(a - b)² - (b - a)

= (a - b)(a - b) + (a - b).1

= (a - b + 1)(a - b)

từng bước đọ

​Từ A,B kẻ đường cao AH,BK (H∈CD, K∈CD)
AB//HK=>ABKH là hình bình hành.
AH⊥DC=>ABKH là hình chữ nhật
=>HK=AB=10
ΔAHD= ΔBKC(ch-gn)
=>DH=HC=(DC-HK)//2=7
ΔKCB vuông tại K =>BC^2=BK^2+KC^2
=>BK=24

Ta có : a + b + c = abc

\(\frac{\Rightarrow\left(a+b+c\right)}{abc}=\frac{abc}{abc}\) 

\(\Rightarrow\frac{1}{ac}+\frac{1}{bc}+\frac{1}{ab}=1\)

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\) 

 \(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=2^2\) 

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=4\)  

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\) 

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\) 

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)

\(\text{Ta có: }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{a+b+c}{abc}=\frac{abc}{abc}=1\left(\text{vì }a+b+c=abc\right)\)

\(\text{Lại có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2.1=2\left(\text{ vì }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\right)\)

Vậy ... 

Ta có: \(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^3-x^3y^2\) 

\(=\left[\left(x+y\right)^2-2xy\right]\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]-x^2y^2\left(x+y\right)\) 

\(=\left(a^2-2b\right)\left(a^3-3ab\right)-ab^2\) 

 \(=a^5-5a^3b+5ab^2\) 

\(=a^5-5ab\left(a^2-b\right)\)

\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^3-x^3y^2\)

\(=\left(\left(x+y\right)^2-2xy\right)\left(\left(x+y\right)^3-3xy\left(x+y\right)\right)-x^2y^2\left(x+y\right)\)

\(=\left(a^2-2b\right)\left(a^3-3ab\right)-ab^2\)

\(=a^5-5ab\left(a^2-b\right)\)

Gọi 2 số tự nhiên lẻ đó làn lượt là a và a + 2

Ta có: ( a + 2 )² - a² = 200

a² + 4a + 4 - a² = 200

4a = 196

a = 49 

a + 2 = 51

Vậy 2 số tự nhiên lẻ cần tìm là 49 và 51

gọi 2 số lẻ liên tiếp cần tìm là \(2k-1\)và \(2k+1\).

Vì 2k+1 > 2k-1 nên ta có \(\left(2k+1\right)^2-\left(2k-1\right)^2=200\)

\(\Leftrightarrow4k^2+4k+1-\left(4k^2-4k+1\right)=200\)

\(\Leftrightarrow8k=200\)\(\Leftrightarrow k=\frac{200}{8}=25\)

Thay k=25 vào 2k-1 và 2k+1 ta được 2 số cần tìm là 49 và 51.

Very long !

1) \(\left(a^2+4\right)^2-16a^2\)

\(=\left(a^2+4-4a\right)\left(a^2+4+4a\right)\)

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

2) \(\left(a^2+9\right)^2-36a^2\)

\(=\left(a^2+9-6a\right)\left(a^2+9+6a\right)\)

\(=\left(a-3\right)^2\cdot\left(a+3\right)^2\)

3) \(\left(a^2+4b^2\right)^2-16a^2b^2\)

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

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

4) \(36a^2-\left(a^2+9\right)^2\)

\(=\left(6a-a^2-9\right)\left(6a+a^2+9\right)\)

\(=\left(6a-a^2-9\right)\left(a+3\right)^2\)

5) \(100a^2-\left(a^2+25\right)^2\)

\(=\left(10a-a^2-25\right)\left(10a+a^2+25\right)\)

\(=\left(10a-a^2-25\right)\left(a+5\right)^2\)

\(\left(x+2\right)^2-\left(x-3\right)\left(3+x\right)=2x-8\)

\(\Leftrightarrow\left(x+2\right)^2-\left(x-3\right)\left(x+3\right)=2x-8\)

\(\Leftrightarrow x^2+4x+4-x^2+9=2x-8\)

\(\Leftrightarrow4x+13=2x-8\)

\(\Leftrightarrow4x-2x=-13-8\)

\(\Leftrightarrow2x=-21\)

\(\Leftrightarrow x=10,5\)

\(1\frac{1}{9}x-\frac{5}{18}=\frac{5}{6}\)

\(\Leftrightarrow\frac{10x}{9}=\frac{5}{6}+\frac{5}{18}=\frac{10}{9}\)

\(\Rightarrow x=1\)