Question

Tìm x, biết

a)   (x+1)³-x(x-2)²+x-1=0

b)   (x-2)³-x²(x-6)=4

Cám ơn trước nha~~ ^.^

Answer 1

(x+1)^3-x(x-2)^2+x-1=0

⇔x^3+3x^2+3x+1-x(x^2-4x+4)+x-1=0

⇔x^3+3x^2+3x+1-x^3+4x^2-4x+x-1=0

⇔7x^2=0

⇔x^2=0

⇔x=0

Vậy x=0

b,(x-2)^3-x^2(x-6)=4

⇔x^3-6x^2+12x-8-x^3+6x^2=4

⇔12x-8=4

⇔12x=12

⇔x=1

Vậy x=1

Answer 2

trả lời

a, x=1

chúc bn 

học tốt

Answer 3

\(a,(x+1)^3-x(x-2)^2+x-1=0\)

\(\Rightarrow x^3+3x^2+3x+1-x^3-4x^2+4x+x-1=0\)

\(\Rightarrow(x^3-x^3)+(3x^2-4x^2)+(3x+4x)+1+x-1=0\)

\(\Rightarrow-x^2+7x+1+x-1=0\)

\(\Rightarrow-x^2+7x+1+x=1\)

\(\Rightarrow-x^2+7x+x+1=1\)

\(\Rightarrow-x^2+8x+1=1\)

\(\Rightarrow-x^2+8x=0\)

\(\Rightarrow-x(x+8)=0\)

\(\Rightarrow\orbr{\begin{cases}-x=0\\x+8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-8(ktm)\end{cases}}\)

Vậy x = 0

Answer 4

\(b,\)\(\left(x-2\right)^3-x^2\left(x-6\right)=4\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+6x^2=4\)

\(\Rightarrow12x-8=4\)

\(\Rightarrow12x=12\)

\(\Rightarrow x=1\)