\(\frac{x}{4x^2y^3}\)và \(\frac{5}{6x^3y}\)
\(\frac{5}{6-2x}\)và\(\frac{3}{x^2-9}\)
\(\frac{2}{x^2+2xy}\)và\(\frac{1}{xy+2y^2}\)
các anh chị giúp em nha mài em phải làm bài rồi
Bài 1:
a, \(\frac{x}{4x^2y^3}và\frac{5}{6x^3y}\)
b, \(\frac{2}{x^2+2xy}và\frac{1}{xy+2y^2}\)
c, \(\frac{5}{6-2x}và\frac{3}{x^2-9}\)
\(a.\frac{x}{4x^2y^3}và\frac{5}{6x^3y}\)
- MTC: 12x3y3
- NTP: 3x ; 2y2
\(\frac{x}{4x^2y^3}=\frac{x.3x}{4x^2y^3.3x}=\frac{3x^2}{12x^3y^3}\\ \frac{5}{6x^3y}=\frac{5.2y^2}{6x^3y.2y^2}=\frac{10y^2}{12x^3y^3}\)
\(b.\frac{2}{x^2+2xy}và\frac{1}{xy+2y^2}\)
- Ta có : x2 + 2xy = x ( x + 2y ) ; xy + 2y2 = y (x + 2y ).
-MTC : xy(x+2y)
- NTP : y ; x.
\(\frac{2}{x^2+2xy}=\frac{2.y}{y\left(x^2+2xy\right)}=\frac{2y}{x^2y+2xy^2}=\frac{2y}{xy\left(x+2y\right)}\\ \frac{1}{xy+2y^2}=\frac{1.x}{x\left(xy+2y^2\right)}=\frac{x}{x^2y+2xy^2}=\frac{x}{xy\left(x+2y\right)}\)
\(c.\frac{5}{6-2x}và\frac{3}{x^2-9};\frac{-5}{2x-6}và\frac{3}{x^2-9}\)
- Ta có : 2x - 6 = 2 ( x - 3 ) ; x2 - 9 = ( x + 3 ) ( x - 3 )
- MTC : 2 ( x - 3 ) ( x + 3 )
- NTP : x + 3 ; 2.
\(\frac{-5}{2x-6}=\frac{-5\left(x+3\right)}{\left(2x-6\right)\left(x+3\right)}=\frac{-5x-15}{2\left(x-3\right)\left(x+3\right)}\\ \frac{3}{x^2-9}=\frac{3.2}{2\left(x^2-9\right)}=\frac{6}{2\left(x-3\right)\left(x+3\right)}\)
Câu hỏi là gì thế bạn??????
Giải hpt:
1, \(\left\{{}\begin{matrix}x^2+y+x^3y+x^2y+xy=\frac{-5}{4}\\x^4+y^2+xy\left(1+2x\right)=\frac{-5}{4}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^4+2x^2y+x^2y^2=-2x+9\\x^2+2xy=6x+6\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}x-\frac{1}{x}=y-\frac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
3) ta xét phương trình thứ nhất
\(x-\frac{1}{x}=y-\frac{1}{y}\)
<=>\(x-y-\frac{1}{x}+\frac{1}{y}=0\)
<=>\(x-y-\left(\frac{1}{x}-\frac{1}{y}\right)=0\)
<=>\(x-y-\left(\frac{y-x}{xy}\right)=0\)
<=>\(\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\)
<=>\(x=y\) hoặc xy=-1
Với x=y thay vào phương trình thứ hai ta có
\(2x=x^3+1
\)
<=> \(x^3-2x+1=0\)
<=>\(x^3-x^2+x^2-x-x+1=0\)
<=>\(\left(x-1\right)\left(x^2+x-1\right)=0\)
<=> \(x=1\) hoặc \(x^2+x-1=0\)
\(x^2+x-1=0\) <=> \(x=\frac{-1+\sqrt{5}}{2}\)
hoặc \(x=\frac{-1-\sqrt{5}}{2}\)
Đối với xy=-1 thì y=-1/x thay vào phương trình 2 giải bình thường
Thực hiện các phếp tính
1)\(\frac{4x-1}{3x^2y}-\frac{7x-1}{3x^2y}\)
2)\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
3)\(\frac{1}{1-x}+\frac{2x}{x^2-1}\)
4)\(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}\)
Giai giúp giùm mình mai thi rồi
Đúng tích cho nha
Thank you!!!!!
a)= \(\frac{-1}{xy}\)
b)\(\frac{3}{2x+6}\) - \(\frac{x-6}{2x^2+6x}\)= \(\frac{3x}{2x\left(x+3\right)}\)- \(\frac{x-6}{2x\left(x+3\right)}\)= \(\frac{2x+6}{2x\left(x+3\right)}\)= \(\frac{2\left(x+3\right)}{2x\left(x+3\right)}\)= \(\frac{1}{x}\)
c)\(\frac{1}{xy-x^2}\)- \(\frac{1}{y^2-xy}\)= \(\frac{1}{x\left(x-y\right)}\)- \(\frac{1}{-y\left(x-y\right)}\)= \(\frac{y}{xy\left(x-y\right)}\)- \(\frac{-x}{xy\left(x-y\right)}\)= \(\frac{y+x}{xy\left(x-y\right)}\)
nhớ tick nhé
TÍNH:
a) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}\)
b) \(\frac{3-2x}{x^2-9}+\frac{1}{x^2-6}\)
c) \(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}\)
d) \(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}\)
a) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}=\frac{6.2}{2x\left(x+4\right)}+\frac{3x}{2x\left(x+4\right)}=\frac{12+3x}{2x\left(x+4\right)}=\frac{3\left(x+4\right)}{2x\left(x+4\right)}=\frac{3}{2x}\)
c) \(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}=\frac{-5.y}{2y\left(y+2\right)}+\frac{2\left(y-2\right)}{2y\left(y+2\right)}=\frac{-5y+2y-4}{2y\left(y+2\right)}=\frac{-3y-4}{2y\left(y+2\right)}\)
d) \(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}=\frac{x-1}{x\left(x-2y\right)}-\frac{3}{x\left(x-2y\right)}=\frac{x-1-3}{x\left(x-2y\right)}=\frac{x-4}{x\left(x-2y\right)}\)
bài 1 : thu gọn đa thức , tìm bậc , hệ số cao nhất
A = 15x^2y^3 + 7x^2 - 8x^3y^2 - 12x^2 + 11x^3y^2 - 12x^2y^3
B = 3x^5y + \(\frac{1}{3}\)xy^4 + \(\frac{3}{4}\)x^2y^3 - \(\frac{1}{2}\)x^5y + 2xy^4 - x^2y^3
bài 2 : tính giá trị biểu thức
A = 3x^3y + 6x^2y^2 + 3xy^3 tại x = \(\frac{1}{2}\); y = -\(\frac{1}{3}\)
B = x^2y^2 + xy +x^3 + y^3 tại x = -1 ; y = 3
bài 3 : cho đa thức
P(x) = x^4 + 2x^2 + 1
Q(x) = x^4 + 4x^3 + 2x^2- 4x + 1
tính P(-1); P(\(\frac{1}{2}\)) ; q(-2);Q(1)
bài 4 : tìm hệ số a của đa thức M(x)= ax^2 + 5x - 3 , tại M (-3) = 0
bài 5 : tìm các hệ số a , b của đa thức f(x) = ax + b , biết f(2) = 3 ; f(-1) = 9
Tính:
\(\frac{4x-1}{2x^2y}-\frac{7x-1}{3x^2y}\)
\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
\(\)Thank you so much!
a.\(\frac{4x-1}{2x^2y}-\frac{7x-1}{3x^2y}\) MTC=6x2y
\(=\frac{3\left(4x-1\right)}{6x^2y}-\frac{2\left(7x-1\right)}{6x^2y}\)
\(=\frac{12x-3-\left(14x-2\right)}{6x^2y}\)
\(=\frac{12x-3-14x+2}{6x^2y}\)
\(=\frac{-2x-1}{6x^2y}=\frac{2\left(-x-1\right)}{6x^2y}=-\frac{x-1}{3x^2y}\)
b.\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) MTC= 2x (x + 3)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{3x-\left(x-6\right)}{2x\left(x+3\right)}\)
\(=\frac{3x-x+6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)
c.\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)MTC= xy (x+2y).(x-2y)
\(=\frac{2xy\left(x-2y\right)}{xy\left(x+2y\right)\left(x-2y\right)}+\frac{xy\left(x+2y\right)}{xy\left(x+2y\right)\left(x-2y\right)}+\frac{4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\frac{2x^2y-4xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\frac{3x^2y-2xy^2+4xy}{xy\left(x-2y\right)\left(x+2y\right)}=\frac{xy\left(3x-2y+4\right)}{xy\left(x-2y\right)\left(x+2y\right)}=\frac{3x-2y+4}{\left(x-2y\right)\left(x+2y\right)}\)
Chọn mk nha!
1/ Xác định hệ số a và b sao cho \(\left(x^4+ax^3+b\right)⋮\left(x^2-1\right)\)
2/ Tìm \(n\inℕ\)để \(-7x^{n+1}y^6⋮4x^5y^n\)
3/ Tìm x và y biết: \(\frac{\left(x-2y\right)\left(x-7y\right)-x^2-4y^2}{x-2y}=18\)
4/ CMR: Giá trị biểu thức A không âm với mọi \(x\ne0\)của x và y: \(A=\frac{75x^5y^2-45x^4y^3}{3x^3y^2}-\frac{\frac{5}{2}x^2y^4-2xy^5}{\frac{1}{2}xy^2}\)
5/ Tìm GTNN của thương: \(\frac{4x^5+4x^4+4x^3-x-1}{2x^3+x-1}\)
6/ Tìm các \(x\inℤ\)để thương \(\frac{2x^5+4x^4-7x^3-44}{2x^3-7}\)có giá trị nguyên.
7/ CMR: Không tồn tại số \(n\inℕ\)để \(\left(n^6-n^4-2n+9\right)⋮\left(n^4+n^2\right)\)
Các bạn giúp mình một trong 7 bài này cũng được nhen. Giúp mình nhen! Mình sắp đi học rồi.
Thực hiện phép tính:
1,\(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)
2,\(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}\)
3,\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
4,\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
5,\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
Làmmmm
1/ \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)(ĐKXĐ:x\(\ne0\), x\(\ne\frac{1}{2}\))
= \(\frac{\left(1-2x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\frac{4x^2}{\left(2x-1\right)2x}-\frac{1}{2x\left(2x-1\right)}\)
\(=\frac{2x-1-4x^2+2x+4x^2-1}{2x\left(2x-1\right)}\)
\(=\frac{4x-2}{2x\left(2x-1\right)}=\frac{2\left(2x-1\right)}{2x\left(2x-1\right)}=\frac{1}{x}\)
KL:..............
2/\(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}\)(ĐKXĐ : x\(\ne1\))
\(=\frac{x^2+2}{x^3-1}+\frac{2x-2}{x^3-1}-\frac{x^2+x+1}{x^3-1}\)
\(=\frac{x^2+2+2x-2-x^2-x-1}{x^3-1}=\frac{x-1}{x^3-1}=\frac{1}{x^2+x+1}\)
Kl:....................
3/ \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)(x\(\ne\pm2y\))
= \(\frac{x^2+2xy}{x^2-4y^2}+\frac{x^2-2xy}{x^2-4y^2}-\frac{4xy}{x^2-4y^2}=\frac{2x^2-4xy}{x^2-4y^2}=\frac{2x\left(x-2y\right)}{x^2-4y^2}=\frac{2x}{x+2y}\)
Kl:................
5,thực hiện phép tính
1,\(\frac{4y^2}{11x^4}.\left(-\frac{3x^2}{8y}\right)\)
2,\(\frac{4x^2}{5y^2}:\frac{6x}{5y}:\frac{2x}{3y}\)
3,\(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}\)
4,\(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)
5,\(\frac{x^2-36}{2x+10}.\frac{3}{6-x}\)
6,\(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6y}\)
7,\(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}\)
1, \(\frac{4y^2}{11x^4}.\left(-\frac{3x^2}{8y}\right)\)\(=\frac{4y.y}{11x^2.x^2}.\frac{-3x^2}{2.4y}\)\(=\frac{y}{11x^2}.\frac{-3}{2}=\frac{-3y}{22x^2}\)
2, \(\frac{4x^2}{5y^2}:\frac{6x}{5y}:\frac{2x}{3y}\)\(=\frac{4x^2}{5y^2}.\frac{5y}{6x}.\frac{3y}{2x}\)\(=\frac{2x.2x}{5y.y}.\frac{5y}{3.2x}.\frac{3y}{2x}\)\(=\frac{2x}{y}.\frac{1}{3}.\frac{3y}{2x}\)
\(\frac{2x}{3y}.\frac{3y}{2x}=1\)
3, \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}\)\(=\frac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\frac{x+4}{2\left(x-2\right)}\)\(=\frac{\left(x+2\right)}{3}.\frac{1}{2}=\frac{x+2}{6}\)
4, \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\left(-\frac{2\left(x-2\right)}{x+2}\right)=\frac{5}{4}.\frac{-2}{1}=-\frac{5}{2}\)
5, \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{3}{-\left(x-6\right)}=\frac{x+6}{2\left(x+5\right)}.\frac{-3}{1}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)
6, \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6y}=\frac{\left(x-3y\right)\left(x+3y\right)}{\left(xy\right)^2}.\frac{3xy}{2\left(x-3y\right)}=\frac{x+3y}{xy}.\frac{3}{2}=\frac{3\left(x+3y\right)}{2xy}\)
7, \(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}=\frac{3\left(x-y\right)\left(x+y\right)}{5xy}.\frac{5xy.3x}{-2\left(x-y\right)}=\frac{3\left(x+y\right)}{1}.\frac{3x}{-2}=\frac{-9x\left(x+y\right)}{2}\)