Question

Bài 1. Chứng minh rằng :

a) ( x+y).(x³-x²y+xy²+x³) = x⁴-y⁴

b) tÌM x:

1, (x-3).(x-2)-(x+10.(x-5)=0

2, (2x-1).(3-x)+(x-2).(x+3)= (1-x).(x-2)

MONG MỌI NGƯỜI GIÚP ĐỠ Ạ ! :))

Answer 1

Bài 1

Em xem lại đề nhé

a. Ta có VP=\(x^4-y^4=\left(x^2\right)^2-\left(y^2\right)^2\)

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

\(=\left(x+y\right)\left(x^3+xy^2-x^2y-y^3\right)\)

\(=VT\)

b. 

1.\(\left(x-3\right)\left(x-2\right)-\left(x+10\right)\left(x-5\right)=0\)

\(\Leftrightarrow x^2-5x+6-\left(x^2+5x-50\right)=0\)

\(\Leftrightarrow-10x=-56\Rightarrow x=\frac{56}{10}\)

2.\(\left(2x-1\right)\left(3-x\right)+\left(x-2\right)\left(x+3\right)=\left(1-x\right)\left(x-2\right)\)

\(=-2x^2+7x-3+x^2+x-6=-x^2+3x-2\)

\(\Leftrightarrow5x=7\Leftrightarrow x=\frac{7}{5}\)

Answer 2

Em cảm ơn chị ạ! :))