\(D=\frac{1}{a^2+b^2-2a-2b+2}+\frac{1}{ab-a-b+1}+4\left(ab-a-b\right)\)

\(=\frac{1}{a^2+b^2-2a-2b+2}+\frac{1}{2ab-2a-2b+2}+\frac{1}{2\left(ab-a-b+1\right)}+4\left(ab-a-b\right)\)

\(\ge\frac{4}{a^2+b^2-4a-4b+2ab+4}+\frac{1}{2\left(ab-a-b+1\right)}+8\left(ab-a-b+1\right)-4\left(ab-a-b+1\right)-4\)

\(\ge\frac{4}{\left(a+b-2\right)^2}+2\sqrt{\frac{1}{2\left(ab-a-b+1\right)}.8\left(ab-a-b+1\right)}-4\left(ab-a-b+1\right)-4\)

\(\ge4+4-4\left(ab-a-b+1\right)-4\)

= 4 ( a + b ) - 4ab 

\(\ge\)4 ( a + b ) - (a + b )² - 4 + 4

=  - ( a + b - 2 )^2 + 4 

\(\ge\)3

Dấu "=" <=> a = b = 3/2

Đưa D về dạng: 

D = \(\frac{1}{\left(a-1\right)^2+\left(b-1\right)^2}+\frac{1}{\left(a-1\right)\left(b-1\right)}+4\left(a-1\right)\left(b-1\right)-4\)

\(\left(a-1\right)+\left(b-1\right)=a+b-2\le1\)

Đặt: a - 1 = x ; b - 1 = y => x + y \(\le\)1

=> \(D=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy-4\)

Tìm min D. Làm như này chắc nhanh hơn. Bạn thử xem nhé!

bạn không nên đưa những câu hỏi linh tinh,vớ  vẩn lên diễn đàn nha

Bài làm:

Ta có: \(C=\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\ge\frac{\left(1+2+3\right)^2}{x+y+z}\ge\frac{6^2}{1}=36\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x=\frac{1}{13}\\y=\frac{4}{13}\\z=\frac{9}{13}\end{cases}}\)

Bài này là áp dụng bđt Cauchy-Schwaz nha bạn.

huyen

\(B=3x+2y+\frac{6}{x}+\frac{8}{y}\)

\(=\frac{3x}{2}+\frac{6}{x}+\frac{3x}{2}+\frac{y}{2}+\frac{8}{y}+\frac{3y}{2}\)

Áp dụng Cauchy ta được :

\(\frac{3x}{2}+\frac{6}{x}\ge2\sqrt{\frac{3x}{2}.\frac{6}{x}}=6\)

\(\frac{y}{2}+\frac{8}{y}\ge2\sqrt{\frac{8y}{2y}}=4\)

\(\Rightarrow B\ge6+4+\frac{3\left(x+y\right)}{2}\ge6+4+9=19\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y=6\\\frac{y}{2}=\frac{8}{y}\\\frac{3x}{2}=\frac{6}{x}\end{cases}\Leftrightarrow x=2;y=4}\)

a¹⁰⁰⁰+\(\frac{1}{a}\)+\(\frac{1}{a}\)+....+\(\frac{1}{a}\) (có 1000 số hạng ) =1001
a⁹⁰⁰+\(\frac{1}{a}\)+...+\(\frac{1}{a}\)(900 số hạng )>=901
a⁹⁰+\(\frac{1}{a}\)+..\(\frac{1}{a}\) >=91
a⁵+\(\frac{1}{a}\)+..+\(\frac{1}{a}\) >=6
\(\Rightarrow\)A>=1001+901+91+6=1999.
" = " khi a¹⁰⁰⁰=\(\frac{1}{a}\); a⁹⁰⁰=\(\frac{1}{a}\) ;a⁹⁰=\(\frac{1}{a}\) và a⁵=\(\frac{1}{a}\)

#Nhi Tiểu Cừu

Mọi người giúp mình nha.

\(P=\frac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

\(=-\frac{2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=\frac{-3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}\)

Trả lời:

\(P=\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(P=\frac{2\sqrt{8}-2\sqrt{3}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(P=\frac{2.\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}.\left(\sqrt{3}-\sqrt{8}\right)}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}.\left(\sqrt{5}+\sqrt{27}\right)}\)

\(P=\frac{-2.\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}.\left(\sqrt{3}-\sqrt{8}\right)}-\frac{1}{\sqrt{6}}\)

\(P=\frac{-2}{\sqrt{6}}-\frac{1}{\sqrt{6}}\)

\(P=\frac{-3}{\sqrt{6}}\)

\(P=\frac{-\sqrt{6}}{2}\)

Học tốt