Question

bài 1:

a) tìm y biết:

(y - 1)³ + 3(y+1)² = (y² - 2y +4) (y+2)

b)tìm x.y.z biết

x² - 2x+ y² + 4y + 5+(2z-3)² =0

c) tìm x

4x² - 12x +9=(5-x)²

bài 2 chứng minh đẳng thức

(a+b+c)³ = a³ + b³ +c³ +3(a+b)(b+c)(c+a)

Answer 1

Câu 1: 

a: \(\left(y-1\right)^3+3\left(y+1\right)^2=\left(y+2\right)\left(y^2-2y+4\right)\)

\(\Leftrightarrow y^3-3y^2+3y-1+3y^2+6y+3=y^3+8\)

\(\Leftrightarrow y^3+9y+2=y^3+8\)

=>9y=6

hay y=2/3

b: \(x^2-2x+y^2+4y+5+\left(2z-3\right)^2=0\)

\(\Leftrightarrow x^2-2x+1+y^2+4y+4+\left(2z-3\right)^2=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=\dfrac{3}{2}\end{matrix}\right.\)

c: \(4x^2-12x+9=\left(5-x\right)^2\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(x-5\right)^2=0\)

\(\Leftrightarrow\left(2x-3-x+5\right)\left(2x-3+x-5\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(3x-8\right)=0\)

hay \(x\in\left\{-2;\dfrac{8}{3}\right\}\)