Question

[Trần Thị Ngọc Ánh]

bài 1 ;thực hiện phép tính

a,(x-1).(x+1).(x+2)

b,\(\dfrac{1}{2}x^2y^2\).(2x+y).(2x-y)

bài 2;tìm x hoặc y biết ;

a,2x.(x-5)-x(2x+3)=26

b,(3y²-y+1).(y-1)+y².(4-3y)-\(\dfrac{5}{2}\)=0

c;2x²+3(x-1).(x+1)=5x(x+1)

Answer 1

[Đời về cơ bản là buồn... cười!!!]

Bài 1:

a) \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)

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

\(=x^3+x-2\)

b) \(\dfrac{1}{2}x^2y^2\left(2x+y\right)\left(2x-y\right)\)

\(=\dfrac{1}{2}x^2y^2\cdot\left(4x^2-y^2\right)\)

\(=2x^4y^2-\dfrac{1}{2}x^2y^4\)

Answer 2

[Đời về cơ bản là buồn... cười!!!]

Bài 2:

a) \(2x\cdot\left(x-5\right)-x\left(2x+3\right)=26\)

\(\Rightarrow2x^2-10x-2x^2-3x=26\)

\(\Rightarrow-13x=26\)

\(\Rightarrow x=2\)

b) \(\left(3y^2-y+1\right)\cdot\left(y-1\right)+y^2\cdot\left(4-3y\right)-\dfrac{5}{2}=0\)

\(\Rightarrow3y^3-3y^2-y^2+y+y-1+4y^2-3y^3-\dfrac{5}{2}=0\)

\(\Rightarrow2y+\dfrac{7}{5}=0\)

\(\Rightarrow2y=-1,4\)

\(\Rightarrow y=-0,7\)

c) \(2x^2+3\left(x-1\right)\cdot\left(x+1\right)=5x\left(x+1\right)\)

\(\Rightarrow2x^2+3\left(x^2-1\right)=5x^2+5x\)

\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)

\(\Rightarrow5x^2-5x^2-5x=3\)

\(\Rightarrow-5x=3\)

\(\Rightarrow x=0,6\)