Bài 6:

a) Ta có: \(A=-x^2+6x-11\)

\(=-\left(x^2-6x+11\right)\)

\(=-\left(x-3\right)^2-2\le-2\forall x\)

Dấu '=' xảy ra khi x=3

b) Ta có: \(B=-x^2-8x+5\)

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

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Dấu '=' xảy ra khi x=-4

c) Ta có: \(C=-x^2+4x+1\)

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

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

\(=-\left(x-2\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=2

Bài 7:

a) Ta có: \(x^2-6x+11\)

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

\(=\left(x-3\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x=3

a: A=x^2-6x+9+2=(x-3)^2+2>=2

Dấu = xảy ra khi x=3

b: B=x^2-20x+100+1=(x-10)^2+1>=1

Dấu = xảy ra khi x=10

d: C=x^2-16x+8+3

=(x-4)^2+3>=3

Dấu = xảy ra khi x=4

\(P_1=\frac{3x^2+6x+10}{x^2+2x+3}\)

      \(=3+\frac{1}{x^2+2x+3}\)

Lại có: \(x^2+2x+3\)

          \(=\left(x+1\right)^2+2\ge2\)

\(\Rightarrow P_1\le3+\frac{1}{2}=\frac{7}{2}\)

Dấu = xảy ra khi x=-1

P₂ tương tự

Ta có: \(A=\frac{3x^2+6x+11}{x^2+2x+3}=3+\frac{2}{x^2+2x+3}=3+\frac{2}{\left(x+1\right)^2+2}\)

Đặt \(B=\frac{2}{\left(x+1\right)^2+2}\),để A đạt giá trị lớn nhất thì B lớn nhất.

Mà B lớn nhất khi \(\left(x+1\right)^2+2\) bé nhất. 

Lại có: \(\left(x+1\right)^2\ge0\forall x\Rightarrow\left(x+1\right)^2+2\ge2\) (1)

Từ (1) suy ra: \(B\le\frac{2}{2}=1\Rightarrow A=3+B\le3+1=4\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Vậy \(A_{max}=4\Leftrightarrow x=-1\)

Bài 2:

a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

hay \(x=\dfrac{2}{7}\)

b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow x^3=-8\)

hay x=-2

Bài 1: 

a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)

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

\(=xy\)

=1

b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)

\(=x^2y^2+x^3y^2+x^2-x^3y^2-x^2y^2-y^2\)

\(=x^2-y^2\)

\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)

Bài 2 :

\(A=4x^2-2.2x.2+4+1\)

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

Thấy : \(\left(2x-2\right)^2\ge0\)

\(A=\left(2x-2\right)^2+1\ge1\)

Vậy \(MinA=1\Leftrightarrow x=1\)

\(B=\left(5x\right)^2-2.5x.1+1-4\)

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

Thấy : \(\left(5x-1\right)^2\ge0\)

\(\Rightarrow B=\left(5x-1\right)^2-4\ge-4\)

Vậy \(MinB=-4\Leftrightarrow x=\dfrac{1}{5}\)

\(C=\left(7x\right)^2-2.7x.2+4-5\)

\(=\left(7x-2\right)^2-5\)

Thấy : \(\left(7x-2\right)^2\ge0\)

\(\Rightarrow C=\left(7x-2\right)^2-5\ge-5\)

Vậy \(MinC=-5\Leftrightarrow x=\dfrac{2}{7}\)

\(1.\)

\(A=-x^2-10x+1=-\left(x^2+10x-1\right)\)

\(=-\left(x^2+2.5x+5^2-5^2-1\right)=-\left[\left(x+5\right)^2-26\right]\)

\(=-\left(x+5\right)^2+26\le26\) dấu "=" xảy ra<=>x=-5

\(B=-4x^2-6x-5=-4\left(x^2+\dfrac{6}{4}x+\dfrac{5}{4}\right)\)

\(=-4\left(x^2+2.\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{11}{16}\right)\)\(=-4\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{6}\right]\le-\dfrac{11}{4}\)

\(C=-16x^2+8x-1=-16\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)

\(=-16\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)=-16\left(x-\dfrac{1}{4}\right)^2\le0\)

dấu"=" xảy ra<=>x=1/4

ĐKXĐ: \(x\ge1;y\ge25\)

\(D=\frac{1}{x}\sqrt{\frac{x-1}{\left(x-2\right)^2+25}}+\frac{1}{y}\sqrt{\frac{y-25}{\left(y-50\right)^2+1}}\)

Vì x>=1,y>=25 => x-1>=0,y-25>=0 

=> D >= 0

Dấu "=" xảy ra <=> x=1,y=25

Vậy MinD=0 khi x=1,y=25

Ta có: \(\left(x-2\right)^2+25\ge25;\left(y-50\right)^2+1\ge1\)

=>\(\frac{1}{x}\sqrt{\frac{x-1}{\left(x-2\right)^2+25}}\le\frac{1}{x}\sqrt{\frac{x-1}{25}};\frac{1}{y}\sqrt{\frac{y-25}{\left(y-50\right)^2+1}}\le\frac{1}{y}\sqrt{y-25}\)

=>\(D\le\frac{1}{x}\sqrt{\frac{x-1}{25}}+\frac{1}{y}\sqrt{y-25}\)

Vì x>=1 => x-1>=0. Áp dụng bđt cosi với 2 số dương x-1 và 1 ta có:

\(\sqrt{x-1}=\sqrt{\left(x-1\right).1}\le\frac{x-1+1}{2}=\frac{x}{2}\)

=>\(\frac{1}{x}\sqrt{\frac{x-1}{25}}\le\frac{1}{x}\cdot\frac{x}{2}\cdot\frac{1}{\sqrt{25}}=\frac{1}{10}\)

Vì y>=25 => y-25>=0. ÁP dụng bđt cô si cho 2 số dương 25 và y-25 ta có:

\(\sqrt{y-25}=\frac{\sqrt{25\left(y-25\right)}}{5}\le\frac{25+y-25}{2.5}=\frac{y}{10}\)

=>\(\frac{1}{y}\sqrt{y-25}=\frac{1}{y}\cdot\frac{y}{10}=\frac{1}{10}\)

Suy ra \(D\le\frac{1}{10}+\frac{1}{10}=\frac{1}{5}\)

Dấu "=" xảy ra <=> x=2,y=50

Vậy MaxD = 1/5 khi x=2,y=50