$\\$

Bài `3.`

Áp dụng tính chất dãy tỉ số bằng nhau có :

`a/b=b/c=c/d=d/a=(a+b+c+d)/(a+b+c+d)=1`

Do đó :

`a/b=1=>a=b`

`b/c=1=>b=c`

`c/d=1 =>c=d`

`d/a=1=>d=a`

Do đó : `a=b=c=d`

`A=(2a-b)/(c+d)+(2b-c)/(d+a)+(2c-d)/(a+b) + (2d-a)/(b+c)`

`=>A=(2a-a)/(a+a)+(2b-b)/(b+b)+(2c-c)/(c+c)+(2d-d)/(d+d)`

`=>A=a/(2a)+b/(2b)+c/(2c)+d/(2d)`

`=>A=1/2+1/2+1/2+1/2`

`=>A=2`

Vậy `A=2`

Bài `4.`

`2x^3-1=15`

`=>2x^3=16`

`=> x^3=8`

`=>x=2`

`(y-25)/16 = (2-16)/9 =(-14)/9`

`=> y -25=(-224)/9`

`=>y=1/9`

`(z+9)/25=(2-16)/9 = (-14)/9`

`=> z+9=(-350)/9`

`=>z=(-431)/9`

`B=x+y+z=2+1/9  - 431/9=-412/9`

Vậy `B=-412/9`

Bài `5.`

`a,`

`(a_1c - b_1d)^{2000} + (a_2c-b_2d)^{2000} + (a_3c-b_3d)^{2000}≤0`

Nhận xét :

`(a_1c-b_1d)^{2000} >= 0 (∀a,c,b,d)`

`(a_2c - b_2d)^{2000}>= 0 (∀ a,c,b,d)`

`(a_3c-b_3d)^{2000} >= 0 (∀a,c,b,d)`

`=> (a_1c - b_1d)^{2000} + (a_2c-b_2d)^{2000} + (a_3c-b_3d)^{2000}≥0(∀a,b,c,d)`

Dấu "`=`" xảy ra khi :

`(a_1c-b_1d)^{2000}=0, (a_2c-b_2d)^{2000}=0,(a_3c - b_3d)^{2000}=0`

`<=> a_1c-b_1d=0, a_2c-b_2d=0, a_3c-b_3d=0`

`<=> a_1c = b_1d, a_2c = b_2d, a_3c = b_3d`

`<=> a_1/b_1 = d/c, a_2/b_2 = d/c, a_3/b_3=d/c`

`<=> a_1/b_1=a_2/b_2=a_3/b_3=d/c(1)`

Áp dụng tính chất dãy tỉ số bằng nhau có :

`a_1/b_1=a_2/b_2=a_3/b_3=(a_1+a_2+a_3)/(b_1+b_2+b_3)(2)`

`(1)(2) => (a_1+a_2+a_3)/(b_1+b_2+b_3)=d/c`

`b,`

Bài toán tổng quát như sau :

`(a_1c - b_1d)^{2n} + (a_2c - b_2d)^{2n}+....+ (a_mc- b_md)^{2n} (m,n∈NN)`

Do đó : `(a_1+a_2+...+a_m)/(b_1+b_2+...+b_m)  = d/c(m,n∈NN)`

$\\$

Bài `3.` Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

`a/b=b/c=c/d=d/a=(a+b+c+d)/(a+b+c+d)=1`

Do đó:

`a/b=1=>a=b`

`b/c=1=>b=c`

`c/d=1 =>c=d`

`d/a=1=>d=a`

Do đó: `a=b=c=d`

`A=(2a-b)/(c+d)+(2b-c)/(d+a)+(2c-d)/(a+b) + (2d-a)/(b+c)`

$\Rightarrow$ `A=(2a-a)/(a+a)+(2b-b)/(b+b)+(2c-c)/(c+c)+(2d-d)/(d+d)`

$\Rightarrow$ `A=a/(2a)+b/(2b)+c/(2c)+d/(2d)`

$\Rightarrow$ `A=1/2+1/2+1/2+1/2`

$\Rightarrow$ `A=2`

Vậy nên: `A=2`

Bài `4.` $2x^3-1=15$

$\Rightarrow$ `2x^3=16`

$\Rightarrow$ `x^3=8`

$\Rightarrow$ `x=2`

`(y-25)/16 = (2-16)/9 =(-14)/9`

$\Rightarrow$ `y -25=(-224)/9`

$\Rightarrow$ `y=1/9`

`(z+9)/25=(2-16)/9 = (-14)/9`

$\Rightarrow$ `z+9=(-350)/9`

$\Rightarrow$ `z=(-431)/9`

`B=x+y+z=2+1/9  - 431/9=-412/9`

Vậy `B=-412/9`

Bài `5:`

`a)`

`(a_1c - b_1d)^{2000} + (a_2c-b_2d)^{2000} + (a_3c-b_3d)^{2000}≤0`

Nhận xét:.

`(a_1c-b_1d)^{2000} >= 0 (∀a,c,b,d)`

`(a_2c - b_2d)^{2000}>= 0 (∀ a,c,b,d)`

`(a_3c-b_3d)^{2000} >= 0 (∀a,c,b,d)`

`=> (a_1c - b_1d)^{2000} + (a_2c-b_2d)^{2000} + (a_3c-b_3d)^{2000}≥0(∀a,b,c,d)`

Dấu "`=`" sẽ xảy ra khi :

`(a_1c-b_1d)^{2000}=0, (a_2c-b_2d)^{2000}=0,(a_3c - b_3d)^{2000}=0`

`<=> a_1c-b_1d=0, a_2c-b_2d=0, a_3c-b_3d=0`

`<=> a_1c = b_1d, a_2c = b_2d, a_3c = b_3d`

`<=> a_1/b_1 = d/c, a_2/b_2 = d/c, a_3/b_3=d/c`

`<=> a_1/b_1=a_2/b_2=a_3/b_3=d/c(1)`

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

`a_1/b_1=a_2/b_2=a_3/b_3=(a_1+a_2+a_3)/(b_1+b_2+b_3)(2)`

`(1)(2) => (a_1+a_2+a_3)/(b_1+b_2+b_3)=d/c`

`B.` Bài toán tổng quát như sau:

`(a_1c - b_1d)^{2n} + (a_2c - b_2d)^{2n}+....+ (a_mc- b_md)^{2n} (m,n∈NN)`

Do đó: `(a_1+a_2+...+a_m)/(b_1+b_2+...+b_m)  = d/c(m,n∈NN)`