`@` `\text {Ans}`

`\downarrow`

`(-x^4 - x^3) + (x^4 + 2x^3 + 5x^2 + 3x) + (-5x^2 - 3x - x^3)`

`= -x^4 - x^3 + x^4 + 2x^3 + 5x^2 + 3x - 5x^2 - 3x - x^3`

`= (-x^4+x^4) + (-x^3 + 2x^3 - x^3) + (5x^2 - 5x^2) + (3x - 3x)`

`= 0 + 0 + 0 + 0`

`= 0`

Vậy, giá trị của biểu thức trên không phụ thuộc vào giá trị của biến.

`@` `\text {Kaizuu lv uuu}`

\(\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot3^6}{2^5\cdot2^6\cdot3^5}=\dfrac{1}{2^4}\cdot3=\dfrac{3}{16}\)

a) \(P\left(x\right)=3x^3-2x+2x^2+7x+8-x^4)\)

   \(P\left(x\right)=3x^3(-2x+7x)+2x^2+8-x^4)\)

   \(P\left(x\right)=3x^3+5x+2x^2+8-x^4)\)

   \(P\left(x\right)=-x^4+3x^3+2x^2+5x+8\)

  \(Q\left(x\right)=2x^2-3x^3+3x^2-5x^4\)

  \(Q\left(x\right)=(2x^2+3x^2)-3x^3-5x^4\)

  \(Q\left(x\right)=5x^2-3x^3-5x^4\)

  \(Q\left(x\right)=-5x^4-3x^2+5x^2\)

b)

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

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

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

Vậy: \(R\left(x\right)\) \(=5x+7x^2+8-6x^4\)

c. \(R\left(x\right)\) \(=5x+7x^2+8-6x^4\)

\(=5x+7x^2+4+4-6x^4\)

\(=\) \((12x-4)^2+4\ge4-6x^4\)

Câu c MIK KHÔNG CHẮC LÀ ĐÚNG 

ko có biết

a: \(P\left(x\right)=-5x^3+3x^2+2x+5\)

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

b: \(H\left(x\right)=Q\left(x\right)+P\left(x\right)=-10x^3+9x^2+3x+10\)

Khi x=1/2 thì \(H\left(x\right)=-10\cdot\dfrac{1}{8}+\dfrac{9}{4}+\dfrac{3}{2}+10=\dfrac{25}{2}\)

\(A=x^2-20x+101\)

\(A=x^2-2\cdot x\cdot10+100+1\)

\(A=\left(x-10\right)^2+1\ge1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=10\)

___

\(B=4a^2+4a+2\)

\(B=4a^2+4a+1+1\)

\(B=\left(2a+1\right)^2+1\ge1\forall a\)

Dấu "=" xảy ra \(\Leftrightarrow a=\frac{-1}{2}\)

___

\(C=x^2-4xy+5y^2+10x-22y+28\)

\(C=x^2-4xy+4y^2+y^2+10x-22y+28\)

\(C=\left(x-2y\right)^2+2\cdot\left(x-2y\right)\cdot5+25+y^2-2y+1+2\)

\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

___

\(D=4x-x^2+3\)

\(D=-\left(x^2-4x-3\right)\)

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

\(D=-\left[\left(x-2\right)^2-7\right]\)

\(D=7-\left(x-2\right)^2\le7\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=2\)

___

\(E=x-x^2\)

\(E=-\left(x^2-x\right)\)

\(E=-\left(x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\right)\)

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

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

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2}\)

a, \(A=x^2-20x+101=x^2-2.x.10+10^2+1\)

\(=\left(x-10\right)^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-10\right)^2=0\)

\(\Leftrightarrow x-10=0\)

\(\Leftrightarrow x=10\)

Vậy : \(A_{min}=1\Leftrightarrow x=10\)

b) \(B=4a^2+4a+2=\left(2a\right)^2+2.2a.1+1^2+1\)

\(=\left(2a+1\right)^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow\left(2a+1\right)^2=0\)

\(\Leftrightarrow2a+1=0\)

\(\Leftrightarrow2a=-1\)

\(\Leftrightarrow a=-\frac{1}{2}\)

Vậy : \(B_{min}=1\Leftrightarrow x=-\frac{1}{2}\)

a: \(P\left(x\right)=-5x^3+3x^2+2x+5\)

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

b: \(H\left(x\right)=P\left(x\right)+Q\left(x\right)=-10x^3+9x^2+4x+10\)

\(H\left(\dfrac{1}{2}\right)=-10\cdot\dfrac{1}{8}+\dfrac{9}{4}+2+10=13\)

c: Q(x)-P(x)=6

\(\Leftrightarrow3x^2=6\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

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