a) \(81-\left(3x+2\right)^2=9^2-\left(3x+2\right)^2=\left(9-3x-2\right)\left(9+3x+2\right)=\left(7-3x\right)\left(11+3x\right)\)

b) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)

\(=\left(5x-5\right)\left(9x-3\right)=15\left(x-1\right)\left(3x-1\right)\)

c) \(9\left(x-5y\right)^2-16\left(x+y\right)^2=\left[3\left(x-5y\right)-4\left(x+y\right)\right]\left[3\left(x-5y\right)+4\left(x+y\right)\right]\)

\(=\left(-x-19y\right)\left(7x-11y\right)\)

`(2x-5)^2-64=(2x-5)^2-8^2=(2x-5+8)(2x-5-8)=(2x+3)(2x-13)`

k cho mik

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k cho mik

kb nữa nha

thanks

hok tốt

đây nhé

Ta có a(a + 1) + 1  = a² + a + 1 = \(a^2+2.\frac{1}{2}a+\frac{1}{4}+\frac{3}{4}=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)(đpcm) 

a, \(\left(2x-1\right)^2-\left(3x-1\right)^2=\left(2x-1-3x+1\right)\left(2x-1+3x-1\right)=-x\left(5x-2\right)\)

b, \(\left(x+1\right)^2-9=\left(x+1-3\right)\left(x+1+3\right)=\left(x-2\right)\left(x+4\right)\)

c, \(\left(4x-1\right)^2-9x^2=\left(4x-1-3x\right)\left(4x-1+3x\right)=\left(x-1\right)\left(7x-1\right)\)

d, \(x^2-9=\left(x-3\right)\left(x+3\right)\); e, \(x^2-25=\left(x-5\right)\left(x+5\right)\)

f, \(\left(x+2\right)^2-\left(3x-1\right)^2=\left(x+2-3x+1\right)\left(x+2+3x-1\right)=\left(-2x+3\right)\left(4x+1\right)\)

i, \(x^6-y^4=\left(x^3\right)^2-\left(y^2\right)^2=\left(x^3-y^2\right)\left(x^3+y^2\right)\)

bài nào hả bạn

Cái đè bài đâu ?

Cute thế.

a) Ta có x + y + z = 0

=> x + y = -z

=> (x + y)³ = (-z)³

=> x³ + y³ + 3xy(x + y) = -z³

=> x³ + y³ + z³ = -3xy(x + y) 

=> x³ + y³ + z³ = -3xy(-z)

=> x³ + y³ + z³ = 3xyz (đpcm) 

Trả lời:

\(-x^2+4x-5=-\left(x^2-4x+5\right)=-\left(x^2-4x+4+1\right)=-\left[\left(x-2\right)^2+1\right]\)

\(=-\left(x-2\right)^2-1\le-1< 0\forall x\)

Dấu "=" xảy ra khi x - 2 = 0 <=> x = 2

Vậy - x² + 4x - 5 < 0 với mọi x

Ta có : \(-x^2+4x-5=-\left(x^2-4x+5\right)=-\left[\left(x-2\right)^2+1\right]=-\left(x-2\right)^2-1\)

Vì ( x-2)² > 0 Với mọi x và 1 > 0

Nên \(-\left(x-2\right)^2-1< 0\forall x\)

Vậy.................

(x + 2)² - (x + 3)(x - 3) = 5

<=> x² + 4x + 4 - x² + 9 = 5

<=> 4x = -8

<=> x = -2

Trả lời:

( x + 2 )² - ( x + 3 ) ( x - 3 ) = 5

<=> x² + 4x + 4 - ( x² - 9 ) = 5

<=> x² + 4x + 4 - x² + 9 = 5

<=> 4x + 13 = 5

<=> 4x = - 8

<=> x = - 2

Vậy x = - 2 là nghiệm của pt.

Trả lời:

Bài 7:

a, \(A=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=" xảy ra khi x - 1 = 0 <=> x = 1

Vây GTNN của A = 4 khi x = 1

b, \(B=x^2-x+1=x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

Dấu "=" xảy ra khi x - 1/2 = 0 <=> x = 1/2

Vậy GTNN của B = 3/4 khi x = 1/2

c, \(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

\(=\left(x^2+6x-x-6\right)\left(x^2+3x+2x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-6^2=\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu "=" xảy ra khi \(x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

Vậy GTNN của C = - 36 khi x = 0; x = - 5

d, \(D=x^2+5y^2-2xy+4y+3=x^2+y^2+4y^2-2xy+4y+1+2\)

\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-y=0\\2y+1=0\end{cases}\Leftrightarrow\hept{x=y=-\frac{1}{2}}}\)

Vậy GTNN của D = 2 khi x = y = - 1/2

Bài 10. 

\(n\left(n+1\right)\left(n+2\right)\left(n+3\right)+1\)

\(=\left[n\left(n+3\right)\right]\left[\left(n+1\right)\left(n+2\right)\right]+1\)

\(=\left(n^2+3n\right)\left(n^2+3n+2\right)+1\)

\(=\left(n^2+3n+1-1\right)\left(n^2+3n+1+1\right)+1\)

\(=\left(n^2+3n+1\right)^2-1^2+1\)

\(=\left(n^2+3n+1\right)^2\)

Ta có đpcm.