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\(\left(3x^{n-1}y^6-5x^{n+1}y^4\right):2x^3y^n=\frac{3}{2}x^{n-4}y^{6-n}-\frac{5}{2}x^{n-2}y^{4-n}\)
Để A chia hết cho B thì tất cả số mũ của phần biến phải không âm
\(n-4\ge0\)\(\Leftrightarrow\)\(n\ge4\)
\(6-n\ge0\)\(\Leftrightarrow\)\(n\le6\)
\(n-2\ge0\)\(\Leftrightarrow\)\(n\ge2\)
\(4-n\ge0\)\(\Leftrightarrow\)\(n\le4\)
Từ những dữ kiện trên \(\Rightarrow\)\(4\le n\le4\)\(\Rightarrow\)\(n=4\)
Vậy \(n=4\)
Chúc bạn học tốt ~
\(\left(3x^{n-1}y^6-5x^{n+1}y^4\right):2x^3y^n=\frac{3}{2}x^{n-4}y^{6-n}-\frac{5}{2}x^{n-2}y^{4-n}\)
Để \(\left(3x^{n-1}y^6-5x^{n+1}y^4\right)⋮2x^3y^n\) thì các số mũ của phần biến phải không âm, do đó :
\(n-4\ge0\)\(\Leftrightarrow\)\(n\ge4\)
\(6-n\ge0\)\(\Leftrightarrow\)\(n\le6\)
\(n-2\ge0\)\(\Leftrightarrow\)\(n\ge2\)
\(4-n\ge0\)\(\Leftrightarrow\)\(n\le4\)
\(\Rightarrow\)\(4\le n\le4\)\(\Rightarrow\)\(n=4\)
\(\left(7x^{n-1}y^5-5x^3y^4\right):5x^2y^n=\frac{7}{5}x^{n-3}y^{5-n}-xy^{4-n}\)
Để \(\left(7x^{n-1}y^5-5x^3y^4\right)⋮5x^2y^n\) thì các số mũ của phần biến phải không âm, do đó :
\(n-3\ge0\)\(\Leftrightarrow\)\(n\ge3\)
\(5-n\ge0\)\(\Leftrightarrow\)\(n\le5\)
\(4-n\ge0\)\(\Leftrightarrow\)\(n\le4\)
\(\Rightarrow\)\(3\le n\le4\)\(\Rightarrow\)\(n\in\left\{3;4\right\}\)
Chúc bạn học tốt ~
a: \(\dfrac{A}{B}=\dfrac{3}{2}x^{n-4}y^{6-n}-\dfrac{5}{2}x^{n-2}y^{4-n}\)
Để A chia hết cho B thì n-4>=0; 6-n>=0; n-2>=0; 4-n>=0
=>n=4
b: \(\dfrac{A}{B}=\dfrac{7}{5}x^{n-3}y^{5-n}-xy^{4-n}\)
Để A chia hết cho B thì n-3>=0; 5-n>=0; 4-n>=0
=>n>=3; n<=4
=>3<=n<=4
a: Để A chia hết cho B thì \(\left\{{}\begin{matrix}n+1-5>0\\2-4>0\left(loại\right)\end{matrix}\right.\Leftrightarrow n\in\varnothing\)
b: \(\dfrac{A}{B}=\dfrac{5x^3y^{n+2}-3x^2y^2}{-3x^{n-1}y^n}=-\dfrac{5}{3}x^{4-n}y^2+x^{3-n}y^{2-n}\)
Để A chia hết cho B thì \(\left\{{}\begin{matrix}4-n>=0\\3-n>=0\\2-n>=0\end{matrix}\right.\Leftrightarrow n< =2\)
c: \(\dfrac{A}{B}=\dfrac{3x^6\left(2x+5\right)^{n+3}}{2x^2\left(2x+5\right)^{n-1}}=\dfrac{3}{2}x^4\left(2x+5\right)^{n+3-n+1}=\dfrac{3}{2}x^4\left(2x+5\right)^4\)
=>Với mọi N thì A chia hết cho B
d. \(\left(x-3y\right)\left(3x^2+y^2+5xy\right)\)
\(=3x^3+xy^2+5x^2y-9x^2y-3y^3-15xy^2\)
\(=3x^3-14xy^2-4x^2y-3y^3\)
Bài 2:
a. \(x^2-y^2-5x+5y\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x+y-5\right)\left(x-y\right)\)
b. \(x^3-x^2-4x^2+8x-4\)
\(=x^2\left(x-1\right)-4\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)-4\left(x-1\right)^2\)
\(=\left(x-1\right)\left[x^2-4\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x^2-4x+4\right)\)
\(=\left(x-1\right)\left(x-2\right)^2\)
Bài 3:
\(87^2+26.87+13^2\)
\(=\left(87+ 13\right)^2\)
\(=100^2\)
\(=10000\)
Bài 1:
a. \(3x^2\left(5x^2-4x+3\right)\)
\(=15x^4-12x^3+9x^2\)
b. \(-5xy\left(3x^2y-5xy-y^2\right)\)
\(=-15x^3y^2+25x^2y^2+5xy^3\)
c. \(\left(5x^2-4x\right)\left(x-3\right)\)
\(=5x^3-19x^2-4x^2+12x\)
a, \(\left(x^2-y^2\right)-\left(5x+5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
b, \(5x^3-5x^2y-10x^2+10xy\)
\(=5x^2\left(x-y\right)-10x\left(x-y\right)\)
\(=\left(5x-10x\right)\left(x-y\right)=5x\left(x-2\right)\left(x-y\right)\)
c, \(2x^2-5x=x\left(2x-5\right)\)
f, \(3x^2-7x-10=3x^2+3x^2-10x-10\)
\(=3x^2\left(x+1\right)-10\left(x+1\right)=\left(3x^2-10\right)\left(x+1\right)\)
d, \(x^3-3x^2+1-3x=x^3-3x^2-3x+1\)
\(=x^3+x^2-4x^2-4x+x+1\)
\(=x^2\left(x+1\right)-4x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x^2-4x+1\right)\left(x+1\right)\)
e, \(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-4z^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
g, \(x^4+1-2x^2=\left(x^2-1\right)^2\)
h, \(3x^2-3y^2-12x+12y=3\left(x^2-y^2\right)-12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)
\(=\left(x-y\right)\left(3x+3y-12\right)\)
\(=3\left(x-y\right)\left(x+y-4\right)\)
j, \(x^2-3x+2=x^2-2x-x+2=x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-1\right)\left(x-2\right)\)
a. \(\left(x^2-y^2\right)-5\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
b. \(5x^3-5x^2y-10x^2+10xy\)
\(=5\left[\left(x^3-x^2y\right)-\left(2x^2-2xy\right)\right]\)
\(=5\left[x^2\left(x-y\right)-2x\left(x-y\right)\right]\)
\(=5x\left(x-y\right)\left(x-2\right)\)
c. \(2x^2-5x=x\left(2x-5\right)\)
d. \(x^3-3x^2+1-3x\)
\(=\left(x^3+1\right)-\left(3x^2+3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left[x^2-x+1-3x\right]\)
\(=\left(x+1\right)\left[x^2-4x+1\right]\)
\(=\left(x+1\right)\left[x^2-2.x.2+2^2-2^2+1\right]\)
\(=\left(x+1\right)\left[\left(x-2\right)^2-3\right]\)
\(=\left(x+1\right)\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)\)
e. \(3x^2-6xy+3y^2-12z^2\)
\(=3\left[x^2-2xy+y^2-4z^2\right]\)
\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y+2z\right)\left(x-y-2z\right)\)
f. \(3x^2-7x-10\)
\(=3x^2-7x-7-3\)
\(=\left(3x^2-3\right)-\left(7x+7\right)\)
\(=3\left(x^2-1\right)-7\left(x+1\right)\)
\(=3\left(x+1\right)\left(x-1\right)-7\left(x+1\right)\)
\(=\left(x+1\right)\left[3\left(x-1\right)-7\right]\)
\(=\left(x+1\right)\left(3x-8\right)\)
g. \(x^4+1-2x^2=\left(x^2\right)^2-2.x^2+1=\left(x^2-1\right)^2\)
\(=\left(x+1\right)^2\left(x-1\right)^2\)
h. \(3x^2-3y^2-12x+12y\)
\(=3\left(x^2-y^2\right)-12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)
\(=\left(x-y\right)\left[3\left(x+y\right)-12\right]\)
\(=\left(x-y\right).3.\left(x+y-4\right)\)
j. \(x^2-3x+2=x^2-x-2x+2\)
\(=x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x-2\right)\)
P/s: ( Có j sai ns nha nhiều số quá tui rối đầu )
Để A chia hết cho B thì
\(\hept{\begin{cases}2\le n-1\\4\ge n\end{cases}}\)
<=> \(3\le n\le4\)
Vậy n cần tìm là 3
Để A : B thì (7xn-1y5-5x3y4): x2yn => 7xn-1y5 : x2yn và 5x3y4:x2yn
=>
*)n-1>=2; 5>=n;
nên n>=3; 5>=n hay 3<=n<=5(1)
*)4>=n(2)
Từ (1);(2) => 3<=n<=4 mà n lẻ nên n=3
Vậy để A : B thì n=3
a: \(\dfrac{A}{B}=\dfrac{-5}{3}x^{3-n+1}y^{n+2-n}+x^{2-n+1}y^{2-n}\)
\(=\dfrac{-5}{3}x^{2-n}y^2+x^{3-n}y^{2-n}\)
Để A chia hết cho B thì \(\left\{{}\begin{matrix}2-n\ge0\\3-n\ge0\end{matrix}\right.\Leftrightarrow n\le2\)
b: Vì n+3>n-1
nên A chia hết cho B với mọi n
\(a,A=4x^{n+1}y^2;B=3x^3y^{n-1}\)
Để \(A⋮B\) thì:
\(\left\{{}\begin{matrix}n+1\ge3\\n-1\le2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n\ge2\\n\le3\end{matrix}\right.\Leftrightarrow2\le n\le3\)
Vậy....
\(b,A=7x^{n-1}y^5-5x^3y^4;B=5x^3y^n\)
Để \(A⋮Bthì:\)
\(\left\{{}\begin{matrix}n-1\ge2\\\\\\n\le4\end{matrix}\right.\)
\(\Leftrightarrow....\Leftrightarrow3\le n\le4\)
Vậy....
De ma bn
De \(A⋮B\)thi cac so mu o B phai nho hon hoac bang so mu o A
the la tim dc n