\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)

\(\Rightarrow\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=4\)

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

=>\(a=b\cdot k;c=d\cdot k\)

\(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\)

Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)

2: \(\dfrac{2a+b}{a-2b}=\dfrac{2\cdot bk+b}{bk-2b}=\dfrac{b\left(2k+1\right)}{b\left(k-2\right)}=\dfrac{2k+1}{k-2}\)

\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{d\left(2k+1\right)}{d\left(k-2\right)}=\dfrac{2k+1}{k-2}\)

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

3: \(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\cdot\left(k-1\right)}=\dfrac{k+1}{k-1}\)

\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)

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

4: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5\cdot bk+3b}{5dk+3d}=\dfrac{b\left(5k+3\right)}{d\left(5k+3\right)}=\dfrac{b}{d}\)

\(\dfrac{5a-3b}{5c-3d}=\dfrac{5\cdot bk-3b}{5\cdot dk-3d}=\dfrac{b\left(5k-3\right)}{d\left(5k-3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)

1-b ; 2-e ; 3-d ; 4-a ; 5- c

2.Giải:

Theo bài ra ta có:

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{d}{5}\) và a + b + c + d = -42

Theo tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{d}{5}=\frac{a+b+c+d}{2+3+4+5}=\frac{-42}{14}=-3\)

+) \(\frac{a}{2}=-3\Rightarrow a=-6\)

+) \(\frac{b}{3}=-3\Rightarrow b=-9\)

+) \(\frac{c}{4}=-3\Rightarrow c=-12\)

+) \(\frac{d}{5}=-3\Rightarrow d=-15\)

Vậy a = -6

        b = -9

        c = -12

        d = -15

Bài 3:

Ta có:\(\frac{a}{2}=\frac{b}{3}\Leftrightarrow\frac{a}{10}=\frac{b}{15}\); \(\frac{b}{5}=\frac{c}{4}\Leftrightarrow\frac{b}{15}=\frac{c}{12}\)

\(\Rightarrow\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\)

Áp dụng tc dãy tỉ:

\(\frac{a}{10}=\frac{b}{15}=\frac{c}{20}=\frac{a+b+c}{10+15+12}=\frac{-49}{37}\)

Với \(\frac{a}{10}=\frac{-49}{37}\Rightarrow a=10\cdot\frac{-49}{37}=\frac{-490}{37}\)

Với \(\frac{b}{15}=\frac{-49}{37}\Rightarrow b=15\cdot\frac{-49}{37}=\frac{-735}{37}\)

Với \(\frac{c}{12}=\frac{-49}{37}\Rightarrow c=12\cdot\frac{-49}{37}=\frac{-588}{37}\)

Bài 2:

a : b : c : d = 2 : 3 : 4 : 5 \(\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{d}{5}\)

Áp dụng tc dãy tỉ:

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{d}{5}=\frac{a+b+c+d}{2+3+4+5}=\frac{-42}{14}=-3\)

Với \(\frac{a}{2}=-3\Rightarrow a=-6\)

Với \(\frac{b}{3}=-6\Rightarrow b=-18\)

Với \(\frac{c}{4}=-6\Rightarrow c=-24\)

Với \(\frac{d}{5}=-6\Rightarrow d=-30\)

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

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

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

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

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