Trả lời:

\(I=x^4-6x^3+11x^2-12x+20\)

\(=x^4-6x^3+9x^2+2x^2-12x+18+2\)

\(=\left(x^4-6x^3+9x^2\right)+\left(2x^2-12x+18\right)+2\)

\(=\left[\left(x^2\right)^2-2.x^2.3x+\left(3x\right)^2\right]+2\left(x^2-6x+9\right)+2\)

\(=\left(x^2-3x\right)^2+2\left(x-3\right)^2+2\ge2\forall x\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x^2-3x=0\\x-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0;x=3\\x=3\end{cases}\Leftrightarrow}\hept{x=3}}\)

Vậy GTNN của I = 2 khi x = 3

\(A=x^4-6x^3+10x^2-6x+9\)

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

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

\(=\left(x^2-3x\right)^2+\left(x-3\right)^2\ge0\forall x\)

Dấu "=" xảy ra khi x = 3  (giống ý trên)

Vậy GTNN của A = 0 khi x = 3

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

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

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

Để \(f\left(x\right)>0\Leftrightarrow\left(x^2+3\right)\left(x-1\right)>0\)

Mà \(x^2\ge0\forall x\Leftrightarrow x^2+3>0\)

\(\Rightarrow x-1>0\Leftrightarrow x=1\)

\(h\left(x\right)=4x^3-14x^2+6x-21< 0\)

\(\Leftrightarrow0\left(x-\frac{7}{2}\right)\left(4x^2+6\right)< 0\)

Mà \(4x^2+6>0\forall x\Leftrightarrow h\left(x\right)< 0\Leftrightarrow x-\frac{7}{2}< 0\Leftrightarrow x< \frac{7}{2}\)

f(x)=x3−x2+3x−3f(x)=x3−x2+3x−3

=x2(x−1)+3(x−1)=x2(x−1)+3(x−1)

=(x2+3)(x−1)=(x2+3)(x−1)

Để f(x)>0⇔(x2+3)(x−1)>0f(x)>0⇔(x2+3)(x−1)>0

Mà x2≥0∀x⇔x2+3>0x2≥0∀x⇔x2+3>0

⇒x−1>0⇔x=1⇒x−1>0⇔x=1

h(x)=4x3−14x2+6x−21<0h(x)=4x3−14x2+6x−21<0

⇔0(x−72)(4x2+6)<0⇔0(x−72)(4x2+6)<0

Mà 4x2+6>0∀x⇔h(x)<0⇔x−72<0⇔x<72

5x ( x - 2 ) = x - 2 

5x ( x - 2 ) - ( x - 2 ) = 0 

( x - 2 ) ( 5x - 1 ) = 0 

x - 2 = 0 hoặc 5x - 1 = 0 

x = 2 hoặc x = 1/5 

các bạn làm giúp mình luôn cái

(2x+3y)²=2x²+12xy+3y²

(5x-y)²=25x²-10xy+y²

(x+1/4)²=x²+1/2x+1/16

(1/3x-1/2y)²=1/9x²-1/3xy+1/4y²

(3x+1)(3x-1)=3x²-1

(x²+2/5y)(x²-2/5y)=x⁴-4/25y²

\(2\left(a-b\right)\left(c-b\right)+2\left(b-a\right)\left(c-a\right)+2\left(b-c\right)\left(a-c\right)\)

\(=2\left(ac-ab-bc+b^2\right)+2\left(bc-ab-ac+a^2\right)+2\left(ab-bc-ac+c^2\right)\)

\(=2ac-2ab-2bc+2b^2+2bc-2ab-2ac+2a^2+2ab-2bc-2ac+2c^2\)

\(=2a^2+2b^2+2c^2-2ab-2bc-2ac\)

\(=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\)

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

Trả lời:

2x - 5 4x^2 - 14x + a 2x - 2 4x^2 - 10x - - 4x + a - 4x + 10 - a - 10

Để A chia hết cho B thì a - 10 = 0 <=> a = 10

Vậy để đa thức A chia hết cho đa thức B thì a = 10

Trả lời:

( x + 3 )² + ( x - 2 )² = 2x² 

<=> x² + 6x + 9 + x² - 4x + 4 = 2x² 

<=> 2x² + 2x + 13 = 2x² 

<=> 2x² + 2x + 13 - 2x² = 0

<=> 2x + 13 = 0

<=> 2x = - 13

<=> x = - 13/2

Vậy x = - 13/2 là nghiệm của pt.