\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)

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

\(\Leftrightarrow\frac{a+b}{c}+\frac{c}{c}=\frac{b+c}{a}+\frac{a}{a}=\frac{a+c}{b}+\frac{b}{b}\)

\(\Leftrightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)

\(\Rightarrow a=b=c\) 

\(\Rightarrow\frac{a}{b}=1;\frac{b}{c}=1;\frac{c}{a}=1\)

\(\Rightarrow M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

a) a = 2 , b = 3, c = 6

Câu3: Ký hiệu [a,b] và (a,b) là gì ? Bạn.

Câu 1:

\(B=\frac{1}{199}+1+\frac{2}{198}+1+\frac{3}{197}+1+...+\frac{198}{2}+1+\frac{199}{1}+1-199\)

\(=\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+1=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+...+\frac{200}{2}\)

\(=200\cdot\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)=200\cdot A\)

Vậy, \(\frac{A}{B}=\frac{1}{200}\).

sửa lại

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

\(=a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\)

áp dụng bđt cauchy ta có:

\(b^2+1\ge2b;c^2+1\ge2c;a^2+1\ge2a\)

\(\Rightarrow a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\ge a-\frac{ab^2}{2b}+b-\frac{bc^2}{2b}+c-\frac{ca^2}{2a}\)

\(=a+b+c-\frac{ab+bc+ca}{2}\)

áp dụng cauchy ta có:

\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)

\(\Rightarrow a+b+c-\frac{ab+bc+ca}{2}\ge3-\frac{3}{2}=\frac{3}{2}\)

\(\Rightarrow\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\left(Q.E.D\right)\)

dấu bằng xảy ra khi a=b=c=1

đặt \(A=\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}=a-\frac{ab^2}{1+b^2}+b-\frac{bc^2}{1+c^2}+c-\frac{ca^2}{1+a^2}\)

\(=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\right)\le3-\left(\frac{ab^2}{2b}+\frac{bc^2}{2c}+\frac{ca^2}{2a}\right)=3-\left(\frac{ab+bc+ca}{2}\right)\ge3-\frac{\left(a+b+c\right)^2}{6}=\frac{3}{2}\left(Q.E.D\right)\)

Bài 1 :

\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}< 1\left(1\right)\)

\(B=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)\)\(>\frac{1}{10}+\frac{1}{100}.90=1\left(2\right)\)

Từ (1) và ( 2) ta có \(A< 1\) \(B>1\)NÊN \(A< B\)

Bài 2:

\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(=\frac{\left(a+b+c\right)-\left(b+c\right)}{b+c}+\)\(\frac{\left(a+b+c\right)-\left(c+a\right)}{c+a}\)\(+\frac{\left(a+b+c\right)-\left(a+b\right)}{a+b}\)

\(=\frac{7-\left(b+c\right)}{b+c}+\frac{7-\left(c+a\right)}{c+a}+\frac{7-\left(a+b\right)}{a+b}\)

\(=7.\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

\(=7.\frac{7}{10}-3\)\(=\frac{49}{10}-3=\frac{19}{10}\)

\(S=\frac{19}{10}>\frac{19}{11}=1\frac{8}{11}\)

Chúc bạn học tốt ( -_- )

Bài 1:

ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}< 1\)

\(\Rightarrow A< 1\)(1) 

ta có: \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};...;\frac{1}{99}>\frac{1}{100}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\) ( có 90 số 1/100)

                                                                               \(=\frac{90}{100}=\frac{9}{10}\)

\(\Rightarrow B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{10}+\frac{9}{10}=1\)

\(\Rightarrow B>1\)(2)

Từ (1);(2) => A<B

1, \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)

Do đó \(\left\{{}\begin{matrix}3a=b+c+d\left(1\right)\\3b=a+c+d\left(2\right)\\3c=a+b+d\left(3\right)\\3d=a+b+c\left(4\right)\end{matrix}\right.\)

Từ (1) và (2) \(\Rightarrow3\left(a+b\right)=a+b+2c+2d\Leftrightarrow2\left(a+b\right)=2\left(c+d\right)\Leftrightarrow a+b=c+d\Leftrightarrow\dfrac{a+b}{c+d}=1\)

Tương tự cũng có: \(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)

\(\Rightarrow A=4\)

2, Có \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)\(\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)

Do đó \(\dfrac{x^2}{4}=\dfrac{1}{4};\dfrac{y^2}{16}=\dfrac{1}{4};\dfrac{z^2}{36}=\dfrac{1}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(1;2;3\right),\left(-1;-2;-3\right)\)

Bài 2 :

a, Ta có : \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)

\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{1}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)

Vậy ...

b, Ta có : \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{5+7}=\dfrac{2x+3y-1}{6x}\)

\(\Rightarrow6x=12\)

\(\Rightarrow x=2\)

\(\Rightarrow y=3\)

Vậy ...

\(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

    \(=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1-3\)

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

    \(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

     \(=7.\frac{7}{10}-3=\frac{49}{10}-3=\frac{19}{10}\)

Ta có:\(1\frac{8}{11}=\frac{19}{11}< \frac{19}{10}\left(đpcm\right)\)

V...

Câu 2)

Ta có \(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{3}\)

\(\Rightarrow\frac{b+1+a+1}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)

Ta có \(a+b=1\)

\(\Rightarrow\frac{3}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)

\(\Rightarrow\frac{3}{\left(a+1\right)b+a+1}\ge\frac{4}{3}\)

\(\Rightarrow\frac{3}{ab+b+a+1}\ge\frac{4}{3}\)

Ta có \(a+b=1\)

\(\Rightarrow\frac{3}{ab+2}\ge\frac{4}{3}\)

\(\Leftrightarrow9\ge4\left(ab+2\right)\)

\(\Rightarrow9\ge4ab+8\)

\(\Rightarrow1\ge4ab\)

Do \(a+b=1\Rightarrow\left(a+b\right)^2=1\)

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

\(\Rightarrow a^2+2ab+b^2\ge4ab\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

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

Câu 3)

Ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)

Mà \(a+b+c=1\)

\(\Rightarrow\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\ge9\)

\(\Rightarrow a+b+c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

Áp dụng bất đẳng thức Cô-si

\(\Rightarrow\left\{\begin{matrix}a+b+c\ge3\sqrt[3]{abc}\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\end{matrix}\right.\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc}\sqrt[3]{\frac{1}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.\sqrt[3]{\frac{abc}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (điều này luôn luôn đúng)

\(\Rightarrow\) ĐPCM

a) 

\(A=\left(\frac{19}{24}-\frac{7}{24}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)\)

\(A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}\)

\(A=\frac{1}{3}\)

\(B=\left(\frac{7}{12}-\frac{5}{12}\right)+\left(\frac{5}{6}+\frac{1}{4}-\frac{3}{7}\right)\)

\(B=\left(\frac{1}{6}+\frac{5}{6}\right)+\frac{1}{4}-\frac{3}{7}\)

\(B=\frac{5}{4}-\frac{3}{7}\)

\(B=\frac{23}{28}\)

b)

\(x=A-B\)

\(x=\frac{1}{3}-\frac{23}{28}\)

\(x=\frac{-41}{84}\)