1,\(\frac{\frac{3}{3}-\frac{3}{5}+\frac{3}{11}+\frac{3}{13}}{\frac{7}{3}-\frac{7}{5}+\frac{5}{11}+\frac{5}{13}}\)
2,a,tim x biet:
a,\(\left(3x-2\right)^{10}=\left(3x-2\right)^6\) b\(\left(2x^2-13\right)^4=\left(-5\right)^4\)
3,tim so nguyen n de phan so sau co gia tri la 1 so nguyen va tinh gia tri do
a,A=\(\frac{3n+9}{n-4}\) b,B=\(\frac{6n+5}{2n-1}\)
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1, \(=\frac{3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}+\frac{1}{13}\right)}{7\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}+\frac{1}{13}\right)}=\frac{3}{7}\)
2, a, \(\Leftrightarrow\left(3x-2\right)^{10}-\left(3x-2\right)^6=0\)
\(\Leftrightarrow\left(3x-2\right)^6\left[\left(3x-2\right)^4-1\right]=0\)
TH1: (3x-2)^6=0 <=> 3x-2=0 <=> x=2/3
TH2: (3x-2)^4-1=0 <=> (3x-2)^4=1
<=> 3x-2 = 1 hoặc 3x-2=-1
<=>x=1 hoặc x=-1/3
Vậy x=2/3 hoặc x=1 hoặc x=-1/3
b, \(\Leftrightarrow\orbr{\begin{cases}2x^2-13=-5\\2x^2-13=5\end{cases}\Leftrightarrow\orbr{\begin{cases}2x^2=8\\2x^2=18\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=4\\x^2=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\pm2\\x=\pm3\end{cases}}}\)
3,a, \(A=\frac{3n+9}{n-4}=\frac{3n-12+21}{n-4}=\frac{3\left(n-4\right)+21}{n-4}=3+\frac{21}{n-4}\)
Để \(A\in Z\Leftrightarrow\frac{21}{n-4}\in Z\Leftrightarrow21⋮n-4\Leftrightarrow n-4\inƯ\left(21\right)\)
Ta có bảng
| n-4 | 1 | -1 | 3 | -3 | 7 | -7 | 21 | -21 |
| n | 5 | 3 | 7 | 1 | 11 | -3 | 25 | -17 |
Vậy..
b, tương tự a
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