Question

B1. tìm x :

a. (x+3)³ -x(3x+1)² + (2x+1).(4x²-2x+1)-3x²=42

b. 5x(x+3)²-5(x+1)³+15(x+2)(x-2)=5

B2. tìm cặp x , y

x²(x+3)+y²(y+5)-(x+y)(x²-xy+y²)=0

Answer 1

Bài 1:

a) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\) (1)

\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=42\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=42\)

\(\Leftrightarrow26x+28=42\)

\(\Leftrightarrow26x=42-28\)

\(\Leftrightarrow26x=14\)

\(\Leftrightarrow x=\dfrac{7}{13}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{7}{13}\right\}\)

Answer 2

1b) \(5x\left(x+3\right)^2-5\left(x+1\right)^3+15\left(x+2\right)\left(x-2\right)=5\Leftrightarrow5x\left(x^2+6x+9\right)-5\left(x^3+3x^2+3x+1\right)+15\left(x^2-4\right)=5\Leftrightarrow30x-65=5\Leftrightarrow30x=70\Leftrightarrow x=\dfrac{7}{3}\)

Answer 3

2) \(x^2\left(x+3\right)+y^2\left(y+5\right)-\left(x+y\right)\left(x^2-xy+y^2\right)=0\Leftrightarrow x^3+3x^2+y^3+5y^2-x^3-y^3=0\Leftrightarrow3x^2+5y^2=0\)Do \(3x^2\ge0\) và \(5y^2\ge0\) => 3x²+5y²\(\ge\)0.Dấu "=" xảy ra khi x=y=0