\(A=\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\)

\(A=1+\frac{1}{b}+\frac{1}{a}+\frac{1}{ab}\)

\(A=1+\frac{a+b}{ab}+\frac{a+b}{ab}\)

\(A=1+\frac{2}{ab}\)

Ta có : 

\(\left(a-b\right)^2\ge0\)

\(\Rightarrow a^2+b^2\ge2ab\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

\(\Rightarrow ab\le\frac{1}{4}\)

\(\Rightarrow A\ge1+\frac{2}{\frac{1}{4}}=9\)

" = " \(\Leftrightarrow a=b=0,5\)

Chúc bạn học tốt !!!

A= 4x² -4x +1 -4

= (2x -1)²- 2²

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

=(2x-1).(2x-3)

= \(\left(ab+a\right)+\left(b+1\right)\) 

= \(a\left(b+1\right)+\left(b+1\right)\)

= \(\left(a+1\right)\left(b+1\right)\)

chuc ban hoc tot

\(ab+1+a+b\)

\(=a\left(b+1\right)+\left(b+1\right)\)

\(=\left(a+1\right)\left(b+1\right)\)

(x + 3)^2 + (4 + x)(4 - x) = 10

x^2 + 6x + 9 + 16 - x^2 = 10

6x + 25 = 10

6x = 10 - 25

6x = -15

x = -15/6 = -5/2

 Ta có ( x+3)^2 +(4+x)(4-x) =10

   => x^2 + 6x +9 +16- x^2 =10

   => 6x+25 =10

   => 6x= -15

   => x=-5/2

Vậy x=-5/2

Đặt H \(=x^4-5x^3+7x^2-6\)

Gỉa sử : \(H=\left(x^2+ax+b\right)\left(x^2+cx+d\right)\)

                   \(=x^4+cx^3+dx^2+ax^{3\:}+acx^2+adx+bx^2+bcx+bd\)

                      \(=x^4+\left(a+c\right)x^3+\left(ac+b+d\right)x^2+\left(ad+bc\right)x+bd\)

       \(\Leftrightarrow\hept{\begin{cases}a+c=-5\\ac+b+d=7\\ad+bc=0\end{cases}}\)

                 \(\left\{bd=6\right\}\)

           \(\Leftrightarrow\hept{\begin{cases}a=-3\\b=3\\c=-2\end{cases}}\)

                   \(\left\{d=-2\right\}\)

\(\Rightarrow H=\left(x^2-3x+3\right)\left(x^2-2x-2\right)\)

Chúc bạn học tốt !!!

Bình phương ba vế suy ra \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)

Sau đó chứng minh tương tự bunhiacopxki

4x(x-2y)+8y(2y-x)

=4x(x-2y)-8y(x-2y)

=(4x-8y)(x-2y)

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

=4(x-2y)^2

\(4x\left(x-2y\right)+8y\left(2y-x\right)\)

\(=\left(x-2y\right)\left(4x-8y\right)\)

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

\(\left(x+1\right)^4+\left(x^2+x+1\right)^2\)

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

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

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

\(5x+6y=4\Rightarrow x=\frac{4-6y}{5}\)

\(P=\left(\frac{4-6y}{5}\right)^2+2y^2=\frac{16-48y+36y^2+50y^2}{25}=\frac{86y^2-48y+16}{25}\)

      \(=\frac{86\left(y^2-\frac{24}{43}y+\frac{8}{43}\right)=86\left(y^2-2.y.\frac{12}{43}+\frac{144}{1849}\right)+\frac{400}{43}}{25}\)

           \(=\frac{86\left(y-\frac{12}{43}\right)^2+\frac{400}{43}}{25}\ge\frac{400}{\frac{23}{25}}=\frac{16}{43}\)

\(Min_P=\frac{12}{43}\Leftrightarrow x=\frac{20}{43};y=\frac{12}{43}\)

Chúc bạn học tốt !!!

Ta có :
\(\sqrt{6-x^2}\le\sqrt{6}\)

\(\Rightarrow-2\sqrt{6-x^2}\ge-2\sqrt{6}\)

\(\Rightarrow5-2\sqrt{6-x^2}\ge5-2\sqrt{6}\)

\(\Rightarrow A=\frac{1}{5-2\sqrt{6-x^2}}\le\frac{1}{5-2\sqrt{6}}=5+2\sqrt{6}\)

\(Max_A=5+2\sqrt{6}\Leftrightarrow x=0\)

Chúc bạn học tốt !!!