\cdot...\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)

\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\cdot...\cdot\left(5^{128}+1\right)+5^{256}-1\)

\(=\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\left(5^{128}+1\right)+5^{256}-1\)

\(=\left(5^{64}-1\right)\left(5^{64}+1\right)\left(5^{128}+1\right)+5^{256}-1\)

\(=\left(5^{128}-1\right)\left(5^{128}+1\right)+5^{256}-1=2\left(5^{256}-1\right)\)

alo

ko ai trả lời à   

ai tra lời cho 2 like

Bài 1:

\(a.A=23,12+45,56+76,88+54,44\\ =\left(23,12+76,88\right)+\left(45,56+54,44\right)\\ =100+100\\ =200\\ b.201,5\cdot9+201,5\cdot2-201,5\\ =201,5\cdot\left(9+2-1\right)\\ =201,5\cdot10\\ =2015\\ c.C=\dfrac{1}{2}:0,5+\dfrac{1}{4}:0,25-\dfrac{1}{8}:0,125+2014\\ =\dfrac{1}{2}\cdot2+\dfrac{1}{4}\cdot4+\dfrac{1}{8}\cdot8+2014\\ =1+1+1+2014\\ =2017\\ d.D=2\dfrac{2}{3}+\dfrac{5}{9}+\dfrac{4}{9}:\left(30\%-\dfrac{1}{10}\right)-\dfrac{2}{9}\\ =\dfrac{8}{3}+\left(\dfrac{5}{9}-\dfrac{2}{9}\right)+\dfrac{4}{9}:\left(\dfrac{3}{10}-\dfrac{1}{10}\right)\\ =\dfrac{8}{3}+\dfrac{1}{3}+\dfrac{4}{9}:\dfrac{1}{5}\\ =\dfrac{9}{3}+\dfrac{4}{45}\\ =3+\dfrac{4}{9}\cdot5=3+\dfrac{20}{9}=\dfrac{47}{9}\)

Bài 2: 

a: y+2=2017

=>y=2017-2=2015

b: \(3y-2\dfrac{2}{7}=3\dfrac{5}{7}\)

=>\(3y=3+\dfrac{5}{7}+2+\dfrac{2}{7}=6\)

=>\(y=\dfrac{6}{3}=2\)

c: \(1\dfrac{3}{4}-\dfrac{3}{4}y=75\%\)

=>\(\dfrac{7}{4}-\dfrac{3}{4}y=\dfrac{3}{4}\)

=>7-3y=3

=>3y=7-3=4

=>\(y=\dfrac{4}{3}\)

d: \(\dfrac{2}{3}+\dfrac{1}{3}y+3\dfrac{2}{3}y=\dfrac{8}{3}\)

=>\(4y=\dfrac{8}{3}-\dfrac{2}{3}=\dfrac{6}{3}=2\)

=>\(y=\dfrac{2}{4}=\dfrac{1}{2}\)

e: \(y-14=25\)

=>y=25+14=39

f: \(5y-25=35\)

=>5y=25+35=60

=>y=60/5=12

g: 9,34-y=1,28

=>y=9,34-1,28=8,06

h: y:1,2=2,4

=>\(y=2,4\cdot1,2=2,88\)

i: 2,4:y=0,2

=>y=2,4:0,2=12

k: (y+1)+(y+3)=24

=>y+1+y+3=24

=>2y=20

=>y=10

        Bài 6:

\(\dfrac{9^5.9^7}{3^{22}}\) = \(\dfrac{3^{15}.3^{21}}{3^{22}}\) = \(\dfrac{3^{36}}{3^{22}}\)  = 3¹⁴

  Bài 7:

\(\dfrac{9^{16}.8^{11}}{6^{33}}\) =  \(\dfrac{3^{32}.2^{33}}{3^{33}.2^{33}}\) = \(\dfrac{1}{3}\) 

a; (\(\dfrac{1}{x}\) - 5)(\(\dfrac{1}{x}\) + 5)

= (\(\dfrac{1}{x}\))^2 -  5²

=  \(\dfrac{1}{x^2}\) - 25

b; (\(\dfrac{x}{3}\) - \(\dfrac{y}{4}\))(\(\dfrac{x}{3}\) + \(\dfrac{y}{4}\))

 = \(\left(\dfrac{x}{3}\right)^2\) - \(\left(\dfrac{y}{4}\right)^2\)

  =  \(\dfrac{x^2}{9}\) - \(\dfrac{y^2}{16}\)

d; (\(\dfrac{x}{y}\) - \(\dfrac{2}{3}\) (\(\dfrac{x}{y}\)+\(\dfrac{2}{3}\))

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

= \(\dfrac{x^2}{y^2}\) - \(\dfrac{4}{9}\)

e; (2\(x\) - \(\dfrac{2}{3}\))(\(\dfrac{2}{3}\) + 2\(x\))

= (2\(x\))² - (\(\dfrac{2}{3}\))²

= 4\(x^2\) - \(\dfrac{4}{9}\)