Question

1)Tìm x, bt

a.(2x + 1/3)^2 - (1- 2x)^2=16/9

b.(x+2)^3 - x.(x-1). (x+1)=6x^2+21

2)c/m:biểu thức sau đây ko phụ thuộc vào GT của biến

(x^2y - 3)^2 - (2x - y)^3 + xy^2.(6-x^3) + 8x^3 - 6xy - y^3

Answer 1

Bài 1:

a) \(\left(2x+\frac{1}{3}\right)^2-\left(1-2x\right)^2=\frac{16}{9}\)

\(\Leftrightarrow4x^2+\frac{4x}{3}+\frac{1}{9}-1+4x-4x^2-\frac{16}{9}=0\)

\(\Leftrightarrow\frac{16x-8}{3}=0\)

\(\Leftrightarrow16x-8=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

b) \(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2+21\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3+x-6x^2-21=0\)

\(\Leftrightarrow13x-13=0\)

\(\Leftrightarrow x=1\)

Bài 2:

\(\left(x^2y-3\right)^2-\left(2x-y\right)^3+xy^2\left(6-x^3\right)+8x^3-6xy-y^3\)

\(=x^4y^2-6x^2y+9-8x^3+12x^2y-6xy^2+y^3+6xy^2-x^4y^2+8x^3-6xy-y^3\)

\(=6x^2y-6xy+9\)

Sai đề ko bạn ?