Rut gọn biểu thức
\(P=\sqrt{1+10^{2n}+\frac{10^{2n}}{\left(10^n+1\right)^2}}+\frac{10^n}{10^n+1}\)
Những câu hỏi liên quan
Rút gọn biểu thức :
a) \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{4}\right).\left(1+\frac{1}{16}\right)...\left(1+\frac{1}{2^{2n}}\right)\)
b) \(\left(10+1\right).\left(10^2+1\right)\left(10^3+1\right)...\left(10^{2n}+1\right)\)
cho biểu thức Afrac{2sqrt{x}+1}{x+sqrt{x}} và Bleft(1-frac{2sqrt{x}}{3sqrt{x}+1}+frac{sqrt{x}+1}{9x-1}right):frac{3}{3sqrt{x}+1} với x0, x≠frac{1}{9}
1, tính giá trị của A khi xleft(1+frac{10+sqrt{10}}{1+sqrt{10}}right)left(frac{10-sqrt{10}}{sqrt{10}-1}-1right)
2, rút gọn biểu thức B
3, đặt PA.B. tìm các giá trị nguyên của x để P có giá trị nguyên
Đọc tiếp
cho biểu thức A=\(\frac{2\sqrt{x}+1}{x+\sqrt{x}}\) và B=\(\left(1-\frac{2\sqrt{x}}{3\sqrt{x}+1}+\frac{\sqrt{x}+1}{9x-1}\right):\frac{3}{3\sqrt{x}+1}\) với x>0, x≠\(\frac{1}{9}\)
1, tính giá trị của A khi x=\(\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\)
2, rút gọn biểu thức B
3, đặt P=A.B. tìm các giá trị nguyên của x để P có giá trị nguyên
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 \(P\in Z\Leftrightarrow x\in\left\{1\right\}\)
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1: Ta có: \(x=\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\cdot\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\)
\(=\left(\frac{1+\sqrt{10}+10+\sqrt{10}}{1+\sqrt{10}}\right)\cdot\left(\frac{10-\sqrt{10}-\left(\sqrt{10}-1\right)}{\sqrt{10}-1}\right)\)
\(=\frac{1+2\sqrt{10}\cdot1+\left(\sqrt{10}\right)^2}{1+\sqrt{10}}\cdot\frac{\left(\sqrt{10}\right)^2-2\cdot\sqrt{10}\cdot1+1}{\sqrt{10}-1}\)
\(=\left(1+\sqrt{10}\right)\cdot\left(\sqrt{10}-1\right)\)
\(=10-1=9\)
Thay x=9 vào biểu thức \(A=\frac{2\sqrt{x}+1}{x+\sqrt{x}}\), ta được:
\(A=\frac{2\cdot\sqrt{9}+1}{9+\sqrt{9}}=\frac{2\cdot3+1}{9+3}=\frac{7}{12}\)
Vậy: \(\frac{7}{12}\) là giá trị của biểu thức \(A=\frac{2\sqrt{x}+1}{x+\sqrt{x}}\) tại \(x=\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\cdot\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\)
2: Ta có: \(B=\left(1-\frac{2\sqrt{x}}{3\sqrt{x}+1}+\frac{\sqrt{x}+1}{9x-1}\right):\frac{3}{3\sqrt{x}+1}\)
\(=\left(\frac{9x-1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}-\frac{2\sqrt{x}\left(3\sqrt{x}-1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}+\frac{\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right)\cdot\frac{3\sqrt{x}+1}{3}\)
\(=\frac{9x-1-6x+2\sqrt{x}+\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}+2}{9\sqrt{x}-3}\)
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a) Rút gọn biểu thức:
\(P=\frac{5+\sqrt{10}+\sqrt{17}}{2}\left(\frac{5+\sqrt{10}+\sqrt{17}}{2}-5\right)\left(\frac{5+\sqrt{10}+\sqrt{17}}{2}-\sqrt{10}\right)\left(\frac{5+\sqrt{10}+\sqrt{17}}{2}-\sqrt{17}\right).\)
b) Giải phương trình: \(\frac{x+2}{2x-1}+|\frac{4x-2}{x+2}|+1=0\)
a) Tính giá trị biểu thức:
N=\(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
b)Rút gọn biểu thức:
A=\(\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}-2}{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}+2}\),trị x>2
Rút gọn biểu thức :
a) \(\left(1+\dfrac{1}{2}\right).\left(1+\dfrac{1}{4}\right).\left(1+\dfrac{1}{16}\right)...\left(1+\dfrac{1}{2^{2n}}\right)\)
b) \(\left(10+1\right).\left(10^2+1\right)\left(10^3+1\right)...\left(10^{2n}+1\right)\)
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 ...
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C/M rằng
a) \(\frac{1.3.5.....39}{21.22.23.......40}=\frac{1}{2^{10}}\)
b) \(\frac{1.3.5.....\left(2n-1\right)}{\left(n+1\right).\left(n+2\right)......2n}=\frac{1}{2^2}\)
a)n-13/n+7=5/7
b)2n-5/3=n+4/2
c)n+10/2n-8 thuộc Z
d)n+3/2n-2 thuộc Z
e)n+10/n+1 rut gọn được
Rút gọn biểu thức sau với n là số tự nhiên:
\(\left(1+\frac{2}{4}\right).\left(1+\frac{2}{10}\right).\left(1+\frac{2}{18}\right)...\left(1+\frac{2}{n^2+3n}\right)\)
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 = \(\frac{3.\left(n+1\right)}{n+2}\)
20 tháng 3 2016 lúc 18:39
Tìm n để biểu thức sau là số nguyên :
\(A=\frac{2n+1}{n+2}-\frac{n+1}{n+2}+\frac{3n+5}{2n+4}+\frac{4n+6}{3n+6}-\frac{10n+12}{5n+10}-\frac{12n+3}{4n+8}\)
