a, CMR : \(4n^2.\left(n+2\right)+4n.\left(n+2\right)⋮24\) với mọi n
b, Tìm \(n\in Z\) để \(6n^2+n-1⋮3n+2\)
Tìm các giới hạn sau:
\(a,\dfrac{4n^5-3n^2}{\left(3n^2-2\right)\left(1-4n^3\right)}\)
\(b,\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}\)
\(b,lim\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}\)
\(=lim\dfrac{\left(1+\dfrac{1}{n^2}\right)\left(\dfrac{1}{n}-\dfrac{10}{n^2}\right)^2}{\left(1+\dfrac{1}{n}\right)\left(\dfrac{3}{n^2}-\dfrac{3}{n^3}\right)}=0\)
\(a,lim\dfrac{4n^5-3n^2}{\left(3n^2-2\right)\left(1-4n^3\right)}\)
\(=lim\dfrac{4-\dfrac{3}{n^3}}{\left(3-\dfrac{2}{n^2}\right)\left(\dfrac{1}{n^3}-4\right)}\)
\(=\dfrac{4-0}{\left(3-0\right)\left(0-4\right)}=\dfrac{4}{-12}=-\dfrac{1}{3}\)
\(\lim\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}=\lim\dfrac{\left(1+\dfrac{1}{n^2}\right)\left(1-\dfrac{10}{n}\right)^2}{\left(1+\dfrac{1}{n}\right)\left(3-\dfrac{3}{n}\right)^3}=\dfrac{1.1^2}{1.3}=\dfrac{1}{3}\)
Tìm \(n\in N\), sao cho
\(a,\left(4n^2-3n-1\right)⋮\left(4n-1\right)\)
\(b,\left(4n^2-3n-1\right)⋮\left(n-1\right)\)
Giúp Em vs ạ =)))))
a)4n2-3n-1 chia hết cho 4n-1
<=>4n2-n-2n-1 chia hết cho 4n-1
<=>n(4n-1)-(2n+1) chia hết cho 4n-1
<=>2n+1 chia hết cho 4n-1
<=>2(2n+1) chia hết cho 4n-1
<=>4n-1+3 chia hết cho 4n-1
<=>3 chia hết cho 4n-1
=>4n-1 thuộc Ư(3)
=>Ư(3)={-1;1;-3;3}
Ta có bảng sau:
4n-1 | -1 | 1 | -3 | 3 |
n | 0 | 1/2 | -1/2 | 1 |
KL | tm | loại | loại | tm |
Vậy n thuộc {0;1}
b)4n2-3n-1 chia hết cho n-1
<=>4n2-4n+n-1 chia hết cho n-1
<=>4n(n-1)+n-1 chia hết cho n-1
<=>(4n+1)(n-1) chia hết cho n-1
<=>n thuộc N với mọi gtrị
P/s: "chia hết cho" thì viết kí hiệu vô
Is that T :))
CMR: vs mọi n thuộc Z thì
a) \(\left(n^2-3n+1\right)\left(n+2\right)-n^3+2⋮5\)
b)\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-10\right)⋮2\)
a: \(=n^3+2n^2-3n^2-6n+n+2-n^3+2\)
\(=-n^2+5n\)
Cái này nếu n=1 thì ko thỏa mãn nha bạn
b: \(=6n^2+30n+n+5-6n^2+30n-10n+50\)
\(=49n+55\)
Nếu n là số lẻ thì 49n+55 chia hết cho 2
Còn nếu n là số chẵn thì 49n+55 ko chia hết cho 2 nha bạn
CMR:Với mọi số tự nhiên n \(\ne\)0 ta đều có:
a.\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left(3n-1\right)\cdot\left(3n+2\right)}=\frac{n}{6n+4}\)
b.\(\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{\left(4n-1\right)\cdot\left(4n+3\right)}=\frac{5n}{4n+3}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)
b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)
\(=\frac{5}{4}.\frac{4n}{12n+9}\)
\(=\frac{5n}{12n+9}\)
( sai đề )
CMR:Với mọi số tự nhiên n \(\ne\)0 ta đều có:
a.\(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+...+\frac{1}{\left(3n-1\right)\times\left(3n+2\right)}=\frac{1}{6n+4}\)
b.\(\frac{5}{3\times7}+\frac{5}{7\times11}+\frac{5}{11\times15}+...+\frac{5}{\left(4n-1\right)\times\left(4n+3\right)}=\frac{5n}{4n+3}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)
\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)
\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)
b)\(VT=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left[\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{1}{3}-\frac{1}{4n+3}\right]=\frac{5}{4}\cdot\left[\frac{4n+3}{3\left(4n+3\right)}-\frac{3}{3\left(4n+3\right)}\right]\)
\(=\frac{5}{4}\cdot\left[\frac{4n+3-3}{12n+9}\right]\)\(=\frac{5}{4}\cdot\frac{4n}{12n+9}=\frac{5n}{12n+9}\)
CMR: với mọi số tự nhiên n thì:
a)\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\) chia hết cho 5
b)\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)chia hết cho 2
a, Ta có: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+3n^2-n+2n^2+6n-2-n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
\(\Rightarrowđpcm\)
b, \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+31n+5-6n^2-7n+5\)
\(=24n+10=2\left(12n+5\right)⋮2\)
\(\Rightarrowđpcm\)
a)
= n3 + 2n2 + 3n2 + 6n - n - 2 + 2
= 5n2 + 5n
= 5(n2 + n ) chia hết cho 5
b)
= 2(12n +5) chia hết cho 2
1, x,y,z∈N*. CMR x+3z-y là hợp số biết `x^2+y^2=z^2`
2,Tìm n∈N* để \(\left(4n^3+n+3\right)⋮\left(2n^2+n+1\right)\)
3, CMR:\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\forall x\ne y,xy\ne0\)
2.
\(4n^3+n+3=4n^3+2n^2+2n-2n^2-n-1+4=2n\left(2n^2+n+1\right)-\left(2n^2+n+1\right)+4\)-Để \(\left(4n^3+n+3\right)⋮\left(2n^2+n+1\right)\) thì \(4⋮\left(2n^2+n+1\right)\)
\(\Leftrightarrow2n^2+n+1\in\left\{1;-1;2;-2;4;-4\right\}\) (do n là số nguyên)
*\(2n^2+n+1=1\Leftrightarrow n\left(2n+1\right)=0\Leftrightarrow n=0\) (loại) hay \(n=\dfrac{-1}{2}\) (loại)
*\(2n^2+n+1=-1\Leftrightarrow2n^2+n+2=0\) (phương trình vô nghiệm)
\(2n^2+n+1=2\Leftrightarrow2n^2+n-1=0\Leftrightarrow n^2+n+n^2-1=0\Leftrightarrow n\left(n+1\right)+\left(n+1\right)\left(n-1\right)=0\Leftrightarrow\left(n+1\right)\left(2n-1\right)=0\)
\(\Leftrightarrow n=-1\) (loại) hay \(n=\dfrac{1}{2}\) (loại)
\(2n^2+n+1=-2\Leftrightarrow2n^2+n+3=0\) (phương trình vô nghiệm)
\(2n^2+n+1=4\Leftrightarrow2n^2+n-3=0\Leftrightarrow2n^2-2n+3n-3=0\Leftrightarrow2n\left(n-1\right)+3\left(n-1\right)=0\Leftrightarrow\left(n-1\right)\left(2n+3\right)=0\)\(\Leftrightarrow n=1\left(nhận\right)\) hay \(n=\dfrac{-3}{2}\left(loại\right)\)
-Vậy \(n=1\)
1. \(x^2+y^2=z^2\)
\(\Rightarrow x^2+y^2-z^2=0\)
\(\Rightarrow\left(x-z\right)\left(x+z\right)+y^2=0\)
-TH1: y lẻ \(\Rightarrow x-z;x+z\) đều lẻ.
\(x+3z-y=x+z-y+2x\) chia hết cho 2. \(\Rightarrow\)Hợp số.
-TH2: y chẵn \(\Rightarrow\)1 trong hai biểu thức \(x-z;x+z\) chia hết cho 2.
*Xét \(\left(x-z\right)⋮2\):
\(x+3z-y=x-z+4z-y\) chia hết cho 2. \(\Rightarrow\)Hợp số.
*Xét \(\left(x+z\right)⋮2\):
\(x+3z-y=x+z+2z-y\) chia hết cho 2 \(\Rightarrow\)Hợp số.
Tìm các giới hạn sau:
\(a,\dfrac{\left(2n+1\right)\left(3n-2\right)^2}{n^3+n-1}\)
\(b,\dfrac{2n-1}{3n^2+4n-1}\)
\(\lim\dfrac{\left(2n+1\right)\left(3n-2\right)^2}{n^3+n-1}=\lim\dfrac{n\left(2+\dfrac{1}{n}\right).n^2.\left(3-\dfrac{2}{n}\right)^2}{n^3\left(1+\dfrac{1}{n^2}-\dfrac{1}{n^3}\right)}\)
\(=\lim\dfrac{\left(2+\dfrac{1}{n}\right)\left(3-\dfrac{2}{n}\right)^2}{1+\dfrac{1}{n^2}-\dfrac{1}{n^3}}=\dfrac{2.3^2}{1}=18\)
\(\lim\dfrac{2n-1}{3n^2+4n-1}=\lim\dfrac{n\left(2-\dfrac{1}{n}\right)}{n^2\left(3+\dfrac{4}{n}-\dfrac{1}{n^2}\right)}=\lim\dfrac{2-\dfrac{1}{n}}{n\left(3+\dfrac{4}{n}-\dfrac{1}{n^2}\right)}=\dfrac{2}{+\infty}=0\)
Tìm \(n\inℕ\), sao cho
\(a,\left(4n^2-3n-1\right)⋮\left(4n-1\right)\)
\(b,\left(4n^2-3n-1\right)⋮\left(4n-1\right)\)
a, 4n2 - 3n -1 chia hết 4n - 1
=> n(4n - 1 ) -2n -1 chia hết 4n - 1
=> 2n -1 chia hết 4n - 1
=> 4n - 1 + 2n chia hết 4n - 1
=> 2n chia hết 4n - 1
Mà 2n - 1 chia hết 4n - 1
=> 2n - (2n - 1) chia hết 4n - 1
=> 1 chia hết 4n - 1
=> 4n - 1 = 1
=> 4n = 2
=> n = \(\frac{1}{2}\)
Mà n thuộc N
Vậy không có giá trị của n
b, 4n2 -3n -1 chia hết n - 1
=> 4n (n - 1) + n - 1 chia hết n - 1
=> n - 1 thuộc N
=> n thuộc N
Vậy n thuộc N