Bài 1:ta có BĐt \(a^3+b^3\ge ab\left(a+b\right)\)vì nó tương đương với \(\left(a+b\right)\left(a-b\right)^2\ge0\)(luôn đúng với a,b>0)

Áp dụng vào bài toán:

\(\dfrac{a^3+b^3}{2ab}+\dfrac{b^3+c^3}{2bc}+\dfrac{c^3+a^3}{2ac}\ge\dfrac{ab\left(a+b\right)}{2ab}+\dfrac{bc\left(b+c\right)}{2bc}+\dfrac{ca\left(c+a\right)}{2ac}=a+b+c\)dấu = xảy ra khi a=b=c

bài 2:

cần chứng minh \(\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge0\)

hay \(\dfrac{a-b}{b+c}+1+\dfrac{b-c}{c+d}+1+\dfrac{c-d}{d+a}+1+\dfrac{d-a}{a+b}+1\ge4\)

\(\Leftrightarrow\dfrac{a+c}{b+c}+\dfrac{b+d}{c+d}+\dfrac{c+a}{d+a}+\dfrac{d+b}{a+b}\ge4\)

xét \(VT=\left(a+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+d}\right)+\left(b+d\right)\left(\dfrac{1}{c+d}+\dfrac{1}{a+b}\right)\)

Áp dụng BĐT cauchy dạng phân thức:

\(\dfrac{1}{b+c}+\dfrac{1}{a+d}\ge\dfrac{4}{a+b+c+d};\dfrac{1}{c+d}+\dfrac{1}{a+b}\ge\dfrac{4}{a+b+c+d}\)

do đó \(VT\ge\dfrac{4\left(a+c\right)}{a+b+c+d}+\dfrac{4\left(b+d\right)}{a+b+c+d}=4\)

dấu = xảy ra khi a=b=c=d

`3/4 + 5/6 = 9/12 + 10/12 = 19/12`

`1/2 + 7/12 = 6/12 + 7/12 = 13/12`

`2/3 xx 3/4 = 2/4 = 1/2`

`7/4 : 2 = 7/4 xx 1/2 = 7/8`

\(a,\dfrac{3}{4}+\dfrac{5}{6}=\dfrac{18}{24}+\dfrac{20}{24}=\dfrac{38}{24}=\dfrac{19}{12}\)

\(b,\dfrac{1}{2}+\dfrac{7}{12}=\dfrac{6}{12}+\dfrac{7}{12}=\dfrac{13}{12}\)

\(c,\dfrac{2}{3}x\dfrac{3}{4}=\dfrac{2}{4}\)

\(d,\dfrac{7}{4}:2=\dfrac{7}{4}x\dfrac{1}{2}=\dfrac{7}{8}\)

day la cac tinh chat ma

a/

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{c^3}{d^3}\)

Áp dụng tỉ lệ thức ta có:

\(\frac{a^3}{b^3}=\frac{c^3}{d^3}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^3+b^3}{c^3+d^3}=\frac{a^3-b^3}{c^3-d^3}\)

Vậy \(\frac{a^3+b^3}{c^3+d^3}=\frac{a^3-b^3}{c^3-d^3}\)

a) Ta có: \(\dfrac{39}{-65}=\dfrac{-39}{65}=\dfrac{-39:13}{65:13}=\dfrac{-3}{5}\)

\(\dfrac{-3}{5}=\dfrac{-3}{5}\)

Do đó: \(\dfrac{-3}{5}=\dfrac{39}{-65}\)

b) Ta có: \(\dfrac{-9}{27}=\dfrac{-9:9}{27:9}=\dfrac{-1}{3}\)

\(\dfrac{-41}{123}=\dfrac{-41:41}{123:41}=\dfrac{-1}{3}\)

Do đó: \(\dfrac{-9}{27}=\dfrac{-41}{123}\)

c) Ta có: \(\dfrac{-3}{4}=\dfrac{-3\cdot5}{4\cdot5}=\dfrac{-15}{20}\)

\(\dfrac{4}{-5}=\dfrac{-4}{5}=\dfrac{-4\cdot4}{5\cdot4}=\dfrac{-16}{20}\)

mà \(\dfrac{-15}{20}>\dfrac{-16}{20}\)

nên \(\dfrac{-3}{4}>\dfrac{4}{-5}\)

d) Ta có: \(\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot7}{3\cdot7}=\dfrac{-14}{21}\)

\(\dfrac{-5}{7}=\dfrac{-5\cdot3}{7\cdot3}=\dfrac{-15}{21}\)

mà \(\dfrac{-14}{21}>\dfrac{-15}{21}\)

nên \(\dfrac{2}{-3}>\dfrac{-5}{7}\)

1. \(a=\dfrac{1}{3};b=1\)

2. \(a=\dfrac{1}{4};b=1;c=-1\)

3. \(b=1;c=\dfrac{1}{4};d=1\)

4. \(a=1;b=1;c=1;d=1\)

Áp dụng công thức : 

\(l=\dfrac{\pi Rn}{180}=\dfrac{\pi.4.30^o}{180^o}=\dfrac{2}{3}\pi cm\\ =>B\)

\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=t\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=t^3\\\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}=t^3\end{matrix}\right.\)

Ta có đpcm

Ta có :

\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)

\(c^2=bd\Leftrightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}\)

Áp dụng t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(3\right)\)

Lại có :

\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(4\right)\)

Từ \(\left(3\right)+\left(4\right)\Leftrightarrowđpcm\)

Bài 1:

Giả sử \(a\ge b\ge c \ge d \ge e\)

\(\Leftrightarrow ab+bc+cd+de \leq a.(b+c+d+e)\)

\(\Leftrightarrow ab+bc+cd+de \leq a.(1-a)\)

\(\Leftrightarrow ab+bc+cd+de \leq -(a-\frac{1}{2})^2 + \frac{1}{4}\)

Đẳng thức xảy ra khi có ít nhất 2 số bằng 0 thì 2 số còn lại bằng \(\frac{1}{2}\) giả sử \(a=b=\dfrac{1}{2};c=d=0\)

Bài 2:

\(BDT\LeftrightarrowΣ\dfrac{3\left(a+b\right)^2+\left(a-b\right)^2}{\left(a-b\right)^2}\ge9\)

\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{\left(a-b\right)^2}+\dfrac{\left(b+c\right)^2}{\left(b-c\right)^2}+\dfrac{\left(c+a\right)^2}{\left(c-a\right)^2}\ge2\)

\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{\left(a-b\right)^2}-1+\dfrac{\left(b+c\right)^2}{\left(b-c\right)^2}-1+\dfrac{\left(c+a\right)^2}{\left(c-a\right)^2}-1\ge-1\)

\(\Leftrightarrow\dfrac{4ab}{\left(a-b\right)^2}+\dfrac{4bc}{\left(b-c\right)^2}+\dfrac{4ca}{\left(a-c\right)^2}\ge-1\)

\(\Leftrightarrow\dfrac{3ab}{\left(a-b\right)^2}+\dfrac{3bc}{\left(b-c\right)^2}+\dfrac{3ca}{\left(a-c\right)^2}\ge-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{3ab}{\left(a-b\right)^2}+1+\dfrac{3bc}{\left(b-c\right)^2}+1+\dfrac{3ca}{\left(a-c\right)^2}+1\ge3-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{a^2+ab+b^2}{\left(a-b\right)^2}+\dfrac{b^2+bc+c^2}{\left(b-c\right)^2}+\dfrac{c^2+ac+c^2}{\left(a-c\right)^2}\ge\dfrac{9}{4}\)

\(\Leftrightarrow\dfrac{a^3-b^3}{\left(a-b\right)^3}+\dfrac{b^3-c^3}{\left(b-c\right)^3}+\dfrac{c^3-a^3}{\left(a-c\right)^3}\ge\dfrac{9}{4}\)(Đúng)

P/s: Ok, xong tưởng dễ ai dè ngốn mất 2 tiếng

mình nghĩ bài 1 P=ab+bc+cd+de thôi bn à cơ sở dựa vào bài Câu hỏi của Mai Thành Đạt - Toán lớp 8 | Học trực tuyến, và 1 số chỗ