\(x^2+5x+7=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\)\(\Rightarrow\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy GTNN của bt trên = 3/4 <=> x = - 5/2

Trả lời :

\(x^2+5x+7=x^2+2.\frac{5}{2}x+\frac{28}{4}=\left(x^2+2.\frac{5}{x}+\frac{25}{4}\right)+\frac{3}{4}\)

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

Mà \(\left(x+\frac{5}{2}\right)^2\ge0\Rightarrow\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra <=> \(\left(x+\frac{5}{2}\right)^2=0\Rightarrow x=\frac{-5}{2}\)

Vậy GTNN của biểu thức là \(\frac{3}{4}\Leftrightarrow x=\frac{-5}{2}\)

a. \(\left(2+xy\right)^2=x^2y^2+4xy+4\)

b. \(\left(5-x^2\right)\left(5+x^2\right)=25+5x^2-5x^2-x^4=-x^4+25\)

c. \(\left(2x-y\right)\left(4x^2+2xy+y^2\right)=8x^3+4x^2y+2xy^2-4x^2y-2xy^2-y^3\)

\(=8x^3-y^3\)

d. \(\left(5-3x\right)^2=25-30x+9x^2\)

e. \(\left(5x-1\right)^3=125x^3-75x^3+15x-1\)

f. \(\left(x+3\right)\left(x^2-3x+9\right)=x^3-3x^2+9x+3x^2-9x+27=x^3+27\)

h. \(\left(2x^2+3y\right)^2=4x^4+12x^2y+9y^2\)

a) (2+xy)² = 2²+4xy+(xy)² = 4 + 4xy +x²y²

b)  ( 5 - x^2 ) . ( 5 + x^2 ) = 5²-x⁴=25-x⁴

c) ( 2x - y ) . ( 4x^2 + 2xy + y^2 )  = 8x³-y³

d)(5-3x)²=5²-2.5.3x+9x²=25-30x+9x²

e) (5x-1)³=(5x)³-3.(5x)².1+3.5x.1-1 =125x³-75x²+15x-1

f) (x+3)(x²-3x+9)=(x+3)(x²-3x+3²)=x³+27

g) -x³+3x²-3x+1 =(−x+1)(x−1)(x−1)= -(x-1)³

h) (2x²+3y)²=4x⁴+2.2x².3y+9y²=4x⁴+12x²y+9y²

\(A=2x\left(x^2-4x\right)-x^2\left(3x-2\right)+2x^2+7\)

\(\Leftrightarrow A=2x^3-8x^2-3x^3+2x^2+2x^2+7\)

\(\Leftrightarrow A=-x^3-4x^2+7\)

\(B=3x^2\left(x^3-2x+5\right)-x^3\left(3x^2+6x-7\right)+6x^4-8x^3\)

\(\Leftrightarrow B=3x^5-6x^3+15x^2-3x^5-6x^4+7x^3+6x^4-8x^3\)

\(\Leftrightarrow B=-7x^3+15x^2\)

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

Bài làm:

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

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

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

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

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

( x² + y² )² - ( 2xy )² 

= ( x² + y² - 2xy )( x² + y² + 2xy )

= ( x - y )²( x + y )²

=> đpcm

Bài làm:

Ta có: \(\left(a+b\right)^2=4\)

\(\Leftrightarrow a^2+2ab+b^2=4\)

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

Thay vào ta tính được:

\(a^3+b^3+6ab\)

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

\(=2\left(4-2ab-ab\right)+6ab\)

\(=8-6ab+6ab=8\)

=> đpcm

Ta có a + b = 2 

=> (a + b)³ = 8

=> a³ + b³ + 3a²b + 3ab² = 8

=> a³ + b³ + 3ab(a + b) = 8

=> a³ + b³ + 3ab.2 = 8

=> a³ + b³ + 6ab = 8 (đpcm)

( a² - 2a + 3 )( a² + 2a - 3 )

= [ a² - ( 2a - 3 ) ][ a² + ( 2a - 3 ) ]

= ( a² )² - ( 2a - 3 )²

= a⁴ - ( 4a² -  12a + 9 )

= a⁴ - 4a² + 12a - 9 

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

\(=a^4+2a^3-3a^2-2a^3-4a^2+6a+3a^2+6a-9\)

\(=a^4-4a^2+12a-9\)