Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)
\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)
\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)
\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)
\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
3.15:
a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)
b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3.16
\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)
\(\Leftrightarrow-14m+35-2m^2+8=0\)
\(\Leftrightarrow-14m-2m^2+43=0\)
\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)
\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)
\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)
\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)
pt vô nghiệm
`|5x| = - 3x + 2`
Nếu `5x>=0<=> x>=0` thì phương trình trên trở thành :
`5x =-3x+2`
`<=> 5x +3x=2`
`<=> 8x=2`
`<=> x= 2/8=1/4` ( thỏa mãn )
Nếu `5x<0<=>x<0` thì phương trình trên trở thành :
`-5x = -3x+2`
`<=>-5x+3x=2`
`<=> 2x=2`
`<=>x=1` ( không thỏa mãn )
Vậy pt đã cho có nghiệm `x=1/4`
__
`6x-2<5x+3`
`<=> 6x-5x<3+2`
`<=>x<5`
Vậy bpt đã cho có tập nghiệm `x<5`
\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
a: =>4x^2-24x+36-4x^2+4x-1<10
=>-20x<10-35=-25
=>x>=5/4
b: =>x(x^2-25)-x^3-8<=3
=>x^3-25x-x^3-8<=3
=>-25x<=11
=>x>=-11/25
\(a,=>x^3-2x^2+4x+2x^2-4x+8-x^3+2x-15=0\)
\(< =>2x-7=0< =>x=\dfrac{7}{2}\)
b,\(=>x\left(x^2-25\right)-\left(x+2\right)\left(x^2-2x+4\right)-3=0\)
\(< =>x^3-25x-x^3+2x^2-4x-2x^2+4x-8-3=0\)
\(< =>-25x-11=0\)
\(< =>x=-0,44\)
a: \(\Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\)
=>(4x+14+3x+9)(4x+14-3x-9)=0
=>(7x+23)(x+5)=0
=>x=-23/7 hoặc x=-5
\(a,\\ \Leftrightarrow7x^2+58x+115=0\\ \Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x+5=0\\7x+23=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
\(b,\\ \Leftrightarrow\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]=0\\ \Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=0\\ \LeftrightarrowĐặt.x^2+6x+5=a\\ \Leftrightarrow a=a\left(a+3\right)=10\\ \Leftrightarrow a^2+3a-10=0\\ \Leftrightarrow\left(a+5\right)\left(a-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=-5\\a=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2+6x+5=-5\\x^2+6x+5=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+6x+10=0\\x^2+6x+3=0\end{matrix}\right.\\ \left(Vô.n_o\Delta=36-40=-4< 0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{6}\\x=-3-\sqrt{6}\end{matrix}\right.\)
a, \(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\left(ĐKXĐ:x\ne\pm2\right)\)
\(\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+2}+\frac{3}{x-2}\)
\(\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
Khử mẫu : \(9=\left(x-1\right)\left(x-2\right)+3\left(x+2\right)\)
Đến đây nhường bn, rất dễ =))
b, \(\frac{1}{x-5}-\frac{3}{x^2-6x+5}=\frac{5}{x-1}\)
\(\frac{1}{x-5}-\frac{3}{\left(x-5\right)\left(x-1\right)}=\frac{5}{\left(x-1\right)}\)
\(\frac{\left(x-1\right)}{x-5}-\frac{3}{\left(x-5\right)\left(x-1\right)}=\frac{5\left(x-5\right)}{\left(x-1\right)\left(x-5\right)}\)
Khử mẫu \(x-1-3=5\left(x-5\right)\)
Tự lm nốt mà cho mk hỏi, đề bài có bpt mà bpt đâu
\(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\left(ĐKXĐ:x\ne2;-2\right)\)
\(< =>\frac{9}{x^2-2^2}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(< =>\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3x+6}{\left(x+2\right)\left(x-2\right)}\)
\(< =>9=x^2-2x-x+2+3x+6\)
\(< =>x^2-\left(2x+x-3x\right)+\left(2+6-9\right)=0\)
\(< =>x^2-2=0\)\(< =>x^2=2\)
\(< =>x=\pm\sqrt{2}\left(tmđk\right)\)
Vậy tập nghiệm của phương trình trên là \(\pm\sqrt{2}\)
