Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)

Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3

\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)

\(\frac{2}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\)\(\Leftrightarrow\)\(\frac{1}{a^2b}\le\frac{2}{3a^3}+\frac{1}{3b^3}\)

\(\Rightarrow\)\(\Sigma\frac{a^2}{b^5}\ge\Sigma\left(\frac{5}{3b^3}-\frac{2}{3a^3}\right)=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)

a) 1 + 1 - 2 = 2 - 2 = 0

b) 2 + 6 - 2 = 8 - 2 = 6

c) 5 + 1 - 4 = 6 - 4 = 2

d) 3 + 5 - 2 = 8 - 2 = 6

a) 1 + 1 - 2 = 0.

b) 2 + 6 - 2 = 6.

c) 5 + 1 - 4 = 2.

d) 3 + 5 - 2 = 6.

Tick cho mình đi bạn!

tth_new             

\(a^3+b^3+c^3=\left(a+b+c\right)^3\)nha !

Học tốt !

toán lớp 1 gì mà ảo diệu quá...

\(\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}=\frac{a^4}{ab+ac}+\frac{b^4}{ab+bc}+\frac{c^4}{ac+bc}\)

\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}\)

\(=\frac{a^2+b^2+c^2}{2}=\frac{1}{2}\)

Dạng này dùng hệ số bât định làm gì cho mệt?

ơ. tự trả lời luôn r còn đâu :|

What??!!!!!!!

Đây là bài toán lớp 1 ???

Bn có nhầm ko z??

đoán xem. :))

Hoàng Lê Bảo Ngọc            alibaba nguyễn Thắng Nguyễn giup e vs

Dốt

bài 1 a, hình như có thêm đk là a+b+c=3

Bài 4 nha

Áp dụng BĐT cô si ta có

\(\frac{1}{x^2}+x+x\ge3\sqrt[3]{\frac{1}{x^2}.x.x}=3.\)

Tương tự với y . \(A\ge6\)dấu = xảy ra khi x=y=1

Vì a,b,c,d có vai trò như nhau

Giả sử \(a\ge b\ge c\ge d\)

=>\(a^2\ge b^2\ge c^2\ge d^2\)

=>\(\frac{1}{a^2}\le\frac{1}{b^2}\le\frac{1}{c^2}\le\frac{1}{d^2}\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\le\frac{1}{d^2}+\frac{1}{d^2}+\frac{1}{d^2}+\frac{1}{d^2}\)

=>\(1\le4.\frac{1}{d^2}\)

=>\(\frac{1}{4}\le\frac{1}{d^2}\)

=>\(4\ge d^2\)

=>\(2\ge d\)

Vì d là số tự nhiên khác 0

=>d=1,2

-Xét d=1

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}=1\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{1^2}=1\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1=1\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=0\)

Vì\(\frac{1}{a^2}>0,\frac{1}{b^2}>0,\frac{1}{c^2}>0=>\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}>0\)

=>Vô lí

-Xét d=2

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{2^2}=1\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{4}=1\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{3}{4}\)

Vì \(a\ge b\ge c\)

=>\(a^2\ge b^2\ge c^2\)

=>\(\frac{1}{a^2}\le\frac{1}{b^2}\le\frac{1}{c^2}\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{c^2}\)

=>\(\frac{3}{4}\le3.\frac{1}{c^2}\)

=>\(\frac{1}{4}\le\frac{1}{c^2}\)

=>\(4\ge c^2\)

=>\(2\ge c\)

Vì \(c\ge d=>c\ge2\)

=>c=2

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{3}{4}\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{2^2}=\frac{3}{4}\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{4}=\frac{3}{4}\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}=\frac{2}{4}\)

Vì \(a\ge b\)

=>\(a^2\ge b^2\)

=>\(\frac{1}{a^2}\le\frac{1}{b^2}\)

=>\(\frac{1}{a^2}+\frac{1}{b^2}\le\frac{1}{b^2}+\frac{1}{b^2}\)

=>\(\frac{2}{4}\le\frac{2}{b^2}\)

=>\(\frac{1}{4}\le\frac{1}{b^2}\)

=>\(4\ge b^2\)

=>\(2\ge b\)

Vì \(b\ge c=>b\ge2\)

=>b=2

=>\(\frac{1}{a^2}+\frac{1}{b^2}=\frac{2}{4}\)

=>\(\frac{1}{a^2}+\frac{1}{2^2}=\frac{2}{4}\)

=>\(\frac{1}{a^2}+\frac{1}{4}=\frac{2}{4}\)

=>\(\frac{1}{a^2}=\frac{1}{4}\)

=>\(a^2=4=>a=2\)

Vậy a=2,b=2,c=2,d=2