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

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

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

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

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

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

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

ta sẽ giết ngươi kí tên dép đờ kiu lờ

\(a^2+b^2=\left(a+b\right)^2-2ab=7^2-2.10=29\)

\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=133\)

\(a^4+b^4=\left(a^2+b^2\right)^2-2\left(ab\right)^2=641\)

\(a^5+b^5=\left(a^2+b^2\right)\left(a^3+b^3\right)-\left(ab\right)^2\left(a+b\right)=3157\)

\(a-b=\pm\sqrt{\left(a-b\right)^2}=\pm\sqrt{\left(a+b\right)^2-4ab}=\pm3\)

a, `A = a^2 + b^2 = (a + b)^2 - 2ab`

Thay `a + b = 7 ; ab = 10` vào A ta được:

`A = 7^2 - 2 . 10 = 29`

Vậy `A = 29` tại `a + b = 7 ; ab = 10`

b, `B = a^3 + b^3 = (a + b)^3 - 3ab (a + b)`

Thay `a + b = 7 ; ab = 10` vào B ta được:

`B = 7^3 - 3 . 10 . 7 = 133`

Vậy `B = 133` tại `a + b = 7 ; ab = 10`

c, Ta có: `a^2 + b^2 = 29` (chứng minh câu a)

`=> (a^2 + b^2)^2 = 29^2`

`=> a^4 + 2a^2b^2 + b^4 = 841`

Thay `ab = 10` vào biểu thức trên ta được:

`a^4 + 2 . 10^2 + b^4 = 841`

`=> a^4 + b^4 = 841 - 2 . 10^2 = 641`

hay `C = 641`

d, Ta có: `(a^3 + b^3) (a^2 + b^2) `

`= a^5 + a^3b^2 + a^2b^3 + b^5`

`= a^5 + b^5 + a^2b^2 (a + b)`

hay `133 . 29 = a^5 + b^5 + 10^2 . 7`

`=> a^5 + b^5 = 3157`

hay `D = 3157`

e, Ta có: \(E=a-b=\pm\sqrt{\left(a-b\right)^2}=\pm\sqrt{\left(a+b\right)^2-4ab}\)

Thay `a + b = 7` và `ab = 10` vào biểu thức trên ta được:

\(E=\pm\sqrt{7^2-4.10}=\pm3\)

a) \(A=a^2+b^2=\left(a+b\right)^2-2ab=7^2-2\cdot10=29\)

b) \(B=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=7^3-3\cdot7\cdot10=343-210=133\)

Là sao!!!!!???????!!!!!???