ngại làm quá

a;b k cho dieu kien j ma ban ?

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:

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

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

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

Lời giải:

\(A=\frac{(2^3+1)(3^3+1)....(1000^3+1)}{(2^3-1)(3^3-1)....(1000^3-1)}=\frac{(2+1)(2^2-2+1)(3+1)(3^2-3+1)....(1000+1)(1000^2-1000+1)}{(2-1)(2^2+2+1)(3-1)(3^2+3+1)...(1000-1)(1000^2+1000+1)}\)

\(=\frac{(2+1)(3+1)...(1000+1)}{(2-1)(3-1)...(1000-1)}.\frac{(2^2-2+1)(3^2-3+1)...(1000^2-1000+1)}{(2^2+2+1)(3^2+3+1)...(1000^2+1000+1)}\)

\(=\frac{1000.1001}{2}.\frac{(2^2-2+1)(3^2-3+1)....(1000^2-1000+1)}{(2^2+2+1)(3^2+3+1)....(1000^2+1000+1)}\)

Ta thấy: \(n^2-n+1=(n^2-2n+1)+n=(n-1)^2+(n-1)+1\)

\(\Rightarrow 3^2-3+1=2^2+2+1\)

\(4^2-4+1=3^2+3+1\)

......

\(1000^2-1000+1=999^2+999+1\)

\(\Rightarrow (3^2-3+1)(4^2-4+1)...(1000^2-1000+1)=(2^2+2+1)(3^2+3+1)...(999^2+999+1)\)

Do đó: \(A=\frac{1000.1001}{2}.\frac{2^2-2+1}{1000^2+1000+1}=\frac{3}{2}.\frac{1000.1001}{1000(1000+1)+1}=\frac{3}{2}.\frac{1000.1001}{1000.1001+1}< \frac{3}{2}\)

a/ \(N=\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

\(=2x\left(4x^2-2xy+y^2\right)+y\left(4x^2-2xy+y^2\right)\)

\(=8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3\)

\(=8x^3+y^3\)

Thay: \(x=\frac{1}{2}\); \(y=\frac{1}{3}\) vào N ta được

\(8.\left(\frac{1}{2}\right)^3+\left(\frac{1}{3}\right)^3\)

\(=8.\frac{1}{8}+\frac{1}{27}\)

\(=1+\frac{1}{27}=\frac{27}{27}+\frac{1}{27}=\frac{28}{27}\)

b/ \(P=2\left(x+1\right)\left(x^2-x+1\right)-2\left(x-1\right)\left(x^2+x+1\right)\)

\(=\left(2x+2\right)\left(x^2-x+1\right)-\left[\left(2x-2\right)\left(x^2+x+1\right)\right]\)

\(=2x\left(x^2-x+1\right)+2\left(x^2-x+1\right)-\left[2x\left(x^2+x+1\right)-2\left(x^2+x+1\right)\right]\)

\(=2x^3-2x^2+2x+2x^2-2x+2-\left(2x^3+2x^2+2x-2x^2-2x-2\right)\)

\(=2x^3-2x^2+2x+2x^2-2x+2-2x^3-2x^2-2x+2x^2+2x+2\)

\(=4\)

c/ \(Q=\left(2x-1\right)\left(2x+1\right)-4\left(x-1\right)\left(x+1\right)\)

\(=\left(2x\right)^2-1^2-4.\left(x^2-1^2\right)\)

\(=4x^2-1-4x^2+4\)

\(=3\)

P/s: Sao 2 câu cuối ko phụ thuôc vào giá trị của x vậy? Ko chắc!

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