= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz 

= xy(x + y) + yz(y + z + x) + xz(x + z + y) 

= xy(x + y) + z(x + y + z)(y + x) 

= (x + y)(xy + zx + zy + z²) 

= (x + y)[x(y + z) + z(y + z)] 

= (x + y)(y + z)(z + x)

Bài nào vậy bạn?

a) \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

b) \(x^2-11=\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)\)

c: \(x-2=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)

d: \(x^2-2\sqrt{5x}+5=\left(x-\sqrt{5}\right)^2\)

Bài 3:

a: =>(2x-7)(x-2)=0

=>x=7/2 hoặc x=2

b: =>(x-1)(x+2)=0

=>x=1 hoặc x=-2

d: =>2x+3=0

hay x=-3/2

a )   x 2   -   3   =   x 2   -   ( √ 3 ) 2   =   ( x   -   √ 3 ) ( x   +   √ 3 )     b )   x 2   -   6   =   x 2   -   ( √ 6 ) 2   =   ( x   -   √ 6 ) ( x   +   √ 6 )     c )   x 2   +   2 √ 3   x   +   3   =   x 2   +   2 √ 3   x   +   ( √ 3 ) 2     =   ( x   +   √ 3 ) 2       d )   x 2   -   2 √ 5   x   +   5   =   x 2   -   2 √ 5   x   +   ( √ 5 ) 2     =   ( x   -   √ 5 ) 2

Tính nhanh : 

a) 25² - 15² = (25 + 15)(25 - 15) = 40 . 10 = 400

b) 87² + 73² - 27² - 13² = (87² - 13²) + (73² - 27²)

= (87 + 13)(87 - 13) + (73 + 27)(73 - 27)

= 100 . 74 + 100 . 26 = 100 . (74 + 26) = 100 . 100 = 10000

Bài 1:

a)\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x\cdot2y=2\left(x+y\right)\)

b) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\\ =\left(4x+2\right)\cdot2x=4x\left(2x+1\right)\)

Bài 2:

a) \(25^2-15^2=\left(25-15\right)\left(25+15\right)=10\cdot40=400\)

b) \(87^2+73^2-27^2-13^2=\left(87^2-27^2\right)+\left(73^2-13^2\right)\\ =\left(87-27\right)\left(87+27\right)+\left(73-13\right)\left(73+13\right)\)

\(=60\cdot114+60\cdot86=60\cdot\left(114+86\right)=60\cdot200=12000\)

Bài 2:

a) \(x^3-0,25\cdot x=0\)

\(\Leftrightarrow x^2\left(x-0,25\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-0,25=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=0,25\end{array}\right.\)

b) \(x^2-10=-25\)

\(\Leftrightarrow x^2=-15\) (vô nghiệm0

c) \(4x^2-4x=-1\)

\(\Leftrightarrow4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\)

\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)

d) \(8x^3+12x^2+6x+1=0\)

\(\Leftrightarrow\left(2x+1\right)^3=0\)

\(\Leftrightarrow2x+1=0\Leftrightarrow x=-\frac{1}{2}\)

Phân tích thành nhân tử : 

a) (x + y)² - (x - y)² = [(x + y) + (x - y)][(x + y) - (x - y)]

= 2x . 2y = 4xy

b) (3x + 1)² - (x + 1)² = [(3x + 1) + (x + 1)][(3x + 1) - (x - 1)]

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

Đáp án A

Phương pháp : Sử dụng phương pháp đổi biến, đặt x = a – t.

Cách giải : Đặt x = a – t => dx = –dt. Đổi cận 

=> 

1)28.122-28.62+30.144

=28(122-62)+30.144

=28.60+30.144

=30.56+30.144

=30(56+144)

=30.200

=6000

2)52.143-52.39-8.26

=52(143-39)-8.26

=52.104-8.26

=8.676-8.26

=8(676-26)

=8.650

=5200

 Bài 2

a)A\(=\)\(\left(x-3\right)\left(x^2+4\right)-4\left(3-x\right)\)

\(=\left(x-3\right)\left(x^2+4x\right)+4\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+4x+4\right)\)

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

b)để biểu thức A=0 ⇔ x-3=0 hoặc x+2=0

                                ⇔ x=3  hoặc x=-2 

a) Ta có: \(x^4-16x^2=0\)

\(\Leftrightarrow x^2\left(x^2-16\right)=0\)

\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

b) Ta có: \(x^8+36x^4=0\)

\(\Leftrightarrow x^4\left(x^4+36\right)=0\)

\(\Leftrightarrow x^4=0\)

hay x=0

c) Ta có: \(\left(x-5\right)^3-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left[\left(x-5\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

d) Ta có: \(5\left(x-2\right)-x^2+4=0\)

\(\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)