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, \(\frac{5}{6-2x}và\frac{3}{x^2-9}\)
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Những câu hỏi liên quan
15 tháng 1 2020
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
12 tháng 7 2019
a
Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)
\(\Rightarrow x=2k+1;y=3k+2;z=4k+3\)
Thay vào,ta được:
\(2\left(2k+1\right)+3\left(3k+2\right)-\left(4k+3\right)=50\)
\(\Leftrightarrow4k+2+9k+6-4k-3=50\)
\(\Leftrightarrow9k+5=50\)
\(\Leftrightarrow9k=45\)
\(\Leftrightarrow k=5\)
12 tháng 7 2019
\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}=\frac{5x-5}{10}=\frac{3y+9}{12}=\frac{4z-20}{24}\)
\(=\frac{5x-5-3y-9-4z+20}{10-12-24}=\frac{\left(5x-3y-4z\right)+\left(20-5-9\right)}{26}=\frac{46+6}{26}=2\)
\(\Rightarrow x=2\cdot2+1=5\)
\(y=4\cdot2-3=5\)
\(z=2\cdot6+5=17\)
Câu c tương tự như câu 1
14 tháng 8 2020
a)\(ĐKXĐ:x\ne0;-1\)
Ta có:\(\frac{x^3+1}{x}.\left(\frac{1}{x+1}+\frac{x-1}{x^2-x+1}\right)=\frac{x^3+1}{x}.\frac{\left(x^2-x+1\right)+\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^3+1}{x}.\frac{x^2-x+1+\left(x^2-1\right)}{x^3+1}=\frac{2x^2-x}{x}=\frac{2x\left(x-1\right)}{x}=2\left(x-1\right)\)
6 tháng 1 2018
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!
14 tháng 2 2020
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
\(a.\frac{x}{4x^2y^3}và\frac{5}{6x^3y}\)
- 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.
\(\frac{-5}{2x-6}=\frac{-5\left(x+3\right)}{\left(2x-6\right)\left(x+3\right)}=\frac{-5x-15}{2\left(x-3\right)\left(x+3\right)}\\ \frac{3}{x^2-9}=\frac{3.2}{2\left(x^2-9\right)}=\frac{6}{2\left(x-3\right)\left(x+3\right)}\)
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