-3x² + 6x + 297 = 0 

<=> -3x² + 6x - 3 + 300 = 0

<=> -3(x² - 2x + 1 - 100) = 0

<=> (x - 1)² - 10² = 0

<=> (x - 11)(x + 9) = 0

<=> \(\orbr{\begin{cases}x-11=0\\x+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-9\end{cases}}\)

b) (x + 1)² + 3(x - 1)² = 0

<=> x² + 2x + 1 + 3x² - 6x + 3 = 0

<=> 4x² - 4x + 4 = 0

<=> (2x - 1)² + 3 = 0 (vô lý) 

=> Phương trình vô nghiệm 

`99^3=(100-1)^3=100^3 - 3 . 100^2 . 1 +3 . 100 . 1^2 - 1^3 = 100 000 - 3 . 100 00 . 1 + 3 . 100 . 1 - 1 = 970299`

số 0 tròn chỉnh

2/11-2/11=0

~HT~

`(2a + b-5) (2a-b+5)`

`= [2a + (b-5)] [2a - (b-5)]`

`= (2a)^2 - (b-5)^2`

`= 4a^2 - (b^2 - 10b + 25)`

`= 4a^2 - b^2 +10b-25`

Trả lời:

\(\left(x-3\right)^3=x^3-3.x^2.3+3.x.3^2-3^3=x^3-9x^2+27x-27\)

...

D = (x² - y²)(x² + y²)(x⁴ + y⁴)(x⁸ + y⁸)

D = (x⁴ - y⁴)(x⁴ + y⁴)(x⁸ + y⁸)

D = (x⁸ - y⁸)(x⁸ + y⁸)

D = x¹⁶ - y¹⁶

`D = (x^2 - y^2)(x^2 + y^2)(x^4 + y^4)(x^8 + y^8)`

`->D = (x^4 - y^4)(x^4 + y^4)(x^8 + y^8)`

`->D = (x^8 - y^8)(x^8 + y^8)`

`->D=x^{16} - y^{16}`

Vậy `D=x^{16}-y^{16}`

Trả lời:

Bài 1:

a, \(4x^2-10x+7\)

\(=4x^2-10x+\frac{25}{2}+\frac{3}{4}\)

\(=\left(4x^2-10x+\frac{25}{2}\right)+\frac{3}{4}\)

\(=\left[\left(2x\right)^2-2.2x.\frac{5}{2}+\left(\frac{5}{2}\right)^2\right]+\frac{3}{4}\)

\(=\left(2x-\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

Vậy đpcm

b, \(2x-4x^2-1\)

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

\(=-\left[4x^2-2x+\frac{1}{4}+\frac{3}{4}\right]\)

\(=-\left[\left(2x\right)^2-2.2x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\right]\)

\(=-\left[\left(2x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\)

\(=-\left(2x-\frac{1}{2}\right)^2-\frac{3}{4}\le-\frac{3}{4}< 0\forall x\)

Vậy đpcm

c, \(4x^2+6x+13\)

\(=4x^2+6x+\frac{9}{4}+\frac{43}{4}\)

\(=\left(4x^2+6x+\frac{9}{4}\right)+\frac{43}{4}\)

\(=\left[\left(2x\right)^2+2.2x.\frac{3}{2}+\left(\frac{3}{2}\right)^2\right]+\frac{43}{4}\)

\(=\left(2x+\frac{3}{2}\right)^2+\frac{43}{4}\ge\frac{43}{4}\forall x\)

Dấu "=" xảy ra khi \(2x+\frac{3}{2}=0\Leftrightarrow x=-\frac{3}{4}\)

Vậy GTNN của bt bằng 43/4 khi x = - 3/4

d, \(-8x-x^2+3\)

\(=-\left(x^2+8x-3\right)\)

\(=-\left(x^2+8x+16-19\right)\)

\(=-\left[\left(x+4\right)^2-19\right]\)

\(=-\left(x+4\right)^2+19\le19\forall x\)

Dấu "=" xảy ra khi x + 4 = 0 <=> x = - 4

Vậy GTLN của bt bằng 19 khi x = - 4

Trả lời:

Bài 2:

a, \(3x^3-18x^2+27x=0\)

\(\Leftrightarrow3x\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow3x\left(x-3\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\\left(x-3\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)

Vậy x = 0; x = 3 là nghiệm của pt.

b, \(6x\left(x-3\right)=x^3-9x\)

\(\Leftrightarrow6x\left(x-3\right)-\left(x^3-9x\right)=0\)

\(\Leftrightarrow6x\left(x-3\right)-x\left(x^2-9\right)=0\)

\(\Leftrightarrow6x\left(x-3\right)-x\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[6x-x\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(6x-x^2-3x\right)=0\)

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}-x=0\\\left(x-3\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)

Vậy x = 0; x = 3 là nghiệm của pt.

c, \(4x^3-6x^2+2x=0\)

\(\Leftrightarrow4x\left(x^2-\frac{3}{2}x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow4x\left(x^2-2.x.\frac{3}{4}+\frac{9}{16}-\frac{1}{16}\right)=0\)

\(\Leftrightarrow4x\left[\left(x-\frac{3}{4}\right)^2-\frac{1}{16}\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x=0\\\left(x-\frac{3}{4}\right)^2-\frac{1}{16}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{3}{4}\right)^2=\frac{1}{16}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1;x=\frac{1}{2}\end{cases}}}\)

Vậy x = 0; x = 1; x = 1/2 là nghiệm của pt.