đề bài sai rồi

Ta cóA=a³+a²-b³+b²+ab-3ab(a-b+1)

=(a³-b³)+(a²+ab+b²)-24ab(do a-b=7)

=(a-b)(a²+ab+b²)+(a²+ab+b²)-24ab

=(a²+ab+b²)(a-b+1)-24ab

mà a-b=7=>A=8a²+8ab+8b²-24ab

=8a²-16ab+8b²

=8(a-b)²=8 . 7²=8 . 49=392

1.

a) \(\frac{16}{24}-\frac{1}{3}=\frac{16}{24}-\frac{8}{24}=\)\(\frac{8}{24}=\frac{1}{3}\)

b) \(\frac{4}{5}-\frac{12}{60}=\frac{48}{60}-\frac{12}{60}=\frac{36}{60}=\frac{9}{15}\)

3.

a)\(\frac{17}{6}-\frac{2}{6}=\frac{17-2}{6}=\frac{15}{6}\)

b) \(\frac{16}{15}-\frac{11}{15}=\frac{16-11}{15}=\frac{5}{15}=\frac{1}{3}\)

c) \(\frac{19}{12}-\frac{13}{12}=\frac{19-13}{12}=\frac{6}{12}=\frac{1}{2}\)

bđt \(\Leftrightarrow\)\(\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)\ge\left(\frac{10}{3}\right)^3abc\) (*) 

đặt \(\left(\sqrt{ab};\sqrt{bc};\sqrt{ca}\right)=\left(x;y;z\right)\)\(\Rightarrow\)\(xyz\le\frac{1}{27}\)

(*) \(\Leftrightarrow\)\(\left(x^2+1\right)\left(y^2+1\right)\left(z^2+1\right)\ge\left(\frac{10}{3}\right)^3xyz\)

\(VT\ge\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\)

Có \(xy+1\ge10\sqrt[10]{\frac{xy}{9^9}}\)

Tương tự với \(yz+1\)\(;\)\(zx+1\)\(\Rightarrow\)\(VT\ge10^3\sqrt[10]{\frac{\left(xyz\right)^2}{9^{27}}}\)

Ta cần CM \(10^3\sqrt[10]{\frac{\left(xyz\right)^2}{9^{27}}}\ge\frac{10^3}{3^3}xyz\) đúng với \(xyz\le\frac{1}{27}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

Đặt \(P=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)\)

Vì a+b+c=1 nên 

\(P=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)=abc+\frac{1}{abc}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\)

Từ BĐt Cosi cho 3 số dương ta có:

\(\frac{1}{3}=\frac{a+b+c}{3}\ge\sqrt[3]{abc}\Rightarrow abc\le\frac{1}{27}\)

đặt x=abc thì \(0< x\le\frac{1}{27}\)

do đó: \(x+\frac{1}{x}-27-\frac{1}{27}=\frac{\left(27-x\right)\left(1-27x\right)}{27x}\ge0\)

=> \(x+\frac{1}{x}=abc+\frac{1}{abc}\ge27+\frac{1}{27}=\frac{730}{27}\)

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

Nên  \(P\ge\frac{730}{27}+10=\frac{1000}{27}=\left(\frac{10}{3}\right)^3\)

Dấu "=" xảy ra khi a=b=c\(=\frac{1}{3}\)

a)\(\frac{42}{15.7}\)= \(\frac{2.3.7}{3.5.7}\)= \(\frac{2}{5}\)

b) \(\frac{35.6}{336}\)= \(\frac{5.7.6}{6.7.8}\)= \(\frac{5}{8}\)

c) \(\frac{4.33}{11.12}\)= \(\frac{4.3.11}{11.3.4}\)= 1

d) \(\frac{9.4+9.11+5.9}{63}\)= \(\frac{9.\left(4+11+5\right)}{9.7}\)= \(\frac{20}{7}\)

#Hk tốt nhé

\(\frac{42}{17.7}\)

\(=\frac{6}{17.1}\)

\(=\frac{6}{17}\)

Bài 1 : 

\(a)\) Ta có : 

\(3x=4y=6z\)

\(\Leftrightarrow\)\(\frac{3x}{12}=\frac{4y}{12}=\frac{6z}{12}\)

\(\Leftrightarrow\)\(\frac{x}{4}=\frac{y}{3}=\frac{z}{2}\)

\(\Leftrightarrow\)\(\frac{2x}{8}=\frac{y}{3}=\frac{5z}{10}\)

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

\(\frac{2x}{8}=\frac{y}{3}=\frac{5z}{10}=\frac{2x-5z}{8-10}=\frac{-36}{-2}=18\)

Do đó : 

\(\frac{x}{4}=18\)\(\Rightarrow\)\(x=18.4=72\)

\(\frac{y}{3}=18\)\(\Rightarrow\)\(y=18.3=54\)

\(\frac{z}{2}=18\)\(\Rightarrow\)\(z=18.2=36\)

Vậy \(x=72\)\(;\)\(y=54\) và \(z=36\)

Chúc bạn học tốt ~ 

2) Ta có: \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)

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

\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2.\left(a+b+c\right)}=\frac{1}{2}\)

\(\Rightarrow\frac{a}{b+c}=\frac{1}{2}\Rightarrow2a=b+c\)

\(\frac{b}{c+a}=\frac{1}{2}\Rightarrow2b=c+a\)

\(\frac{c}{a+b}=\frac{1}{2}\Rightarrow2c=a+b\)

Ta có: \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{b+a}{b}.\frac{c+b}{c}.\frac{a+c}{a}=\frac{2c.2a.2b}{b.c.a}=8\)

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