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c)\(7^{2n}+7^{2n+2}=2450\)
⇒\(7^{2n}+7^{2n}.7^2=2450\)
⇒\(7^{2n}.50=2450\)
⇒\(7^{2n}=49\)\(=7^2\)
⇒2n=2
⇒n=1
a,\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+.....+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=0+0+...0+0+\dfrac{13}{15}=\dfrac{13}{15}\)
câu b và c xem lại đề nha
Chúc bạn học tốt!!!
1. Tìm n, biết:
a) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
\(\Rightarrow\left(-2\right)^n.\left(-2\right)^2=\left(-2\right)^5\)
(-2)n + 2 = (-2)5
n + 2 = 5
n = 5 - 2
n = 3.
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow\dfrac{2^3}{2^n}=2\)
\(\Rightarrow\) 2n . 2 = 23
n + 1 = 3
n = 3 - 1
n = 2.
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
2n - 1 = 3
2n = 3 + 1
2n = 4
n = 4 : 2
n = 2.
2. Tính:
a) \(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^2\right]^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{2}\right)^4\)
\(=\left(\dfrac{1}{2}\right)^7\)
\(=\dfrac{1}{128}\)
b) 273 : 93
= (33)3 : (32)3
= 39 : 36
= 33
= 27
c) 1252 : 253
= (53)2 : (52)3
= 56 : 56
= 1
d) \(\dfrac{27^2.8^5}{6^6.32^3}\)
\(=\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{6^6.\left(2^5\right)^3}\)
\(=\dfrac{3^6.2^{15}}{6^6.2^{15}}\)
\(=\dfrac{3^6}{6^6}\)
\(=\dfrac{1}{64}.\)
B2 :
b) 27\(^3\): 9\(^3\)= (27:9)\(^3\)= 3\(^3\)
c) 125\(^2\): 25\(^3\)= 15625 : 15625 = 1
\(a,5^{n-1}=125\)
\(\Rightarrow5^{n-1}=5^3\)
\(\Rightarrow n-1=3\)
\(\Rightarrow n=3+1\)
\(\Rightarrow n=4\)
a) \(a\left(b+1\right)=3\left(a;b\inℤ\right)\)
\(\Rightarrow a;\left(b+1\right)\in U\left(3\right)=\left\{-1;1;-3;3\right\}\)
\(\Rightarrow\left(a;b\right)\in\left\{\left(-1;-4\right);\left(1;2\right);\left(-3;-2\right);\left(3;0\right)\right\}\)
b) \(2n+7⋮n+1\left(n\inℤ\right)\)
\(\Rightarrow2n+7-2\left(n+1\right)⋮n+1\)
\(\Rightarrow2n+7-2n-2⋮n+1\)
\(\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)
\(\Rightarrow n\in\left\{-2;0;-6;4\right\}\)
c) \(xy+x-y=6\left(x;y\inℤ\right)\)
\(\Rightarrow x\left(y+1\right)-y-1+1=6\)
\(\Rightarrow x\left(y+1\right)-\left(y+1\right)=5\)
\(\Rightarrow\left(x-1\right)\left(y+1\right)=5\)
\(\Rightarrow\left(x-1\right);\left(y+1\right)\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(-0;-6\right);\left(2;4\right);\left(-4;-2\right);\left(6;0\right)\right\}\)
\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)
Để A là số nguyên thì 2n+2-5 chia hết cho n+1
=>\(n+1\in\left\{1;-5;-1;5\right\}\)
=>\(n\in\left\{0;-6;-2;4\right\}\)
A = \(\dfrac{3n+1}{2n+3}\) (n \(\ne\) - \(\dfrac{3}{2}\))
A \(\in\) Z ⇔ 3n + 1 ⋮ 2n + 3
6n + 2 ⋮ 2n + 3
6n + 9 - 7 ⋮ 2n + 3
3.(2n + 3) - 7 ⋮ 2n + 3
7 ⋮ 2n + 3 ⇒ 2n + 3 \(\in\) Ư(7) = { -7; -1; 1; 7}
Lập bảng ta có:
2n+3 | -7 | -1 | 1 | 7 |
n | -5 | -2 | -1 | 2 |
Vậy các số nguyên n thỏa mãn đề bài là:
n \(\in\) { -5; -2; -1; 2}
\(A=\dfrac{3n+1}{2n+3}\inℤ\) \(\left(n\ne-\dfrac{3}{2}\right)\)
\(\Rightarrow3n+1⋮2n+3\)
\(\Rightarrow2\left(3n+1\right)-3\left(2n+3\right)⋮2n+3\)
\(\Rightarrow6n+2-6n-9⋮2n+3\)
\(\Rightarrow-7⋮2n+3\)
\(\Rightarrow2n+3\in\left\{-1;1;-7;7\right\}\)
\(\Rightarrow n\in\left\{-2;-1;-5;2\right\}\)
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