a) = 5( x² - 9y² - 6y - 1 ) = 5[ x² - ( 9y² + 6y + 1 ) ] = 5[ x² - ( 3y + 1 )² ] = 5( x - 3y - 1 )( x + 3y + 1 )

b) = 125x³ - 25x² + 15x² - 3x + 5x - 1 = 25x²( 5x - 1 ) + 3x( 5x - 1 ) + ( 5x - 1 ) = ( 5x - 1 )( 25x² + 3x + 1 )

c) = 5( x - 7 ) + a( x - 7 ) = ( x - 7 )( a + 5 )

d) = ( a - b )² + ( a - b ) = ( a - b )( a - b + 1 )

e) = ax² + a - a²x - x = ax( a - x ) + ( a - x ) = ( a - x )( ax + 1 )

f) = ( 10x )² - ( x² + 25 )² = ( 10x - x² - 25 )( 10x + x² + 25 ) = -( x - 5 )²( x + 5 )²

-c²(a - b) + b²(a - c) - a²(b - c) 

= -c²a + c²b + b²a - b²c - a²b + a²c 

= (a²c - c²a + c²b - b²c) + (b²a - a²b) 

= c(a² - ac + bc - b²) + ab(b - a) 

= c²[(a - b)(a + b) - c(a - b)] - ab(a - b) 

= c²(a - b)(a + b - c) - ab(a - b) 

= (a - b)(c²a + c²b - c³ - ab) 

cảm ơn bạn

Ta có: 

\(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\)

\(\Leftrightarrow a^2+b^2+2ab+\left(\frac{ab+1}{a+b}\right)^2\ge2\left(ab+1\right)\)

\(\Leftrightarrow\left(a+b\right)^2-2\left(a+b\right).\frac{ab+1}{a+b}+\left(\frac{ab+1}{a+b}\right)^2\ge0\)

\(\Leftrightarrow\left(a+b-\frac{ab+1}{a+b}\right)^2\ge0\)

Bất đẳng thức cuối luôn đúng, ta ta biến đổi tương đương nên bất đẳng thức ban đầu cũng đúng. 

Ta có đpcm. 

1) = ( x - 4 )( x + 3 + x - 4 ) = ( x - 4 )( 2x - 1 ) = 2x² - 9x + 4

2) = x³ - 6x² + 12x - 8 - ( 9 - 3x + 3x² - x³ ) = x³ - 6x² + 12x - 8 - 9 + 3x - 3x² + x³ = 2x³ - 9x² + 15x - 17

3) = x² - 14x - 10( x² - 2x + 1 ) = x² - 14x - 10x² + 20x - 10 = -9x² + 6x - 10

4) = ( x + 2 )( 2x - x + 2 ) = ( x + 2 )( x + 2 ) = ( x + 2 )² = x² + 4x + 4

5) = x³ - 27 - x³ + 27x = 27x - 27

6) = x³ + y³ - ( x³ - y³ ) = x³ + y³ - x³ + y³ = 2y³

Trả lời:

91, 8a³ + 12a² + 6a + 1 

= ( 2a )³ + 3.(2a)².1 + 3.2a.1 + 1³

= ( 2a + 1 )³

92, 8a³ - 12a² + 6a - 1

= ( 2a )³ - 3.(2a)² + 3.2a.1 - 1³

= ( 2a - 1 )³

8a3-12a2+6a-1

=8a^3-4a^2-8a^2+4a+2a-1
=4a^2(2a-1)-4a(2a-1)+2a-1
=(2a-1)(4a^2-4a-1)

ta có :

\(P=a+\frac{1}{b\left(a-b\right)}=\left(a-b\right)+b+\frac{1}{b\left(a-b\right)}\ge3\sqrt[3]{\left(a-b\right).b.\frac{1}{b\left(a-b\right)}}=3\)

Vậy m=3

dấu bằng xảy ra khi \(a-b=b=\frac{1}{b\left(a-b\right)}\Leftrightarrow\hept{\begin{cases}a=2\\b=1\end{cases}}\)

vậy \(\hept{\begin{cases}a_1=2\\b_1=1\end{cases}\Rightarrow a_1+b_1+m=2+1+3=6}\)

B = x¹⁵ - 8x¹⁴ + 8x¹³ - 8x² + ... - 8x² + 8x – 5 

= x¹⁵ - ( x + 1 ) x ¹⁴ + ( x + 1 ) x ¹³ - ........- 8x² + 8x - 5 

= x ^ 15 - x ^15 -x^14 + x^14 + x ^ 13  - ........... - x^3 - x^2 + x - 5

= x - 5 = 7 -5 = 2

Khi x = 7 

=> x + 1 = 8

Khi đó B = x¹⁵ - 8x¹⁴ + 8x¹³ - ... - 8x² + 8x - 5

= x¹⁵ - (x + 1)x¹⁴ + (x + 1)x¹³ - ... - (x + 1)x² + (x + 1)x - 5

= x¹⁵ - x¹⁵ - x¹⁴ + x¹⁴ + x¹³ - ... - x³ - x² + x² + x - 5

= x - 5 = 7 - 5 = 2 

Ta có : x² + 2x - 3

= x² - x + 3x - 3

= x(x - 1) + 3(x - 1) 

= (x + 3)(x - 1) 

x2+2x-3 

= ( x + 1 ) ( x - 3 )

\(x^2-x-2x+2+4.\left(3x^2+9x+2x+6\right)=38\)

\(\Leftrightarrow x^2-3x+1+12x^2+44x+24=38\)

\(\Leftrightarrow13x^2+41x-13=0\)

\(\Leftrightarrow x^2+\frac{41}{13}x-1=0\)

\(\Leftrightarrow x^2+2.\frac{41}{26}x+\left(\frac{41}{26}\right)^2-1-\left(\frac{41}{26}\right)^2=0\)

\(\Leftrightarrow\left(x+\frac{41}{26}\right)^2=\frac{2357}{676}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{41}{26}=\frac{\sqrt{2357}}{26}\\x+\frac{41}{26}=-\frac{\sqrt{2357}}{26}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{2357}}{26}-\frac{41}{26}\\x=-\frac{\sqrt{2357}}{26}-\frac{41}{26}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{2357}-41}{26}\\x=\frac{-\sqrt{2357}-41}{26}\end{cases}}\)