Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\dfrac{9}{x^2-9}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\) ( sửa đề \(x^3-9\) thành \(x^2-9\) )
\(=\left(\dfrac{9}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\dfrac{9+x-3}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{3\left(x-3\right)}{3x\left(x+3\right)}-\dfrac{x.x}{3x\left(x+3\right)}\right)\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{\left(x+6\right)3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(3x-9-x^2\right)}\)
\(=\dfrac{3x\left(x+6\right)}{\left(x-3\right)\left(3x-9-x^2\right)}\)
<=>\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
<=>\(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)
Vì \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)
=>x+100=0
<=>x=-100
k nha bạn
\(\Leftrightarrow\frac{37x+1648}{1026}=\frac{11x+556}{272}\Rightarrow\left(37x+1648\right)272=1026\left(11x+556\right)\)
<=>(37x+1648)272=272(37x+1648)
=>272(37x+1648)=1026(11x+556)
=>10064x+448256=11286x+570456
<=>-1222x=122200
=>x=122200:-1222
=>x=-100 ( dễ hiểu chưa hả )
a) \(\dfrac{2\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x}\)
\(a,=\dfrac{2\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x}\\ b,=\dfrac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\\ =\dfrac{\left(1-3x\right)\left(2x-1\right)+\left(2x-1\right)\left(3x-2\right)}{2x\left(2x-1\right)}\\ =\dfrac{\left(2x-1\right)\left(1-3x+3x-2\right)}{2x}=\dfrac{-1}{2x}\)
Lời giải:
$(2x-3)(x^2+1)=0$
\(\Leftrightarrow \left[\begin{matrix} 2x-3=0\\ x^2+1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{3}{2}(\text{chọn})\\ x^2=-1<0(\text{vô lý})\end{matrix}\right.\)
Vậy pt có nghiệm $x=\frac{3}{2}$
\(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{2.\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{1.\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow2x-4-x+1=3x-11\\ \Leftrightarrow x-3=3x-11\\ \Leftrightarrow x-3x=-11+3\\ \Leftrightarrow-2x=-8\\ \Leftrightarrow x=4\)
Vậy tập nghiệm của phương trình là S = { 4 }
Theo đề, ta có:
\(\dfrac{x}{45}+\dfrac{x}{30}=5\)
hay x=90
a: \(Q=\dfrac{2x^2-4x+x-3-6}{\left(x-3\right)\left(x-2\right)}\cdot\dfrac{x-2}{x^2+1}=\dfrac{2x^2-3x-9}{x-3}\cdot\dfrac{1}{x^2+1}\)
\(=\dfrac{2x^2-6x+3x-9}{x-3}\cdot\dfrac{1}{x^2+1}=\dfrac{2x+3}{x^2+1}\)
b: Để Q>0 thì 2x+3>0
hay x>-3/2
a) Để phương trình có nghiệm thì: m2≠0 =>m≠0
b) Vì phương trình có nghiệm bằng -2m
=>-(m+2)2m-m2=0 ⇔-2m2-4m-m2=0 ⇔-3m2-4m=0 ⇔-m(3m+4)=0
⇔m=0 hay m=\(\dfrac{-4}{3}\)mà m phải khác 0 nên m=\(\dfrac{-4}{3}\).
c) -(m+2)x-m2=0 ⇔x=\(\dfrac{m^2}{m+2}\)>0 ⇔m+2>0 ⇔m>-2.
d) -(m+2)x-m2=0 ⇔x=\(\dfrac{m^2}{m+2}\).
Để x nguyên thì m2 ⋮ m+2.
⇔ m2-4+4 ⋮ m+2
⇔ 4 ⋮ m+2
⇔ m∈{-1;-3;0;-4;2;-6} mà m khác 0 nên m∈{-1;-3;-4;2;-6}
a: \(=5x^2-10x-5x^2+7x=-3x\)
b: \(=2x^3+3xy^2-4y-3xy^2=2x^3-4y\)
\(\Leftrightarrow x\cdot\dfrac{18}{5}+x\cdot\dfrac{17}{6}-x\cdot\dfrac{10}{7}+x\cdot\dfrac{1}{8}=\dfrac{15}{8}\)
\(\Leftrightarrow x\cdot\dfrac{4309}{840}=\dfrac{15}{8}\)
=>x=1575/4309