đặt x^1000=m,y^1000=n

m+n=a

m²+n²=2b/3 =>a²=2b/3+2mn =>mn=a²-2b/3

m⁵+n⁵=c/36 <=>(n+m)[(m+n)⁴-5mn(m+n)²+5m²n²]=c/36

<=>a.[a⁴-5(a²-2b/3)a²+5(a²-2b/3)² ]=c/36 <=>a(a⁴-10a²b/3+20b²/9)=c/36 <=>a(9a⁴-30a²b+20b²)=c/4

\(bx^2=ay^2\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)

\(\Rightarrow\left(\dfrac{x^2}{a}\right)^{1000}=\left(\dfrac{y^2}{b}\right)^{1000}=\left(\dfrac{1}{a+b}\right)^{1000}\)

\(\Rightarrow\dfrac{x^{2000}}{a^{1000}}=\dfrac{y^{2000}}{b^{1000}}=\dfrac{1}{\left(a+b\right)^{1000}}\)

\(\Rightarrow\dfrac{x^{2000}}{a^{1000}}+\dfrac{y^{2000}}{b^{1000}}=\dfrac{1}{\left(a+b\right)^{1000}}+\dfrac{1}{\left(a+b\right)^{1000}}=\dfrac{2}{\left(a+b\right)^{1000}}\)

Áp dụng tính chất tỉ lệ thức và tính chất dãy tỉ số bằng nhau:

\(ay^2=bx^2\Leftrightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)

\(\Rightarrow\left(\dfrac{x^2}{a}\right)^{1000}=\left(\dfrac{y^2}{b}\right)^{1000}=\dfrac{1}{\left(a+b\right)^{1000}}\)

\(\Rightarrow\dfrac{x^{2000}}{a^{1000}}+\dfrac{y^{2000}}{b^{1000}}=\dfrac{2}{\left(a+b\right)^{1000}}\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{x^4}{a}+\frac{y^4}{b}\right)(a+b)\geq (x^2+y^2)^2=1\)

\(\Rightarrow \frac{x^4}{a}+\frac{y^4}{b}\geq \frac{1}{a+b}\)

Dấu "=" xảy ra khi \(\frac{x^2}{a}=\frac{y^2}{b}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)

\(\Rightarrow \frac{x^{2000}}{a^{1000}}+\frac{y^{2000}}{b^{1000}}=\left(\frac{x^2}{a}\right)^{1000}+\left(\frac{y^2}{b}\right)^{1000}\)

\(=\frac{1}{(a+b)^{1000}}+\frac{1}{(a+b)^{1000}}=\frac{2}{(a+b)^{1000}}\)

Chu y dua ve bieu thuc dong bac de bien doi nhe

\(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{\left(x^2+y^2\right)^2}{a+b}\)\(\Leftrightarrow\)\(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{x^4+y^4-2x^2y^2}{a+b}\)

\(\Leftrightarrow\dfrac{bx^4\left(a+b\right)+\left(a+b\right)ay^4-ab\left(x^4+y^4-2x^2y^2\right)}{ab\left(a+b\right)}=0\)

\(\Leftrightarrow\dfrac{a^2y^4+b^2x^4-2abx^2y^2}{ab\left(a+b\right)}=0\)\(\Leftrightarrow\left(ay^2-bx^2\right)^2=0\)

\(\Leftrightarrow ay^2=bx^2\Leftrightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)

\(\Leftrightarrow\dfrac{x^{2000}}{a^{1000}}=\dfrac{y^{2000}}{b^{1000}}=\dfrac{1}{\left(a+b\right)^{1000}}\)

-->QED

a. 3,507 x 300 =3,507.3.100=10,521.100=1052,1

b. 0,0036 x 4000 =  0,0036 .4.1000 = 0,0144.100=1,44

c. 0,026 x 2500 =0,026 .25.1000 =0,65.100=65

d. 234,105 x 2000 = 234,105.2.1000=468,21.1000=468210

a. 3,507 x 300 =3,507 x 3x100=10,521x100=1052,1

b. 0,0036 x 4000 = 0,0036 x 4x1000=0,0144x1000=14,4

c. 0,026 x 2500 =0,026 x 25x100=0,65x100=65

d. 234,105 x 2000 = 234,105 x 2x1000=468,21x1000=468210

HT

a. 3,507 x 300 =  3,507 x 3 x 100 = 10,521 x 100 = 1052,1

b. 0,0036 x 4000 = 0,0001x 36 x 4 x 1000 = 0.1 x 144 = 14,4

c. 0,026 x 2500 = 0,001 x 26 x 25 x 100 = 0,1 x 650 = 65

d. 234,105 x 2000 = 234,105 x 2 x 1000 = 468,21 x 1000 = 468210

5000 + 100 = 5100

7400 - 400 = 7000

2000 3 + 600 = 6600

8000 : 2 + 2000 = 6000

6000 - (5000 - 1000) = 2000

6000 - 5000 + 1000 = 2000

7000 - 3000 x 2 = 1000

(7000 - 3000) x 2 = 8000

5000 + 100 = 5100

7400 - 400 = 7000

2000 3 + 600 = 6600

8000 : 2 + 2000 = 6000

6000 - (5000 - 1000) = 2000

6000 - 5000 + 1000 = 2000

7000 - 3000 x 2 = 1000

(7000 - 3000) x 2 = 8000

5000 + 100 = 5100

7400 - 400 = 7000

2000 3 + 600 = 6600

8000 : 2 + 2000 = 6000

6000 - (5000 - 1000) = 2000

6000 - 5000 + 1000 = 2000

7000 - 3000 x 2 = 1000

(7000 - 3000) x 2 = 8000

sai đề bài rùi !!!

Với x, y khác 0

Ta có: 

\(a^2+b^2=1\Leftrightarrow\left(a^2+b^2\right)^2=1\Leftrightarrow a^4+2a^2b^2+b^4=1\)

Từ bài ra ta suy ra:

\(\frac{a^4}{x}+\frac{b^4}{y}=\frac{a^4+2a^2b^2+b^4}{x+y}\)

<=> \(a^4\left(x+y\right)y+b^4\left(x+y\right)x=a^4xy+2a^2b^2xy+b^4xy\)

<=> \(a^4y^2+b^4x^2-2a^2y.b^2x=0\)

<=> \(\left(a^2y-b^2x\right)^2=0\)

<=> \(a^2y-b^2x=0\)

<=> \(a^2y=b^2x\)

Câu b em xem lại đề nhé: Thử \(a=b=\frac{1}{\sqrt{2}};x=y=1\)vào ko thỏa mãn