b) Ta có:

 P(x) + H(x) = x⁴ - x³ + 2x² + x + 1

=> H(x) = x⁴ - x³ + 2x² + x + 1 - P(x)

=> H(x) = (x⁴ - x³ + 2x² + x + 1) - (2x⁴ - x² + x - 2)

=> H(x) = -x⁴ - x³ + 3x² + 3

Vậy H(x) = -x⁴ - x³ + 3x² + 3

`P(x)=\(4x^2+x^3-2x+3-x-x^3+3x-2x^2\)

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

`= 2x^2+3`

`Q(x)=`\(3x^2-3x+2-x^3+2x-x^2\)

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

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

`P(x)-Q(x)-R(x)=0`

`-> P(X)-Q(x)=R(x)`

`-> R(x)=P(x)-Q(x)`

`-> R(x)=(2x^2+3)-(-x^3+2x^2-x+2)`

`-> R(x)=2x^2+3+x^3-2x^2+x-2`

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

`= x^3+x+1`

`@`\(\text{dn inactive.}\)

a: P(x)-Q(x)-R(x)=0

=>R(x)=P(x)-Q(x)

=2x^2+3+x^3-2x^2+x-2

=x^3+x+1

Mình nghĩ đề đổi lại: \(\left(x-3\right)\left(x^3+3x+9\right)\rightarrow\left(x-3\right)\left(x^2+3x+9\right)\)

\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)

\(\Leftrightarrow x^3-27+4x-x^3=1\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

a, Không rõ đề bạn ơi ;-;

b, ĐKXĐ : \(x\ge0\)

Ta có : \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}\right)=4-x=\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)=\left(\sqrt{x}-2\right)\left(2+\sqrt{x}\right)\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-5-\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\sqrt{x}-2=0\)

\(\Leftrightarrow x=4\) ( TM )

Vậy ...

`b)(sqrtx-2)(5-sqrtx)=4-x`

`đk:0<=x`

`pt<=>(sqrtx-2)(sqrtx-5)=x-4`

`<=>x-7sqrtx+10=x-4`

`<=>7sqrtx=14`

`<=>sqrtx=2`

`<=>x=4(tmđk).`

   Ta có :

\(x\div\frac{2}{3}+x\times\frac{3}{2}+x=2\)

\(x\times\frac{3}{2}+x\times\frac{3}{2}+x=2\)

\(x\times(\frac{3}{2}+\frac{3}{2}+1)=2\)

\(x\times4=2\)

\(x=2\div4\)

\(x=\frac{1}{2}\)

  Vậy x = \(\frac{1}{2}\)

\(x\div\frac{2}{3}+x\times\frac{3}{2}+x=2\)

\(x\times\frac{3}{2}+x\times\frac{3}{2}+x=2\)

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

\(x\times4=2\)

\(x=\frac{1}{2}\)

\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)

\(\Leftrightarrow-5x-18=0\)

\(\Leftrightarrow x=-\dfrac{18}{5}\)

Vậy ...

\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)

\(\Leftrightarrow12x+6=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy ...

\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)

\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)

Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)

\(\Rightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy ...

\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)

\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)

Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy ...

sao lại trả lời lại nhỉ ??