(2y-5)(4y^2+10y+25)=2y3-125

(2-y)^3=-y^3+6y^2-12y+8

(3y+4)(9y^2-12y+16)=27y^3+64

(x-3)^3+(2-x)^3= -3x^2+15x-19

(x+y)^3-(x-y)^3= 2y^3+6x^2y

\(x^2+10x+y^2-2y+26+\left(3z-6\right)^2=0\)

\(\Leftrightarrow x^2+10x+25+y^2-2y+1+\left(3z-6\right)^2=0\)

\(\Leftrightarrow\left(x+5\right)^2+\left(y-1\right)^2+\left(3z-6\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x+5=0\\y-1=0\\3z-6=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=1\\z=2\end{cases}}\)

\(\frac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}=\frac{x^2\left(x+3\right)^2-1}{x^4+6x^3+9x^2-2x^2-6x+1}=\)

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

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

I don't know. Mong bạn thông càm, hehehe

a,2x²+8x+20=2(x²+4x)+20

=2(x²+4x+4)+20-4.2

=2(x+2)²+12

Ta có : 2(x+2)² ≥0với∀x≥0với∀x

12 > 0

⇒⇒2(x+2)²+12>0 với ∀x∀x

⇒⇒2x²+8x+20>0 với ∀∀x

b,x⁴-3x²+5

=(x⁴-3x²)+5

=(x⁴-2.3232x²+9494)+5-9494

=(x²-3232)²+114114

Có : (x²-3/2)²≥0với∀x≥0với∀x

114114>0

⇒⇒(x²-3232)²+114>0với∀x114>0với∀x

Thu gọn

\(16-x-x-65-10-x-x-24\)24

Là đáp án chính xác

HT

\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)=0\)

\(\Rightarrow4\left(x^2+2x.1+1^2\right)+[\left(2x\right)^2-2.2x.1+1^2]-8\left(x^2-1^2\right)=0\)

\(\Rightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=0\)

\(\Rightarrow4x+13=0\)

\(\Rightarrow4x=-13\)

\(\Rightarrow x=\frac{-13}{4}\)