Question

a)Cho a,b thuộc N^* và b=a+1

Thu gọn biểu thức:

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

b)Áp dụng:Tính giá trị biểu thức:

\(P=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)

c)Tính tổng:

\(Q=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{2020^2}+\frac{1}{2021^2}}\)

Answer 1

a)

\(P=a\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}+\frac{a}{b}=a\sqrt{\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}\)

      =\(a\sqrt{\frac{a^2\left(a+1\right)^2+2a\left(a+1\right)+1}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}=a\sqrt{\frac{\left[a\left(a+1\right)+1\right]^2}{\left[a\left(a+1\right)\right]^2}}+\frac{a}{a+1}\)

      \(=a.\frac{a\left(a+1\right)+1}{a\left(a+1\right)}+\frac{a}{a+1}=a+\frac{1}{a+1}+\frac{a}{a+1}=a+1\)

Vay P=a+1

phan b,c ap dung phan a la ra

Answer 2

CM bài toán phụ: \(x+y+z=0\) 

CM: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) với x,y,z dương

Ta có: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}\)

\(=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\cdot\frac{x+y+z}{xyz}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)

\(=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Áp dụng vào ta được: \(Q=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2020}-\frac{1}{2021}\)

\(Q=2021-\frac{1}{2021}=...\)

Answer 3

Phần b mượn bài Upin ta có:

\(P=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)

\(P=2020+1\)

\(P=2021\)

Answer 4

a,\(P=\sqrt{1+a^2+\frac{a^2}{b^2}}+\frac{a}{b}=\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(=\sqrt{a^2+2a+1+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(=\sqrt{\left(a+1\right)^2-2a+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(=\sqrt{\left(a+1\right)^2-2.\left(a+1\right).\frac{a}{a+1}+\left(\frac{a}{a+1}\right)^2}+\frac{a}{a+1}\)

\(=\sqrt{\left[\left(a+1\right)-\frac{a}{a+1}\right]^2}+\frac{a}{a+1}=\left(a+1\right)-\frac{a}{a+1}+\frac{a}{a+1}=a+1=b\)

b,\(\sqrt{1+2020^2+\frac{2020^2}{2021^2}}=2020+1=2021\)