a) Ta có: \(\left(a^2+b^2\right)^2-4a^2b^2=\left(a^2+b^2\right)^2-\left(2ab\right)^2\)

\(=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)=\left(a-b\right)^2.\left(a+b\right)^2\)( đpcm )

b) Ta có: \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(=\left(a-b+b-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-b+b-c\right)+\left(c-a\right)^3\)

\(-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(=\left(a-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(=\left(a-c\right)^3+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-c\right)-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(=\left(a-c\right)^3-\left(a-c\right)^3+3\left(a-b\right)\left(b-c\right)\left(c-a\right)-3\left(a-b\right)\left(b-c\right)\left(c-a\right)=0\)

\(\Rightarrow\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)( đpcm )

1) Ta có: \(\left(a^2+b^2\right)^2-4a^2b^2\)

\(=a^4+2a^2b^2+b^4-4a^2b^2\)

\(=a^4-2a^2b^2+b^4\)

\(=\left(a^2-b^2\right)^2\)

\(=\left[\left(a-b\right)\left(a+b\right)\right]^2\)

\(=\left(a-b\right)^2\left(a+b\right)^2\)

2) Ta có: \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)

\(=\left(a-b+b-c\right)\left[\left(a-b\right)^2-\left(a-b\right)\left(b-c\right)+\left(b-c\right)^2\right]+\left(c-a\right)^3\)

\(=\left(a-c\right)\left(a^2-2ab+b^2-ab+ac+b^2-bc+b^2-2bc+c^2\right)+\left(c-a\right)^3\)

\(=-\left(c-a\right)\left(a^2+3b^2+c^2-3ab+ac-3bc\right)+\left(c-a\right)\left(c^2-2ca+a^2\right)\)

\(=\left(c-a\right)\left(c^2-2ca+a^2-a^2-3b^2-c^2+3ab-ac+3bc\right)\)

\(=\left(c-a\right)\left(3ab+3bc-3b^2-3ac\right)\)

\(=3\left(c-a\right)\left(ab-b^2-ac+bc\right)\)

\(=3\left(c-a\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]\)

\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Ta có : (x + 4)(y + 3) = 3 = 1.3 = 3.1 = (-1)(-3) = (-3)(-1)

+) x + 4 = 1 => x = -3 ; y + 3 = 3 => y = 0

+) x + 4 = 3 => x = -1 ;  y + 3 = 1 => y = -2

+) x + 4 = -1 => x = -5 ; y + 3 = -3 => y = -6

+) x + 4 = -3 => x = -7 ; y + 3 = -1 => y = -4

(x + 2)(y - 3) = -3 = (-1).3 = (-3).1 

+) x + 2 = -1 => x = -3 ; y - 3 = 3 => y = 6

+) x + 2 = -3 => x = -5 ; y - 3 = 1 => y = 4

\(\left(x+4\right)\left(y+3\right)=3\)

\(x+4;y+3\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

tương tự 

Bài làm:

a) Tại x = 2 thì giá trị của B là:

\(B=-\frac{10}{2-4}=\frac{-10}{-2}=5\)

b) Ta có:

\(A=\frac{x+2}{x+5}+\frac{-5x-1}{x^2+6x+5}-\frac{1}{1+x}\)

\(A=\frac{x+2}{x+5}-\frac{5x+1}{\left(x+1\right)\left(x+5\right)}-\frac{1}{x+1}\)

\(A=\frac{\left(x+2\right)\left(x+1\right)-5x-1-\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}\)

\(A=\frac{x^2+3x+2-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}\)

\(A=\frac{x^2-3x-4}{\left(x+1\right)\left(x+5\right)}\)

\(A=\frac{\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x+5\right)}\)

\(A=\frac{x-4}{x+5}\)

c) Ta có: \(P=A.B=\frac{x-4}{x+5}\cdot\frac{-10}{x-4}=\frac{-10}{x+5}\)

Để \(-\frac{10}{x+5}\inℤ\Rightarrow\left(x+5\right)\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

=> \(x\in\left\{-15;-10;-7;-6;-4;-3;0;5\right\}\)

a) \(B=\frac{-10}{x-4}\)( ĐKXĐ : \(x\ne4\))

Tại x = 2 ( tmđk ) thì \(B=\frac{-10}{2-4}=\frac{-10}{-2}=5\)

b) \(A=\frac{x+2}{x+5}+\frac{-5x-1}{x^2+6x+5}-\frac{1}{1+x}\)

ĐKXĐ : \(x\ne-5,x\ne-1\)

\(A=\frac{x+2}{x+5}-\frac{5x+1}{\left(x+1\right)\left(x+5\right)}-\frac{1}{x+1}\)

\(A=\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}-\frac{5x+1}{\left(x+1\right)\left(x+5\right)}-\frac{1\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}\)

\(A=\frac{x^2+3x+2-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}\)

\(A=\frac{x^2-3x-4}{\left(x+1\right)\left(x+5\right)}\)

\(A=\frac{\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x+5\right)}=\frac{x-4}{x+5}\)

c) \(P=A\cdot B=\frac{x-4}{x+5}\cdot\frac{-10}{x-4}=\frac{-10}{x+5}\)( ĐKXĐ : \(x\ne-5\))

Để P nguyên => \(\frac{-10}{x+5}\)nguyên

=> -10 chia hết cho x + 5

=> x + 5 thuộc Ư(-10) = { ±1 ; ±2 ; ±5 ; ±10 }

Các giá trị của x đều tmđk

Vậy x = { -4 ; -6 ; -3 ; -7 ; 0 ; -10 ; 5 ; -15 }

đáp án: C-> Can you hold my books for me

b + a + a = b + 2a

Võ : \(\frac{9}{15}.10=6\)(h/s)

bơi:\(\frac{9}{15}.20=12\)(h/s)

Bóng đá :\(\frac{9}{15}.30=18\)(h/s)

Cầu lông : \(\frac{9}{15}.25=15\)(h/s) 

                  Đ/S:........

1, this building has the same as that one high

2, I am as weight as her

3, the bigger the appartment, the higher the rent.

4, the fatter she gets, the tired she feels

5, the more book you read, the more knowledges you will have

6, that book is more eating than this one

7, I am as heigh as her

đó là ý kiến của tui. cung ko chắc là đúng hay ko. nếu bn nào có ý kiến khác thì comment bên dưới nheng!

hok tốt

Cái đầu là tính à?

Ta có: \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}\)

\(=\left(\sqrt{15}\right)^2+2.2\sqrt{3}.\sqrt{15}+\left(2\sqrt{3}\right)^2+12\sqrt{5}\)

\(=15+12\sqrt{5}+12+12\sqrt{5}\)

\(=27+24\sqrt{5}\)

Sau:

Ta thấy: Điều kiện để \(\sqrt{-\left|x+5\right|}\) có nghĩa là \(-\left|x+5\right|\ge0\left(\forall x\right)\)

Mà \(-\left|x+5\right|\le0\left(\forall x\right)\) nên dấu "=" xảy ra khi: \(\left|x+5\right|=0\Rightarrow x=-5\)

Vậy khi x = -5 thì \(\sqrt{-\left|x+5\right|}\) có nghĩa

Làm lại ý 2

\(\sqrt{-\left|x+5\right|}\)có nghĩa

\(\Leftrightarrow-\left|x+5\right|\ge0\)

\(\Leftrightarrow\left|x+5\right|\le0\)

\(\Leftrightarrow x+5\le0\)

\(\Leftrightarrow x\le-5\)