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 1:

\(P=2a^2-2b^2-a^2+2ab-b^2+a^2+2ab+b^2+b^2=2a^2-b^2+4ab\\ Q=\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(2x-3\right)\left(2x+3\right)\\ Q=\left(2x+3-2x+3\right)^2=9^2=81\)

Bài 2:

\(Sửa:A=x^2+2xy+y^2-4x-4y+2=\left(x+y\right)^2-4\left(x+y\right)+4-2\\ A=\left(x+y-2\right)^2-2=\left(3-2\right)^2-2=1-2=-1\)

Bài 1:

           \(A=x^2-6x+13=\left(x-3\right)^2+4\ge4\)

Vậy  \(Min\)\(A=4\)\(\Leftrightarrow\)\(x=3\)

        \(B=2x^2+8x=2\left(x^2+4x+4\right)-8=2\left(x+2\right)^2-8\ge-8\)

Vậy  \(Min\)\(B=-8\)\(\Leftrightarrow\)\(x=-2\)

        \(C=4x^2+20x=\left(2x+5\right)^2-25\ge-25\)

Vậy  \(Min\)\(C=-25\)\(\Leftrightarrow\)\(x=-\frac{5}{2}\)

Bài 3:

a)   \(x^2+12x+39=\left(x+6\right)^2+3>0\) 

b)   \(4x^2+4x+3=\left(2x+1\right)^2+2>0\)

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a: Sửa đề: \(-x^3-12x^2-48x-64\)

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

\(=-\left(-6+4\right)^3=-\left(-2\right)^3=-\left(-8\right)=8\)

b: \(=8x^3-y^3-8x^3+27y^3=26y^3=26\cdot\left(-3\right)^3=-702\)

c: \(=-\left(4x^4-12x^2y+9y^2\right)\)

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

\(=-\left(2x^2-2x-11\right)^2\)

ko có biết