Rut gọn biểu thức
\(P=\sqrt{1+10^{2n}+\frac{10^{2n}}{\left(10^n+1\right)^2}}+\frac{10^n}{10^n+1}\)
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S
28 tháng 6 2020
1/ \(x=\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\)
\(x=\left(1+\frac{\sqrt{10}\left(\sqrt{10}+1\right)}{1+\sqrt{10}}\right)\left(\frac{\sqrt{10}\left(\sqrt{10}-1\right)}{\sqrt{10}-1}-1\right)\)
\(x=\left(1+\sqrt{10}\right)\left(\sqrt{10}-1\right)\)
\(x=10-1=9\)
Thay \(x=9\) vào A:
\(A=\frac{2\sqrt{9}+1}{9+\sqrt{9}}=\frac{7}{12}\)
Vậy với \(x=\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\Leftrightarrow A=\frac{7}{12}\)
2/ \(B=\left(1-\frac{2\sqrt{x}}{3\sqrt{x}+1}+\frac{\sqrt{x}+1}{9x-1}\right):\frac{3}{3\sqrt{x}+1}\)
\(\Leftrightarrow B=\frac{9x-1-2\sqrt{x}\left(3\sqrt{x}-1\right)+\sqrt{x}+1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\cdot\frac{3\sqrt{x}+1}{3}\)
\(\Leftrightarrow B=\frac{9x-1-6x+2\sqrt{x}+\sqrt{x}+1}{3\left(3\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
3/ \(P=A.B=\frac{2\sqrt{x}+1}{x+\sqrt{x}}\cdot\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{2\sqrt{x}+1}{3\sqrt{x}-1}\)
Để \(P\in Z\Leftrightarrow2\sqrt{x}+1⋮3\sqrt{x}-1\)
\(\Leftrightarrow6\sqrt{x}+2⋮3\sqrt{x}-1\)
\(\Leftrightarrow2\left(3\sqrt{x}-1\right)+4⋮3\sqrt{x}-1\)
\(\Leftrightarrow4⋮3\sqrt{x}-1\)
\(\Leftrightarrow3\sqrt{x}-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Leftrightarrow3\sqrt{x}\in\left\{0;2;-1;3;-3;5\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;\frac{2}{3};-\frac{1}{3};1;-1;\frac{5}{3}\right\}\)
\(\Leftrightarrow x\in\left\{0;\frac{4}{9};\frac{1}{9};1;\frac{25}{9}\right\}\)
Loại bỏ những giá trị x < 0 , x \(x\notin Z\)và x không thỏa mãn ĐKXĐ
Vậy để \(P\in Z\Leftrightarrow x\in\left\{1\right\}\)
6 tháng 7 2017
a, \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
\(=\left(1-\dfrac{1}{4}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
\(=\left(1-\dfrac{1}{16}\right)\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right).2\)
...
\(=\left(1-\dfrac{1}{2^{2n}}\right)\left(1+\dfrac{1}{2^{2n}}\right).2=\left(1-\dfrac{1}{2^{4n}}\right).2=2-\dfrac{1}{2^{4n-1}}\)
Vậy ...
b,Sửa đề: \(\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right)\)
Ta có:\(\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right)\)
\(=\left(10-1\right).\left(10+1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
\(=\left(10^2-1\right).\left(10^2+1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
\(=\left(10^4-1\right).\left(10^4+1\right)...\left(10^{2n}+1\right).\dfrac{1}{9}\)
...
\(=\left(10^{2n}-1\right)\left(10^{2n}+1\right).\dfrac{1}{9}=\left(10^{4n}-1\right).\dfrac{1}{9}=\dfrac{10^{4n}}{9}-\dfrac{1}{9}\)
Vậy ...
6 tháng 7 2017
áp dụng hằng đẳng thức (a+b)(a-b)=a^2-b^2 Minh Hoang Hai
5 tháng 4 2020
Đặt A = \(\left(1+\frac{2}{4}\right).\left(1+\frac{2}{10}\right).\left(1+\frac{2}{18}\right).....\left(1+\frac{2}{n^2+3n}\right)\)
Ta có : A = \(\left(1+\frac{2}{4}\right).\left(1+\frac{2}{10}\right).\left(1+\frac{2}{18}\right).....\left(1+\frac{2}{n^2+3n}\right)\)
= \(\frac{6}{4}.\frac{12}{10}.\frac{20}{18}.....\frac{\left(n+1\right).\left(n+2\right)}{n.\left(n+3\right)}\)
= \(\frac{3.2}{4}.\frac{3.4}{2.5}.\frac{4.5}{3.6}.....\frac{\left(n+1\right).\left(n+2\right)}{n.\left(n+3\right)}\)
= \(\frac{3.2.3.4.4.5....n}{2.3.4.5.6.....\left(n+2\right)}\)
= \(\frac{3.\left(n+1\right)}{n+2}\)
Vậy A = \(\frac{3.\left(n+1\right)}{n+2}\)
\(P=\sqrt{\left(10^n+1\right)^2-2.10^n+\left(\frac{10^n}{10^n+1}\right)^2}+\frac{10^n}{10^n+1}\)
\(=\sqrt{\left(10^n+1-\frac{10^n}{10^n+1}\right)^2}+\frac{10^n}{10^n+1}\)
\(=10^n+1-\frac{10^n}{10^n+1}+\frac{10^n}{10^n+1}\left(\text{vì }10^n+1-\frac{10^n}{10^n+1}>0\text{ }\right)\)
\(=10^n+1\)
cut cho