\(\sqrt{\frac{y+1}{\left(x+2\right)^2}}.\sqrt{\frac{\left(x-2\right)^6}{y^2-2y+1}}\)

\(=\frac{\sqrt{y+1}}{x+2}\sqrt{\frac{\left(x-2\right)^6}{\left(y-1\right)^2}}\)

\(=\frac{\sqrt{y+1}}{x+2}.\frac{\left(x-2\right)^6}{y-1}\)

\(=\frac{-\left(x-2\right)^5\sqrt{y+1}}{y-1}\)

a: \(A\left(\dfrac{1}{2}\right)=-2\cdot\dfrac{1}{8}+3\cdot\dfrac{1}{4}+5=\dfrac{11}{2}\)

\(A\left(1\right)=-2+3+5=6\)

\(A\left(-1\right)=2+3+5=10\)

\(A\left(0\right)=-2\cdot0+3\cdot0+5=5\)

\(A\left(-3\right)=-2\cdot\left(-27\right)+3\cdot9+5=86\)

b: Khi x=2 và y=1 thì 

\(B=-3\cdot8\cdot1+2\cdot4-2\cdot2=-20\)

Khi x=-2 và y=1 thì

\(B=-3\cdot\left(-8\right)\cdot1+2\cdot4-2\cdot\left(-2\right)=36\)

\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}-\dfrac{3}{2}+1=\)

\(=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{3}{2}+1=-\dfrac{3}{2}\)

= 4 . -1/8 - 2 . -1/4 + 3 . -1/2 + 1

= -1/2 - -1/2 + -3/2 + 1

= -1/2

Đề có vẻ thiếu điều kiện để tìm min. Bạn xem lại.

Công thức tổng quát:

\(1-\frac{1}{n^2}=\left(\frac{n-1}{n}\right)\left(\frac{n+1}{n}\right)\)

Do đó:

\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{12^2}\right)\)

\(A=\frac{1}{2}.\frac{3}{2}.\frac{2}{3}.\frac{4}{3}.\frac{3}{4}.\frac{5}{4}....\frac{11}{12}.\frac{13}{12}\)

\(A=\frac{1}{2}.\frac{13}{12}=\frac{13}{24}\)

tk mk , mk tk 

\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\)

\(=\sqrt{3}+1+\sqrt{3}-1\)

\(=2\sqrt{3}\)

mũ 2 với căn lớn bên ngoài sẽ triệt tiêu cho nhau
=\(\sqrt{3}+1+1-\sqrt{3}=2\)