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:

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

a)\({-1\over 2}x^2×y^2 - x^2×y^2 +{2\over 3} x^2×y^2 \)

=\(({ -1\over 2}-1+{ 2\over 3})x^2×y^2\)

=\({-5 \over 6}x^2×y^2\)

b)\({1 \over 2}a^3×b^2 +{4 \over 3}3ab^2 × {1 \over 2}a^2\)

=\({1 \over 2}a^3×b^2 +({4 \over 3}× {1 \over 2})3b^2 (a×a^2) \)

=\({1 \over 2}a^3×b^2 +{2 \over 3}3a^3b^2\)

=\(({1 \over 2} +{2 \over 3}3)a^3b^2\)

=\({5 \over 2}a^3b^2\)

c)

a) Ta có: \(A=x^2-2x+5\)

\(=x^2-2x+1+4\)

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

c) Ta có: \(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

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

\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu '=' xảy ra khi x(x+5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

d) Ta có: \(x^2+5y^2-2xy+4y+3\)

\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)

\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)

a)đặt A=\(x^2+5y^2-2xy+4y+3\)

 \(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)

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

ta thấy GTNN của A =2 khi x=y=-1/2

a) \(3x^2-3xy-5x+5y\)

\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)

\(=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x-5\right)\)

b) \(2x^3y-2xy^3-4xy^2-2xy\)

\(=2xy\left(x^2-y^2-2y-1\right)\)

\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)

\(=2xy\left[x^2-\left(y+1\right)^2\right]\)

\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)

c) \(x^2+1+2x-y^2\)

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

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

\(=\left(x+1+y\right)\left(x+1-y\right)\)

d) \(x^2+4x-2xy-4y+y^2\)

\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)

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

\(=\left(x-y\right)\left(x-y+4\right)\)

e) \(x^3-2x^2+x\)

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

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

f) \(2x^2+4x+2-2y^2\)

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

\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)

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

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

a: =3x(x-y)-5(x-y)

=(x-y)(3x-5)

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)

\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

=(x-y)^2+4(x-y)

=(x-y)(x-y+4)

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

=2[(x+1)^2-y^2]

=2(x+1+y)(x+1-y)

\(1.\)

\(a;A=-2x^2+4x-18\)

\(A=-2\left(x^2-4x+18\right)\)

\(A=-2\left(x^2-2.x.2+4+14\right)\)

\(A=-2\left(x-2\right)^2-14\le-14\)

Dấu = xảy ra khi : \(x-2=0\)

                              \(\Rightarrow x=2\)

Vậy A_max =-14 tại x = 2

Các câu còn lại lm tương tự........