\(\dfrac{a}{b}=2\Rightarrow\dfrac{a}{2}=b\Rightarrow\dfrac{a}{6}=\dfrac{b}{3}\)

\(\dfrac{b}{c}=3\Rightarrow\dfrac{b}{3}=c\)

\(\Rightarrow\dfrac{a}{6}=\dfrac{b}{3}=c=k\\ \Rightarrow a=6k;b=3k;c=k\)

\(\dfrac{a+b}{b+c}=\dfrac{6k+3k}{3k+k}=\dfrac{9k}{4k}=\dfrac{9}{4}\)

Đặt \(\left\{{}\begin{matrix}\dfrac{a}{b^2}=x\\\dfrac{b}{c^2}=y\\\dfrac{c}{a^2}=z\end{matrix}\right.\Rightarrow xyz=1;x+y+z=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)

Ta có \(x+y+z=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)

\(\Leftrightarrow x+y+z=xy+yz+zx\)

\(\Leftrightarrow xyz-1+x+y+z-xy-yz-zx=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)

​\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)​

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=1\\z=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b^2}=1\\\dfrac{b}{c^2}=1\\\dfrac{c}{a^2}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b^2\\b=c^2\\c=a^2\end{matrix}\right.\left(đpcm\right)\)

Ta có: \(A=\frac{2}{1\cdot5}+\frac{3}{5\cdot11}+\frac{4}{11\cdot19}+\frac{5}{19\cdot29}+\frac{6}{29\cdot41}\)

\(=\frac12\left(\frac{4}{1\cdot5}+\frac{6}{5\cdot11}+\frac{8}{11\cdot19}+\frac{10}{19\cdot29}+\frac{12}{29\cdot41}\right)\)

\(=\frac12\left(1-\frac15+\frac15-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{41}\right)\)

\(=\frac12\left(1-\frac{1}{41}\right)=\frac12\cdot\frac{40}{41}=\frac{20}{41}\)

Ta có: \(B=\frac{40}{31\cdot39}+\frac{35}{39\cdot46}+\frac{30}{46\cdot52}+\frac{25}{52\cdot57}+\frac{20}{57\cdot61}\)

\(=5\left(\frac{8}{31\cdot39}+\frac{7}{39\cdot46}+\frac{6}{46\cdot52}+\frac{5}{52\cdot57}+\frac{4}{57\cdot61}\right)\)

\(=5\left(\frac{1}{31}-\frac{1}{39}+\frac{1}{39}-\frac{1}{46}+\frac{1}{46}-\frac{1}{52}+\frac{1}{52}-\frac{1}{57}+\frac{1}{57}-\frac{1}{61}\right)\)

\(=5\left(\frac{1}{31}-\frac{1}{61}\right)=5\cdot\frac{30}{61\cdot31}=\frac{150}{1891}\)

\(\frac{A}{B}=\frac{20}{41}:\frac{150}{1891}=\frac{20}{41}\cdot\frac{1891}{150}=\frac{2}{15}\cdot\frac{1891}{41}=\frac{3782}{615}\)

Sửa đề: b>0; d>0

\(\frac{a}{b}<\frac{c}{d}\)

=>\(\frac{a}{b}-\frac{c}{d}<0\)

=>\(\frac{ad-bc}{bd}<0\)

mà bd>0

nên ad-bc<0

=>ad<bc

\(\frac{a}{b}<\frac{a+c}{b+d}\)

=>a(b+d)<b(a+c)

=>ab+ad<ba+bc

=>ad-bc<0(đúng)

=>\(\frac{a}{b}<\frac{a+c}{b+d}\) đúng

\(\frac{a+c}{b+d}<\frac{c}{d}\)

=>\(\frac{a+c}{b+d}-\frac{c}{d}<0\)

=>\(\frac{d\left(a+c\right)-c\left(b+d\right)}{d\left(b+d\right)}<0\)

=>d(a+c)-c(b+d)<0

=>ad+cd-bc-cd<0

=>ad-bc<0(đúng)

=>\(\frac{a+c}{b+d}<\frac{c}{d}\) đúng

Do đó: \(\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}\)

\(\frac{-6}{7}=\frac{-6}{7};\frac{-1}{3}=\frac{-1\cdot6}{3\cdot6}=\frac{-6}{18}\)

=>BA số hữu tỉ lớn hơn -6/7 và nhỏ hơn -1/3 là -6/8; -6/11; -6/15

Theo bài ra ta có \(\dfrac{a}{b}=\dfrac{2}{7};\dfrac{b}{c}=\dfrac{2}{3}\Rightarrow\dfrac{a}{2}=\dfrac{b}{7};\dfrac{b}{2}=\dfrac{c}{3}\Rightarrow\dfrac{a}{4}=\dfrac{b}{14}=\dfrac{c}{21}\)

\(\Rightarrow\dfrac{a}{4}=\dfrac{c}{21}\Leftrightarrow\dfrac{a}{c}=\dfrac{4}{21}\)

Vậy ... 

Tỉ số giữa a và c là:

\(\frac27\cdot\frac23=\frac{4}{21}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{15}=\dfrac{b}{10}=\dfrac{c}{8}=\dfrac{a+b+c}{15+10+8}=\dfrac{11}{33}=\dfrac{1}{3}\)

Do đó: a=5; b=10/3; c=8/3

Câu 1: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a+2b}{a-2b}=\frac{bk+2b}{bk-2b}=\frac{b\left(k+2\right)}{b\left(k-2\right)}=\frac{k+2}{k-2}\)

\(\frac{c+2d}{c-2d}=\frac{dk+2d}{dk-2d}=\frac{d\left(k+2\right)}{d\left(k-2\right)}=\frac{k+2}{k-2}\)

Do đó: \(\frac{a+2b}{a-2b}=\frac{c+2d}{c-2d}\)

BÀi 2:

a: 2x=3y

=>\(\frac{x}{3}=\frac{y}{2}\)

=>\(\frac{x}{21}=\frac{y}{14}\left(1\right)\)

5y=7z

=>\(\frac{y}{7}=\frac{z}{5}\)

=>\(\frac{y}{14}=\frac{z}{10}\left(2\right)\)

Từ (1),(2) suy ra \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

mà 3x+5y-7z=30

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x+5y-7z}{3\cdot21+5\cdot14-7\cdot10}=\frac{30}{63}=\frac{10}{21}\)

=>x=10; \(y=14\cdot\frac{10}{21}=10\cdot\frac23=\frac{20}{3}\) ; \(z=\frac{10}{21}\cdot10=\frac{100}{21}\)

b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}=\frac{-3x-4y+5z+3-12-25}{-3\cdot2-4\cdot4+5\cdot6}=\frac{16}{8}=2\)

=>x-1=4; y+3=8; z-5=12

=>x=5; y=5; z=17

c: \(\frac12x=\frac23y=\frac34z\)

=>\(12\cdot\frac12x=12\cdot\frac23y=12\cdot\frac34z\)

=>6x=8y=9z

=>\(\frac{6x}{72}=\frac{8y}{72}=\frac{9z}{72}\)

=>\(\frac{x}{12}=\frac{y}{9}=\frac{z}{8}\)

mà x-y=15

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{12}=\frac{y}{9}=\frac{z}{8}=\frac{x-y}{12-9}=\frac{15}{3}=5\)

=>\(\begin{cases}x=5\cdot12=60\\ y=5\cdot9=45\\ z=5\cdot8=40\end{cases}\)

_c/m ... a,b,c nha