\(<=> 9x^2-6x+1+(2x+1)^2+2(3x-1)(2x-1)\)

\(<=> 9x^2-6x+1+4x^2+4x+1+(6x-2)(2x-1)\)

 \(<=> 9x^2-6x+1+4x^2+4x+1+12x^2-6x-4x+2\) 

 \(<=> 25x^2-12x+4\)

có bạn nào có thể giúp mình giải câu b và d được không ạ mình cần gấp

\(A=\frac{1}{1\left(2n-1\right)}+\frac{1}{3\left(2n-3\right)}+...+\frac{1}{\left(2n-1\right).1}\)

\(A=\frac{1}{2n}\left[\frac{2n-1+1}{1\left(2n-1\right)}+\frac{2n-3+3}{3\left(2n-3\right)}+...+\frac{1+2n-1}{\left(2n-1\right).1}\right]\)

\(A=\frac{1}{2n}\left[\frac{1}{1}+\frac{1}{2n-1}+\frac{1}{3}+\frac{1}{2n-3}+...+\frac{1}{2n-1}+\frac{1}{1}\right]\)

\(A=\frac{1}{n}\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-3}+\frac{1}{2n-1}\right)\)

\(\Rightarrow\frac{a}{b}=\frac{1}{n}\).

Em con quá non

a: m>n

=>2m>2n

=>2m-2>2n-2

b: m>n

=>-3m<-3n

=>-3m+1<-3n+1

c: m>n

=>2m>2n

=>2m+3>2n+3

mà 2n+3>2n+1

nên 2m+3>2n+1

d: m>n

=>-5m<-5n

=>-5m+3<-5n+3

mà -5n+3<-5n+7

nên -5m+3<-5n+7

a)\(\frac{-2n^3+n^2-5n}{2n+1}\)= \(\frac{-n^2\left(2n+1\right)+n\left(2n+1\right)-6n}{2n+1}\)=\(\frac{\left(2n+1\right)\left(2n-1\right)-6n}{2n+1}\)

=\(\left(n-n^2\right)-\frac{6n}{2n+1}\)=\(\left(n-n^2\right)-\frac{3\left(2n+1\right)-3}{2n+1}\)=\(\left(n-n^2\right)-3-\frac{3}{2n+1}\)

Để (-2n³+n²-5n)⋮(2n+1) thì n∈Z

⇒n∈Z thì (2n+1)∈Ư(3)=\(\left\{-1;-3;1;3\right\}\)

Ta có bảng sau:

Vậy n=(0;1;-1;-2) thì (-2n³+n²-5n) chia hết cho (2n+1).

b)\(\frac{3n^3+10n^2-5}{3n+1}\)=\(\frac{n^2\left(3n+1\right)+3n\left(3n+1\right)-\left(3n+1\right)-4}{3n+1}\)

=\(\frac{\left(3n+1\right)\left(n^2+3n-1\right)-4}{3n+1}\)=\(\left(n^2+3n-1\right)-\frac{4}{3n+1}\)

Để (3n³+10n²-5)⋮(3n+1) thì n∈Z

⇒n∈Z thì (3n+1)∈Ư(4)=\(\left\{1;2;4;-1;-2;-4\right\}\)

Ta có bảng sau:

Vì n∈Z nên ta loại (\(\frac{1}{3}\) ;\(\frac{-2}{3}\); \(\frac{-5}{3}\)) .

Vậy n=(0;1;-1) thì (3n³+10n²-5) chia hết cho (3n+1).

chúc bạn học tốt ^_^

Giúp mình với mn

\(a,d=ƯCLN\left(5n+2;2n+1\right)\\ \Rightarrow2\left(5n+2\right)⋮d;5\left(2n+1\right)⋮d\\ \Rightarrow\left[5\left(2n+1\right)-2\left(5n+2\right)\right]⋮d\\ \Rightarrow-1⋮d\Rightarrow d=1\)

Suy ra ĐPCM

Cmtt với c,d

a) gọi d là \(UCLN\left(5n+2;2n+1\right)\)

\(\Rightarrow\left\{{}\begin{matrix}5n+2⋮d\\2n+1⋮d\end{matrix}\right.\Rightarrow5\left(2n+1\right)-2\left(5n+2\right)=10n+5-10n-4⋮d\Rightarrow1⋮d\Rightarrow d\inƯ\left(1\right)=\left\{\pm1\right\}\\ \RightarrowƯCLN\left(5n+2;2n+1\right)=1\)b) gọi d là \(UCLN\left(7n+10;5n+7\right)\)

\(\Rightarrow\left\{{}\begin{matrix}7n+10⋮d\\5n+7⋮d\end{matrix}\right.\Rightarrow5\left(7n+10\right)-7\left(5n+7\right)=35n+50-35n-49⋮d\Rightarrow1⋮d\Rightarrow d\inƯ\left(1\right)=\left\{\pm1\right\}\\ \RightarrowƯCLN\left(7n+10;5n+7\right)=1\)

d) gọi d là \(UCLN\left(3n+1;5n+2\right)\)

\(\Rightarrow\left\{{}\begin{matrix}3n+1⋮d\\5n+2⋮d\end{matrix}\right.\Rightarrow3\left(5n+2\right)-5\left(3n+1\right)=15n+6-15n-5⋮d\Rightarrow1⋮d\Rightarrow d\inƯ\left(1\right)=\left\{\pm1\right\}\\ \RightarrowƯCLN\left(3n+1;5n+2\right)=1\)

\(P=\frac{n^3+2n^2-1}{n^3+2n^2+2n+1}\)

ĐKXĐ : \(n\ne-1\)

\(=\frac{n^3+n^2+n^2+n-n-1}{n^3+2n^2+2n+1}=\frac{n^2\left(n+1\right)+n\left(n+1\right)-\left(n+1\right)}{\left(n^3+1\right)+2n\left(n+1\right)}\)

\(=\frac{\left(n+1\right)\left(n^2+n-1\right)}{\left(n+1\right)\left(n^2-n+1\right)+2n\left(n+1\right)}=\frac{\left(n+1\right)\left(n^2+n-1\right)}{\left(n+1\right)\left(n^2+n+1\right)}=\frac{n^2+n-1}{n^2+n+1}\)

Với n nguyên, đặt ƯC( n² + n - 1 ; n² + n + 1 ) = d

=> n² + n - 1 ⋮ d và n² + n + 1 ⋮ d

=> ( n² + n + 1 ) - ( n² + n - 1 ) ⋮ d

=> n² + n + 1 - n² - n + 1 ⋮ d

=> 2 ⋮ d => d = 1 hoặc d = 2

Dễ thấy n² + n + 1 ⋮/ 2 ∀ n ∈ Z ( bạn tự chứng minh )

=> loại d = 2

=> d = 1

=> ƯCLN( n² + n - 1 ; n² + n + 1 ) = 1

hay P tối giản ( đpcm )

1/

$10n+4\vdots 2n+7$

$\Rightarrow 5(2n+7)-31\vdots 2n+7$

$\Rightarrow 31\vdots 2n+7$

$\Rightarrow 2n+7\in Ư(31)$

$\Rightarrow 2n+7\in \left\{1; -1; 31; -31\right\}$

$\Rightarrow n\in \left\{-3; -4; 12; -19\right\}$

2/

$5n-4\vdots 3n+1$

$\Rightarrow 3(5n-4)\vdots 3n+1$

$\Rightarroq 15n-12\vdots 3n+1$

$\Rightarrow 5(3n+1)-17\vdots 3n+1$

$\Rightarrow 17\vdots 3n+1$

$\Rightarrow 3n+1\in Ư(17)$

$\Rightarrow 3n+1\in \left\{1; -1; 17; -17\right\}$

$\Rightarrow n\in \left\{0; \frac{-2}{3}; \frac{16}{3}; -6\right\}$

Do $n$ nguyên nên $n\in\left\{0; -6\right\}$

3/

$2n^2+n-6\vdots 2n+1$

$\Rightarrow n(2n+1)-6\vdots 2n+1$

$\Rightarrow 6\vdots 2n+1$

$\Rightarrow 2n+1\in Ư(6)$

Mà $2n+1$ lẻ nên: $2n+1\in \left\{1; -1; 3; -3\right\}$

$\Rightarrow n\in \left\{0; -1; 1; -2\right\}$