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\(a,ĐKXĐ:x-1\ge0\Leftrightarrow x\ge1\)
Đặt \(\hept{\begin{cases}\sqrt[3]{2-x}=a\\\sqrt{x-1}=b\left(b\ge0\right)\end{cases}\Rightarrow}a^3+b^2=2-x+x-1=1\)
Lại có: \(a=1-b\)
Thay vào được
\(\left(1-b\right)^3+b^2=1\)
\(\Leftrightarrow1-3b+3b^2-b^3+b^2-1=0\)
\(\Leftrightarrow-b^3+4b^2-3b=0\)
\(\Leftrightarrow b^3-4b^2+3b=0\)
\(\Leftrightarrow b\left(b^2-4b+3\right)=0\)
\(\Leftrightarrow b\left(b-1\right)\left(b-3\right)=0\)
\(\Leftrightarrow b=0\left(h\right)b=1\left(h\right)b=3\)(T/m ĐK b>0)
*Với b = 0
\(\Leftrightarrow\sqrt{x-1}=0\)
\(\Leftrightarrow x=1\left(TmĐKXĐ\right)\)
*Với b = 1
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(TmĐKXĐ\right)\)
*Với b = 3
\(\Leftrightarrow\sqrt{x-1}=3\)
\(\Leftrightarrow x-1=9\)
\(\Leftrightarrow x=10\)
Vậy \(S\in\left\{1;2;10\right\}\)
em chỉ bt bài 2 nha!
\(A=\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)...\left(1-\frac{2}{2020\cdot2021}\right)\)
\(\frac{2}{3}\cdot\frac{5}{6}\cdot\frac{9}{10}\cdot...\cdot\frac{2020\cdot2021-2}{2020\cdot2021}\left(1\right)\)
Mặt khác:\(2020\cdot2021-2=2020\left(2022-1\right)+2020-2022\)
\(=2020\cdot2022-2022\)
\(=2022\left(2020-1\right)=2019\cdot2022\left(2\right)\)
Từ (1),(2) ta có:
\(A=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot...\cdot\frac{2022\cdot2019}{2020\cdot2021}\)
\(=\frac{\left(4\cdot5\cdot6\cdot...\cdot2022\right)\left(1\cdot2\cdot3\cdot...\cdot2019\right)}{\left(2\cdot3\cdot4\cdot...\cdot2020\right)\left(3\cdot4\cdot5\cdot...\cdot2021\right)}\)
\(=\frac{2021\cdot2022}{2\cdot3}\cdot\frac{1\cdot2}{2020\cdot2021}=\frac{2022}{3\cdot2020}=\frac{2022}{6060}\)
Với mọi \(n\inℕ^∗\) ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n-1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\)
Áp dụng đẳng thức trên ta có:
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)
\(=1-\frac{1}{\sqrt{2019}}\)
\(t\text{ổng}qu\text{át}:\frac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}=\frac{n\sqrt{n-1}-\left(n-1\right)\sqrt{n}}{n^2\left(n-1\right)-\left(n-1\right)^2n}\)
\(=\frac{n\sqrt{n-1}-\left(n-1\right)\sqrt{n}}{\left(n-1\right)n}\)
\(=\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\)
Thay vào A có
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)
\(=1-\frac{1}{\sqrt{2017}}\)
1) Ta có: \(2020^2=\left(2019+1\right)^2=2019^2+2.2019+1.\)
\(\Rightarrow1+2019^2=2020^2-2.2019\)
\(\Rightarrow M=\sqrt{1+2019^2+\frac{2019^2}{2020^2}}+\frac{2019}{2020}=\sqrt{2020^2-2.2019+\frac{2019^2}{2020^2}}+\frac{2019}{2020}\)
\(=\sqrt{2020^2-2.2020.\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2}+\frac{2019}{2020}\)
\(=\sqrt{\left(2020-\frac{2019}{2020}\right)^2}+\frac{2019}{2020}=2020-\frac{2019}{2020}+\frac{2019}{2020}\)
\(=2020\)
Vậy M=2020.
2) Xét : \(k\in N;k\ge2\)ta có:
\(\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{\left(k-1\right)k}-\frac{2}{k}\)
\(=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{k-1}+\frac{2}{k}-\frac{2}{k}\)
\(\Rightarrow\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)
\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=1+\frac{1}{k-1}+\frac{1}{k}\)
Cho \(k=3,4,...,2020.\)Ta có:
\(N=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)
\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2018}-\frac{1}{2019}\right)+\left(1+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=2018+\frac{1}{2}-\frac{1}{2020}=2018\frac{1009}{2020}\)
Vậy \(N=2018\frac{1009}{2020}.\)
EM tham khảo phần đầu ở link: Câu hỏi của Đinh Nguyến Nhật Minh - Toán lớp 8 - Học toán với OnlineMath
Trong 3 số a,b, c có hai số đối nhau g/s 2 số đó là a và b kho đó a=-b
=> \(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{\left(-b\right)^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=-\frac{1}{b^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{c^{2019}}\)
và \(\frac{1}{a^{2019}+b^{2019}+c^{2019}}=\frac{1}{\left(-b\right)^{2019}+b^{2019}+c^{2019}}=\frac{1}{-b^{2019}+b^{2019}+c^{2019}}=\frac{1}{c^{2019}}\)
Do đó: \(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{a^{2019}+b^{2019}+c^{2019}}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)
=> \(A=\frac{1}{1}-\frac{1}{2006}=\frac{2005}{2006}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(A=1-\frac{1}{2006}\)
\(A=\frac{2005}{2006}\)
\(C=1-\frac{2}{2.3}+1-\frac{2}{3.4}+...+1-\frac{2}{2019.2020}\)
\(=2018-2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(=2018-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=2018-2\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(=2018-2.\frac{1009}{2020}\)
\(=2018-\frac{1009}{1010}\)
\(=\frac{2037171}{1010}\)