Bài làm:

Ta có: \(C=3x^2-6x-1\)

\(C=3\left(x^2-2x-\frac{1}{3}\right)\)

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

\(C=3\left(x-1\right)^2-4\ge-4\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(3\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy \(Min_C=-4\Leftrightarrow x=1\)

C = 3x² - 6x - 1

= 3( x² - 2x + 1 ) - 4

= 3( x - 1 )² - 4

\(3\left(x-1\right)^2\ge0x\Rightarrow\forall3\left(x-1\right)^2-4\ge-4\)

Đẳng thức xảy ra <=> x - 1 = 0 => x = 1

=> MinC = -4 <=> x = 1

Đặt \(\hept{\begin{cases}x=a+b\\y=b+2c\\z=c+2a\end{cases}\Rightarrow x+y+z=3a+2b+3c}\)

Khi đó biểu thức đã cho trở thành :

\(\left(x+y+z\right)^3=x^3+y^3+z^3+450\)

\(\Leftrightarrow x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(y+z\right)=x^3+y^3+z^3+450\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=90\)

\(\Leftrightarrow\left(a+2b+2c\right)\left(b+3c+2a\right)\left(3a+c+b\right)=90\) 

Phân tích 90 thành tích của 3 số nguyên dương rồi bạn tìm được \(a,b,c\) tương ứng.

lô mày

Bài làm:

Ta có: \(3x^2+x-1\)

\(=3\left(x^2+\frac{1}{3}x-\frac{1}{3}\right)\)

\(=3\left(x^2+\frac{1}{3}x+\frac{1}{36}\right)-\frac{13}{12}\)

\(=3\left(x+\frac{1}{6}\right)^2-\frac{13}{12}\ge-\frac{13}{12}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(3\left(x+\frac{1}{6}\right)^2=0\Rightarrow x=-\frac{1}{6}\)

Vậy \(Min=-\frac{13}{12}\Leftrightarrow x=-\frac{1}{6}\)

It's khai triển :)

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

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

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

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

e) \(x^2-2x+9=\left(x-1\right)^2+8??\) ko ra gì cả-.-

g) \(\left(x+1\right)\left(x-1\right)=x^2-1\)

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

i) \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)

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

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

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

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

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

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

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

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