Lần sau bạn chú ý viết đầy đủ yêu cầu của đề bài.

a: \(=\left(a+b\right)^2-\left(c+d\right)^2\)

b: \(=\left(a-d\right)^2-\left(b-c\right)^2\)

c: \(=\left(x+3z\right)^2-4y^2\)

d: \(=\left(a^2-9\right)\left(a^2+9\right)=a^4-81\)

e: \(=\left(a-5\right)^2\cdot\left(a+5\right)^2=\left(a^2-25\right)^2\)

\(a.A=5x-x^2\)

\(=-\left(x^2-5x\right)=-\left[\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\right]=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)

\(\Rightarrow Max_A=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)

\(b.B=x-x^2=-\left(x^2-x\right)=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\right]=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

\(\Rightarrow Max_B=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

\(c.C=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4-7\right)=-\left(x-2\right)^2+7\le7\)

\(\Rightarrow Max_C=7\Leftrightarrow x=2\)

a) Ta có:

\(A=5x-x^2\)

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

\(=-\left(x^2-5x\right)-6,25+6,25\)

\(=-\left(x^2-5x+6,25\right)+6,25\)

\(=-\left(x-2,5\right)^2+6,25\)

Ta lại có:

\(\left(x-2,5\right)^2\ge0\)

\(\Rightarrow-\left(x-2,5\right)^2\le0\)

\(\Rightarrow-\left(x-2,5\right)^2+6,25\le6,25\)

\(\Rightarrow A\le6,25\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-2,5\right)^2=0\)

\(\Leftrightarrow x-2,5=0\)

\(\Leftrightarrow x=2,5\)

Vậy Max_A = 6,25 \(\Leftrightarrow x=2,5\)

a: \(A=-3\left(x^2-2x+\dfrac{2}{3}\right)\)

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

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

Dấu '=' xảy ra khi x=1

b: \(B=-\left(16x^2+8x-4\right)\)

\(=-\left(16x^2+8x+1-5\right)\)

\(=-\left(4x+1\right)^2+5< =5\)

Dấu '=' xảy ra khi x=-1/4

d: \(x^2+2x+3=\left(x+1\right)^2+2>=2\)

=>E<=1/2

Dấu '=' xảy ra khi x=-1