Đề sai rồi bạn

a)  \(x^2+4x+4=\left(x+2\right)^2\)

b)  \(9x^2-12x+4=\left(3x-2\right)^2\)

c)  \(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)

d)  kiểm tra lại đề nhé

e)  \(4y^2-9x^2=\left(2x-3x\right)\left(2x+3x\right)\)

f)  \(9y^2-\frac{1}{4}=\left(3y-\frac{1}{2}\right)\left(3y+\frac{1}{2}\right)\)

g)  \(8x^3+8a^3=\left(2x+2a\right)\left(4x^2-4ax+4a^2\right)\)

h)  \(64x^3-27y^3=\left(4x-3y\right)\left(9y^2+12xy+16x^2\right)\)

Cảm ơn bạn nhìu !!

Đặt bthuc = A nhé

ĐKXĐ : \(2x\ne3y\)

\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)

\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)

\(=\dfrac{\left(2x-3y\right)^3}{\left(2x-3y\right)^2}=2x-3y\)

Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3

  bài đây nhé mn nhanh nhanh giúp mik với

a: (x-1/x)^2=x^2-2+1/x^2

b: (1/2x+1/3y)(1/4x^2-1/6xy+1/9y^2)=1/8x^3+1/27y^3

a) \(x^2+4x+4=\left(x+2\right)^2\)

b) \(x^2-8x+16=\left(x-4\right)^2\)

c) \(\left(x+5\right)\left(x-5\right)=x^2-25\)

d) \(x^2+2x+1=\left(x+1\right)^2\)

e) \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)

f) \(\left(2x+3y\right)^2+2\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)

g) \(\left(2+bx^2\right)\left(bx^2-2\right)=\left(bx^2+2\right)\left(bx^2-2\right)=\left(bx^2\right)^2-4=b^2x^4-4\)