Tính \(S=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+...+\frac{1}{2020\sqrt{2021}+2021\sqrt{2020}}\)
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:
\(Q=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{2020^2}+\frac{1}{2021^2}}\)
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
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: \(Q=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2020}-\frac{1}{2021}\)
\(Q=2021-\frac{1}{2021}=...\)
Phần b mượn bài Upin ta có:
\(P=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)
\(P=2020+1\)
\(P=2021\)
giải phương trình :\(\sqrt{x^2+1-2x}+\sqrt{x^2+4x+4}=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)
Đk: \(\forall x\in R\)
Ta có:\(\sqrt{x^2+1-2x}+\sqrt{x^2+4x+4}=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)
<=> \(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}=\sqrt{1+2020^2+2.2020+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)
<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(1+2020\right)^2+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)
<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)
<=> \(\left|x-1\right|+\left|x+2\right|=\frac{2021^2-2020}{2021}+\frac{2020}{2021}\)
<=> \(\left|x-1\right|+\left|x+2\right|=2021\)
Lập bảng xét dầu
x -2 1
x - 1 - | - 0 +
x + 2 - 0 + | -
Xét các TH xảy ra :
TH1: x \(\le\)-2 => pt trở thành: 1 - x - x - 2 = 2021
<=> -2x = 2022 <=> x = -1011 (tm)
TH2: \(-2< x\le1\) => pt trở thành: 1 - x + x + 2 = 2021
<=> 0x = 2018 (vô lí) => pt vô nghiệm
TH3: \(x>1\) => pt trở thành: x - 1 + x + 2 = 2021
<=> 2x = 2020 <=> x = 1010 (tm)
Vậy S = {-1011; 1010}
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:
\(Q=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{2020^2}+\frac{1}{2021^2}}\)
c) Áp dụng công thức \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\),ta được:
\(Q=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2020}-\frac{1}{2021}\)
\(=1+1+1+...+1-\frac{1}{2021}\)
\(=2021-\frac{1}{2021}=\frac{4084440}{2021}\)
Giải phương trình
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)
Giải phương trình:
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)
Please help me, chiều nay 5h tui đi học rùi . Giải giúp tui ik mừ.
Giải phương trình :
\(a,\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)
\(b,\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+10-6\sqrt{x+1}}=2\sqrt{x+2-2\sqrt{x+1}}\)
\(c,\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=x+\frac{3}{2}\)
Tính : \(\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)
Ta có: \(2021^2=\left(2020+1\right)^2=2020^2+2.2020.1+1^2\)
\(\Rightarrow1+2020^2=2021^2-2.2020\)
\(\Rightarrow\sqrt{1+2020^2+\frac{2020^2}{2021}}+\frac{2020}{2021}\)
\(=\sqrt{2021^2-2.2020+\frac{2020^2}{2021}}+\frac{2020}{2021}\)
\(=\sqrt{2021^2-2.2021.\frac{2020}{2021}+\left(\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)
\(=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)
\(=2021-\frac{2020}{2021}+\frac{2020}{2021}=2021\)
Tính giá trị của biểu thức N=x^2019 +3x^2020-2x^2021 với x=\(\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{ }2}\)
\(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)
Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)
=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)
\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
==> \(x=\sqrt{2}-\left(\sqrt{2}+1\right)=-1\)
Do đó: N = (-1)2019 + 3.(-1)2020 - 2.(-1)2021 = -1 + 3 + 2 = 4
So sánh \(\sqrt{1+2020^2+\frac{2020^2}{2021^2}}\) và 2021