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a: Số số hạng là:
(2n-2):2+1=n(số)
Theo đề, ta có:
\(\left(2n+2\right)\cdot\dfrac{n}{2}=210\)
\(\Leftrightarrow n\left(n+1\right)=210\)
\(\Leftrightarrow n=14\)
a) \(2^n=8\)
\(\Rightarrow2^n=2^3\)
\(\Rightarrow n=3\)
b) \(5^{n+1}=125\)
\(\Rightarrow5^{n+1}=5^3\)
\(\Rightarrow n+1=3\)
\(\Rightarrow n=3-1=2\)
c) Mình không rõ đề:
d) \(2\cdot7^{n-1}+3=101\)
\(\Rightarrow2\cdot7^{n-1}=101-3\)
\(\Rightarrow2\cdot7^{n-1}=98\)
\(\Rightarrow7^{n-1}=\dfrac{98}{2}\)
\(\Rightarrow7^{n-1}=49\)
\(\Rightarrow7^{n-1}=7^2\)
\(\Rightarrow n-1=2\)
\(\Rightarrow n=1+2=3\)
e) \(3\cdot5^{2n+1}-6^2=339\)
\(\Rightarrow3\cdot5^{2n+1}=339+36\)
\(\Rightarrow3\cdot5^{2n+1}=375\)
\(\Rightarrow5^{2n+1}=125\)
\(\Rightarrow5^{2n+1}=5^3\)
\(\Rightarrow2n+1=3\)
\(\Rightarrow2n=2\)
\(\Rightarrow n=\dfrac{2}{2}=1\)
Mình mẫu đầu với cuối nhé:
a) Đặt \(ƯCLN\left(3n+4,3n+7\right)=d\)
\(\Rightarrow\left\{{}\begin{matrix}3n+4⋮d\\3n+7⋮d\end{matrix}\right.\)
\(\Rightarrow\left(3n+7\right)-\left(3n+4\right)⋮d\)
\(\Rightarrow3⋮d\)
\(\Rightarrow d\in\left\{1,3\right\}\)
Nhưng do \(3n+4,3n+7⋮̸3\) nên \(d\ne3\Rightarrow d=1\)
Vậy \(ƯCLN\left(3n+4,3n+7\right)=1\) hay \(3n+4,3n+7\) nguyên tố cùng nhau.
e) \(ƯCLN\left(2n+3,3n+5\right)=d\)
\(\Rightarrow\left\{{}\begin{matrix}2n+3⋮d\\3n+5⋮d\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}6n+9⋮d\\6n+10⋮d\end{matrix}\right.\)
\(\Rightarrow\left(6n+10\right)-\left(6n+9\right)⋮d\)
\(\Rightarrow1⋮d\) \(\Rightarrow d=1\)
Vậy \(ƯCLN\left(2n+3,3n+5\right)=1\), ta có đpcm.
a, Ta có : \(\text{n + 5 = (n - 1)+6}\)
Vì \(\text{(n-1) ⋮ n-1}\)
Nên để \(\text{n+5 ⋮ n-1}\)⋮ `n-1`
Thì \(\text{6 ⋮ n-1}\)
\(\Rightarrow\) \(\text{n - 1 ∈ Ư(6)}\)
\(\Rightarrow\) \(\text{n - 1 ∈}\) \(\left\{\text{±1;±2;±3;±6}\right\}\)
\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{0;-1;-2;-5;2;3;4;7}\right\}\) \(\text{( TM )}\)
\(\text{________________________________________________________}\)
b, Ta có : \(\text{2n-4 = (2n+4)- 8 = 2(n+2) - 8}\)
Vì \(\text{2(n+2) ⋮ n+2}\)
Nên để \(\text{2n-4 ⋮ n+2}\)
Thì \(\text{8 ⋮ n+2}\)
\(\Rightarrow\) \(\text{n + 2 ∈ Ư(8)}\)
\(\Rightarrow\) \(\text{n + 2 ∈}\) \(\left\{\text{±1;±2;±4;±8}\right\}\)
\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{-3;-4;-6;-10;-1;0;2;6}\right\}\) ( TM )
\(\text{_________________________________________________________________ }\)
c, Ta có :\(\text{ 6n + 4 = (6n + 3) +1 = 3(2n+1) + 1}\)
Vì \(\text{3(2n+1) ⋮ 2n+1}\)
Nên để\(\text{ 6n+4 ⋮ 2n+1}\)
Thì \(\text{1 ⋮ 2n+1}\)
\(\Rightarrow\) \(\text{2n + 1 ∈ Ư(1)}\)
\(\Rightarrow\) \(\text{2n + 1 ∈}\) \(\left\{\text{±1}\right\}\)
\(\Rightarrow\) \(\text{2n ∈}\) \(\left\{\text{-2;0}\right\}\)
\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{-1;0}\right\}\) ( TM )
\(\text{_______________________________________}\)
Ta có : \(\text{3 - 2n = -( 2n - 3 ) = -( 2n + 2 ) + 5 = -2( n+1)+5}\)
Vì \(\text{-2(n+1) ⋮ n+1}\)
Nên để \(\text{3-2n ⋮ n+1}\)
Thì\(\text{ 5 ⋮ n + 1}\)
\(\Rightarrow\) \(\text{n + 1 ∈}\) \(\left\{\text{±1;±5}\right\}\)
\(\Rightarrow\) \(\text{n ∈}\) \(\text{-2;-6;0;4}\) ( TM )
1 . tìm giá trị x
\(\left(x+1\right)+\left(x+4\right)+....+\left(x+28\right)=115.\)
\(\Rightarrow\left(x+x+x+....x\right)+\left(1+4+..+28\right)=115\)
\(\Rightarrow10x+\left(28+1\right).10:2=115\)
\(\Rightarrow10x+145=115\)
\(\Rightarrow10x=115-145=-30\)
\(\Rightarrow x=-30:10=-3\)
Bai 1
so so hang la: [(28+x)-(x+1)]/3+1= 10 so hang
tong =[(x+1)+(x+28)]*10/2=(2x+29)*10/2=115
(2x+29)*5=115
2x+29=115/5=23
2x=23-29=-6
x=-3
a, \(\dfrac{15}{n-1}\); n∈Z
\(\dfrac{15\left(n-1\right)}{n-1}=\dfrac{15n-15}{n-1}\)
=> Ư(15)={\(\pm1;\pm3;\pm5;\pm15\)}
| n-1 | -15 | -5 | -3 | -1 | 1 | 3 | 5 | 15 |
| n | -14 | -4 | -2 | 0 | 2 | 4 | 6 | 16 |
| Đánh giá | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn |
Vậy n∈{-14;-4;-2;0;2;4;6;16}
b, \(\dfrac{-21}{n+3}\) n∈Z
\(\dfrac{-21\left(n+3\right)}{n+3}=\dfrac{\left(-21n-63\right)}{n+3}\)
Ư(63)={±1;±3;±7;±9;±21;±63}
| n+3 | -63 | -21 | -9 | -7 | -3 | -1 | 1 | 3 | 7 | 9 | 21 | 63 |
| n | -66 | -24 | -12 | -10 | -6 | -4 | -2 | 0 | 4 | 6 | 18 | 60 |
| Đ/gia | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn |
Vậy n∈{-66;-24;-12;-10;-6;-4;-2;0;4;6;18;60}
\(\dfrac{2n+7}{n-2};n\inℤ\\ \Rightarrow\dfrac{\left(2n-4\right)+7+2}{n-2}=\dfrac{2\left(n-2\right)+9}{n-2}=2+\dfrac{9}{n-2}\)
\(\LeftrightarrowƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
Ta có bảng sau:
| n-2 | -9 | -3 | -1 | 1 | 3 | 9 |
| n | -7 | -1 | 1 | 3 | 5 | 11 |
| Đ/gia | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn |
Vậy n={-7;-1;1;3;5;11}
a)Gọi ƯCLN (\(n+3;2n+5\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(n+3\right)⋮d\Rightarrow2\left(n+3\right)⋮d\Rightarrow\left(2n+6\right)⋮d\\\left(2n+5\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(2n+6\right)-\left(2n+5\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)
⇒ƯCLN (\(n+3;2n+5\))=1
\(\Rightarrow\frac{n+3}{2n+5}\)là phân số tối giản(đpcm)
b)Gọi ƯCLN (\(2n+9;3n+14\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(2n+9\right)⋮d\Rightarrow3\left(2n+9\right)⋮d\Rightarrow\left(6n+27\right)⋮d\\\left(3n+14\right)⋮d\Rightarrow2\left(3n+14\right)⋮d\Rightarrow\left(6n+28\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(6n+28\right)-\left(6n+27\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)
⇒ƯCLN (\(2n+9;3n+14\))=1
\(\Rightarrow\frac{2n+9}{3n+14}\) là phân số tối giản.(đpcm)
c)Gọi ƯCLN(\(6n+11;2n+5\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(6n+11\right)⋮d\\\left(2n+5\right)⋮d\Rightarrow3\left(2n+5\right)⋮d\Rightarrow\left(6n+15\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(6n+15\right)-\left(6n+11\right)⋮d\)
\(\Rightarrow4⋮d\)
Mà \(\left(6n+15\right);\left(6n+11\right)⋮̸2\)
\(\Rightarrow d=1\)
⇒ƯCLN(\(6n+11;2n+5\))=1
\(\Rightarrow\frac{6n+11}{2n+5}\)là phân số tối giản (đpcm)
d)Gọi ƯCLN(\(12n+1;30n+2\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(12n+1\right)⋮d\Rightarrow5\left(12n+1\right)⋮d\Rightarrow\left(60n+5\right)⋮d\\\left(30n+2\right)⋮d\Rightarrow2\left(30n+2\right)⋮d\Rightarrow\left(60n+4\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(60n+5\right)-\left(60n+4\right)⋮d\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
⇒ƯCLN(\(12n+1;30n+2\))=1
\(\Rightarrow\frac{12n+1}{30n+2}\) là phân số tối giản (đpcm)
e)Gọi ƯCLN(\(21n+4;14n+3\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(21n+4\right)⋮d\Rightarrow2\left(21n+4\right)⋮d\Rightarrow\left(42n+8\right)⋮d\\\left(14n+3\right)⋮d\Rightarrow3\left(14n+3\right)⋮d\Rightarrow\left(42n+9\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(42n+9\right)-\left(42n+8\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)
⇒ƯCLN(\(21n+4;14n+3\))=1
\(\Rightarrow\frac{21n+4}{14n+3}\)là phân số tối giản (đpcm)
f) Gọi ƯCLN(\(2n+3;n+2\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(2n+3\right)⋮d\\\left(n+2\right)⋮d\Rightarrow2\left(n+2\right)⋮d\Rightarrow\left(2n+4\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(2n+4\right)-\left(2n+3\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)
⇒ƯCLN(\(2n+3;n+2\))=1
\(\Rightarrow\frac{2n+3}{n+2}\)là phân số tối giản (đpcm)
g) Gọi ƯCLN(\(n+1;3n+2\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(n+1\right)⋮d\Rightarrow3\left(n+1\right)⋮d\Rightarrow\left(3n+3\right)⋮d\\\left(3n+2\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(3n+3\right)-\left(3n+2\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)
⇒ƯCLN(\(n+1;3n+2\))=1
\(\Rightarrow\frac{n+1}{3n+2}\) là phân số tối giản (đpcm)
Ta có : 2 + 4 + 6 + ... + 2(n - 1) + 2n = 210
<=> 2[1 + 2 + 3 + ... + (n - 1) + n] = 210
<=> 1 + 2 + 3 + ... + n = 105
<=> [(n - 1) : 1 + 1)(n + 1) : 2 = 105
<=> n(n + 1) = 210
<=> n(n + 1) = 14.15
=> n = 14
Vậy n = 14
b) Ta có : 1 + 3 + 5 + ... + (2n - 1) = 225
<=> [(2n - 1 - 1) : 2 + 1](2n - 1 + 1) : 2 = 225
<=> n2 = 225
<=> n2 = 152
<=> n = 15
Vậy n = 15