Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)
d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)
e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)
f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)
\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)
2)
a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)
b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )
d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)
e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)
a)\(\dfrac{2x^2-10xy}{2xy}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)
\(=\dfrac{2x\left(x-5y\right)}{2xy}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)
\(=\dfrac{x-5y}{y}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)
\(=\dfrac{x\left(x-5y\right)+x\left(5y-x\right)+y\left(x+2y\right)}{xy}\)
\(=\dfrac{x^2-5xy+5xy-x^2+xy+2y^2}{xy}\)
\(=\dfrac{y\left(x+2y\right)}{xy}\)
b) \(\dfrac{x+1}{2x-2}+\dfrac{x^2+3}{2-2x^2}\)
\(=\dfrac{x+1}{2x-2}-\dfrac{x^2+3}{2x^2-2}\)
\(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x^2-1\right)}\)
\(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\) MTC: \(2\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)-\left(x^2+3\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)
e) \(\dfrac{2x^2-xy}{x-y}+\dfrac{xy+y^2}{y-x}+\dfrac{2y^2-x^2}{x-y}\)
\(=\dfrac{2x^2-xy}{x-y}-\dfrac{xy+y^2}{x-y}+\dfrac{2y^2-x^2}{x-y}\)
\(=\dfrac{\left(2x^2-xy\right)-\left(xy+y^2\right)+\left(2y^2-x^2\right)}{x-y}\)
\(=\dfrac{2x^2-xy-xy-y^2+2y^2-x^2}{x-y}\)
\(=\dfrac{x^2-2xy+y^2}{x-y}\)
\(=\dfrac{\left(x-y\right)^2}{x-y}\)
\(=x-y\)
a,\(\frac{x^2+y^2-xy}{x^2-y^2}:\frac{x^3+y^3}{x^2+y^2-2xy} =\frac{x^2+y^2-xy}{(x-y)(x+y)}\frac{(x+y)^2}{(x+y) (x^2-xy+y^2)}=\frac{1}{x-y} \)
b,\(\frac{x^3y+xy^3}{x^4y}:(x^2+y^2)=\frac{xy(x^2+y^2)}{x^4y(x^2+y^2)}=\frac{1}{x^3} \)
c,\(\frac{x^2-xy}{y}:\frac{x^2-xy}{xy+y}:\frac{x^2-1}{x^2+y} =\frac{x(x-y)y(x+y)(x^2+y)}{yx(x-y)(x^2-1)} =\frac{(x^2+y)(x+y)}{x^2-1} \)
d,\(\frac{x^2+y}{y}:(\frac{z}{x^2}:\frac{xy}{x^2y})=\frac{x^2+y}{ y}:(\frac{z}{x^2}\frac{x^2y}{xy})=\frac{x^2+y}{y}\frac{z}{x} \)
a) \(\dfrac{3x-2}{2xy}+\dfrac{7x+2}{2xy}\)
\(=\dfrac{\left(3x-2\right)+\left(7x+2\right)}{2xy}\)
\(=\dfrac{3x-2+7x+2}{2xy}\)
\(=\dfrac{10x}{2xy}\)
\(=\dfrac{5}{y}\)
b) \(\dfrac{5x+y^2}{x^2y}+\dfrac{x^2-5y}{xy^2}\) MTC: \(x^2y^2\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}+\dfrac{x\left(x^2-5y\right)}{x^2y^2}\)
\(=\dfrac{y\left(5x+y^2\right)+x\left(x^2-5y\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3+x^3-5xy}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
c) \(\dfrac{3x-2}{2xy}-\dfrac{7x-y}{2xy}\)
\(=\dfrac{\left(3x-2\right)-\left(7x-y\right)}{2xy}\)
\(=\dfrac{3x-2-7x+y}{2xy}\)
\(=\dfrac{-2-4x+y}{2xy}\)
d) \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\) MTC: \(x^2y^2\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}-\dfrac{x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{y\left(5x+y^2\right)-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
e) \(\dfrac{16xy}{3x-1}.\dfrac{3-9x}{12xy^3}\)
\(=\dfrac{16xy\left(3-9x\right)}{12xy^3\left(3x-1\right)}\)
\(=\dfrac{4\left(3-9x\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-4\left(9x-3\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-4.3\left(3x-1\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-12}{3y^2}\)
\(=\dfrac{-4}{y^2}\)
f) \(\dfrac{8xy}{3x-1}:\dfrac{12xy^3}{5-15x}\)
\(=\dfrac{8xy}{3x-1}.\dfrac{5-15x}{12xy^3}\)
\(=\dfrac{8xy\left(5-15x\right)}{12xy^3\left(3x-1\right)}\)
\(=\dfrac{2\left(5-15x\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-2\left(15x-5\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-2.5\left(3x-1\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-10}{3y^2}\)
c, x^3 - y^3 = xy + 8
1) Nếu x-y <= -1
(x -y)(x^2 + xy + y^2) = xy +8
=> (x -y)(x^2 + xy + y^2) <= -(x^2 + xy +y^2)
=> xy +8 <= -(x^2 + xy +y^2)
=> (x+y)^2 + 8 <=0 => Vô nghiệm
2) Nếu x-y =0 => x=y , Vô nghiệm
3) x- y>=1
=> (x -y)(x^2 + xy + y^2) >= x^2 + xy + y^2
=> xy + 8 >= x^2 + xy + y^2
=> x^2 + y^2 <=8
=> x^2 <=8
=> x=0 => y= -2
=> x= 1 => y + y^3 + 7 =0 (loại)
Đặt \(A=\left[\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-1}{x^2-xy-2y^2}\right):\dfrac{4x^4+4x^2y+y^2-4}{x^2+x+xy+y}\right]:\dfrac{x+1}{2x^2+y+2}\)
\(A=\left[\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-1}{\left(x+y\right).\left(x-2y\right)}\right):\dfrac{\left(2x^2+y+2\right).\left(2x^2+y-2\right)}{\left(x+y\right).\left(x+1\right)}\right]:\dfrac{x+1}{2x^2+y+2}\)
\(A=\left(\dfrac{\left(x-y\right).\left(x+y\right)+x^2+y^2+y-2}{\left(x+y\right).\left(2y-x\right)}.\dfrac{\left(x+y\right).\left(x+1\right)}{\left(2x^2+y+2\right).\left(2x^2+y-2\right)}\right):\dfrac{2x^2+y+2}{x+1}\)
\(A=\left(\dfrac{2x^2+y-2}{2y-x}.\dfrac{x+1}{2x^2+y-2}\right).\dfrac{1}{x+1}\)
\(A=\dfrac{1}{2y-x}\)
Thay \(x=-1,76\) và \(y=\dfrac{3}{25}\) vào biểu thức ta được:
\(A=\dfrac{1}{2.\dfrac{3}{25}-\left(-1,76\right)}\)
\(A=\dfrac{1}{2}\)
Lời giải:
ĐKXĐ: \(y\neq \pm 5; y\neq 0\)
Ta có:
\(\frac{y+1}{y^2-5y}-\frac{y-5}{2y^2+10y}=\frac{y+25}{2y^2-50}\)
\(\Leftrightarrow \frac{y+1}{y(y-5)}-\frac{y-5}{2y(y+5)}-\frac{y+25}{2(y^2-25)}=0\)
\(\Leftrightarrow \frac{2(y+1)(y+5)}{2y(y-5)(y+5)}-\frac{(y-5)(y-5)}{2y(y+5)(y-5)}-\frac{y(y+25)}{2y(y^2-25)}=0\)
\(\Leftrightarrow \frac{2(y^2+6y+5)}{2y(y^2-25)}-\frac{y^2-10y+25}{2y(y^2-25)}-\frac{y^2+25y}{2y(y^2-25)}=0\)
\(\Leftrightarrow \frac{2(y^2+6y+5)-(y^2-10y+25)-(y^2+25y)}{2y(y^2-25)}=0\)
\(\Leftrightarrow \frac{-3(y+5)}{2y(y^2-25)}=0\)
\(\Leftrightarrow -3(y+5)=0\Leftrightarrow y+5=0\Leftrightarrow y=-5\) (không t/m ĐKXĐ)
Vậy PT vô nghiệm.
\(\dfrac{y+1}{y^2-5y}-\dfrac{y-5}{2y^2+10y}=\dfrac{y+25}{2y^2-50}\left(ĐKXĐ:y\ne O;y\ne\pm5\right)\)
\(\Leftrightarrow\dfrac{y+1}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)
\(\Leftrightarrow\dfrac{2\left(y+1\right)\left(y+5\right)-\left(y-5\right)^2}{2y\left(y-5\right)\left(y+5\right)}=\dfrac{y\left(y+25\right)}{2y\left(y-5\right)\left(y+5\right)}\)
\(\Rightarrow2\left(y+1\right)\left(y+5\right)-\left(y-5\right)^2=y\left(y+25\right)\)
\(\Leftrightarrow\left(2y+2\right)\left(y+5\right)-\left(y^2-10y+25\right)=y^2+25y\)
\(\Leftrightarrow2y^2+10y+2y+10-y^2+10y-25=y^2+25y\)
\(\Leftrightarrow y^2+22y-15=y^2+25y\)
\(\Leftrightarrow y^2-y^2+22y-25y=15\)
\(\Leftrightarrow-3y=15\)
\(\Leftrightarrow y=-5\) (ko thỏa mãn ĐKXĐ)
Vậy ....................