\(5x^4y+10x^3y+10x^2y^3+5xy^4\)

\(=5xy.x^3+5xy.2x^3+5xy.2xy^3+5xy.y^3\)

\(=5xy\left(x^3+2x^3+2xy^3+y^3\right)\)

Ht pt

D = \(x^{10}-25x^9+25x^8-25x^7+...+25x^2-25x+25\)với x = 24

thiếu 1 câu

A= với x = 4

= (x+1)(x+1)(x+1)

= x5−x4+x4+x3−x3+x2−x2+x−1

=x−1=4−1=3

Tương tự với các câu B,C,D

=5x.x^2-5x.2x+5x.1 =5x(x^2-2x+1)

`1)x^4 -10x^3 +26x^2 -10x+1=0`
`x=0=>VT=1=>x=0(l)`
Chia 2 vế cho `x^2>0` ta có
`x^2-10x+26-10/x+1/x^2=0`
`=>x^2+1/x^2+26-10(x+1/x)=0`
`=>(x+1/x)^2-10(x+1/x)+24=0`
Đặt `a=x+1/x`
`pt<=>a^2-10a+24=0`
`<=>` $\left[ \begin{array}{l}a=4\\a=6\end{array} \right.$
`a=4<=>x+1/x=4<=>x^2-4x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt3+2\\x=-\sqrt3+2\end{array} \right.$
`a=6<=>x+1/x=6<=>x^2-6x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt8+3\\x=-\sqrt8+3\end{array} \right.$
Vậy `S={\sqrt3+2,-\sqrt3+2,\sqrt8+3,-\sqrt8+3}`

2)Do hệ số chẵn bằng=hệ số lẻ
`=>x=-1`
`pt<=>x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0`
`<=>(x+1)(x^3+4x^2+6x+9)=0`
`<=>(x+1)(x^3+3x^2+x^2+6x+9)=0`
`<=>(x+1)[x^2(x+3)+(x+3)^2]=0`
`<=>(x+1)(x+3)(x^2+x+3)=0`
Do `x^2+x+3=(x+1/2)^2+11/4>0`
`=>` $\left[ \begin{array}{l}x=-3\\x=-1\end{array} \right.$
Vậy `S={-1,-3}`

`-10x(2-x)-5x(x+2)=5x(x+3)`

`<=> -20x + 10x^2 - 5x^2 - 10x = 5x^2 +15x`

`<=> 5x^2 - 30x = 5x^2 + 15x`

`<=> -30x = 15x`

`<=> -45x = 0`

`<=> x = 0`

Vậy `S = {0}`

\(-10x\left(2-x\right)-5x\left(x+2\right)=5x\left(x+3\right)\)

\(\text{⇔}10x\left(x-2\right)+5x\left(x-2\right)=-5x\left(x-3\right)\)

\(\text{⇔}\left(x-2\right)\left(10x+5x\right)=-5x\left(x-3\right)\)

\(\text{⇔}15x\left(x-2\right)=-5x^2+15\)

\(\text{⇔}15x^2-30=-5x^2+15\)

\(\text{⇔}15x^2+5x^2=30+15\)

\(\text{⇔}20x^2=45\)

\(\text{⇔}x=\sqrt{\dfrac{45}{20}}=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

Ta có: \(-10x\left(2-x\right)-5x\left(x+2\right)=5x\left(x+3\right)\)

\(\Leftrightarrow-20x+10x^2-5x^2-10x-5x^2-15x=0\)

\(\Leftrightarrow-45x=0\)

hay x=0

4x⁴+y⁴

\(=\dfrac{x\left(x+1\right)}{5\left(x+1\right)^2}\cdot\dfrac{5x-1}{3\left(x+1\right)}=\dfrac{x\left(5x-1\right)}{15\left(x+1\right)^2}\)

\(\dfrac{x^2+x}{5x^2+10x+5}:\dfrac{3x+3}{5x-1}=\dfrac{x\left(x+1\right)}{5\left(x^2+2x+1\right)}:\dfrac{3\left(x+1\right)}{5x-1}=\dfrac{x\left(x+1\right)}{5\left(x+1\right)^2}.\dfrac{5x-1}{3\left(x+1\right)}=\dfrac{x\left(5x-1\right)}{15\left(x+1\right)^2}\)