DK
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
3 tháng 5 2021
=) vào ngay quả bảng phá dấu GTTĐ, cay thế :<
a, \(3x+\frac{2x}{3}-3=\frac{5}{2}x-2\Leftrightarrow\frac{18x+4x-18}{6}=\frac{15x-12}{6}\)
\(\Rightarrow22x-18=15x-12\Leftrightarrow7x=6\Leftrightarrow x=\frac{6}{7}\)
Vậy pt có nghiệm x = 6/7
b, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)
\(\Leftrightarrow\frac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\frac{x+7}{12}\)
\(\Rightarrow18x+9-10x-6+4x+4=x+7\)
\(\Leftrightarrow12x+7=x+7\Leftrightarrow11x=0\Leftrightarrow x=0\)
Vậy pt có nghiệm là x = 0
c, \(\frac{3x}{x-3}-\frac{x-3}{x+3}=2\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow\frac{3x\left(x+3\right)-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow3x^2+9x-x^2+6x-9=2\left(x^2-9\right)\)
\(\Leftrightarrow2x^2+15x-9=2x^2-18\Leftrightarrow15x+9=0\Leftrightarrow x=-\frac{9}{15}=-\frac{3}{5}\)
Vậy pt có nghiệm là x = -3/5
d, Sửa đề : \(\frac{x+10}{2003}+\frac{x+6}{2007}+\frac{x+2}{2011}+3=0\)
\(\Leftrightarrow\frac{x+10}{2003}+1+\frac{x+6}{2007}+1+\frac{x+2}{2011}+1=0\)
\(\Leftrightarrow\frac{x+2013}{2003}+\frac{x+2013}{2007}+\frac{x+2013}{2011}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2003}+\frac{1}{2007}+\frac{1}{2011}\ne0\right)=0\Leftrightarrow x=-2013\)
Vậy pt có nghiệm là x = -2013
3 tháng 5 2021
e, \(4\left(x+5\right)-3\left|2x-1\right|=10\)
\(\Leftrightarrow4x+20-3\left|2x-1\right|=10\Leftrightarrow-3\left|2x-1\right|=-10-4x\)
\(\Leftrightarrow\left|2x-1\right|=\frac{10+4x}{3}\)
ĐK : \(\frac{10+4x}{3}\ge0\Leftrightarrow10+4x\ge0\Leftrightarrow x\ge-\frac{10}{4}=-\frac{5}{2}\)
TH1 : \(2x-1=\frac{10+4x}{3}\Rightarrow6x-3=10+4x\Leftrightarrow2x=13\Leftrightarrow x=\frac{13}{2}\)( tm )
TH2 : \(2x-1=\frac{-10-4x}{3}\Rightarrow6x-3=-10-4x\Leftrightarrow10x=-7\Leftrightarrow x=-\frac{7}{10}\)( tm )
f, để mình xem lại đã, quên cách phá GTTĐ rồi :v :>
7 tháng 5 2020
a)
\(\left(5x+3\right)\cdot\left(x^2+4\right)\cdot\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x+3=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{5}\\x=4\end{matrix}\right.\)
b)
\(\left(4x-1\right)\cdot\left(x-3\right)-\left(x-2\right)\cdot\left(5x+2\right)=0\\ \Leftrightarrow4x^2-12x-x+3-5x^2-2x+10x+4=0\\ \Leftrightarrow-x^2-5x+7=0\\ \Rightarrow x=\left[{}\begin{matrix}-\frac{5+\sqrt{53}}{2}\\-\frac{5-\sqrt{53}}{2}\end{matrix}\right.\)
c)
\(\left(x+3\right)\cdot\left(x-5\right)+\left(x+3\right)\cdot\left(3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(x-5+3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(4x-9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)
d)
\(\left(x+6\right)\cdot\left(3x-1\right)+x^2-36=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x^2-36\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x+6\right)\cdot\left(x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1+x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(4x-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)
e)
\(0.75x\cdot\left(x+5\right)=\left(x+5\right)\cdot\left(3-1.25x\right)\\ \Leftrightarrow0.75x\cdot\left(x+5\right)-\left(x+5\right)\cdot\left(3-1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(0.75x-3+1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(2x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.\)
HY
8 tháng 2 2020
\(a.1,2-\left(x-0,8\right)=-2\left(0,9+x\right)\\\Leftrightarrow1,2-x+0,8=-1,8-2x\\ \Leftrightarrow-x+2x=-1,2-0,8-1,8\\ \Leftrightarrow x=-3,8\)
Vậy nghiệm của phương trình trên là \(-3,8\)
\(b.2,3x-2\left(0,7+2x\right)=3,6-1,7x\\ \Leftrightarrow2,3x-1,4-4x=3,6-1,7x\\ \Leftrightarrow2,3x-4x+1,7x=1,4+3,6\\ \Leftrightarrow0x=5\)
\(\Rightarrow\)Vô nghiệm
\(c.5-\left(x-6\right)=4\left(3-2x\right)\\ \Leftrightarrow5-x+6=12-8x\\ \Leftrightarrow-x+8x=-5-6+12\\ \Leftrightarrow7x=1\\\Leftrightarrow x=\frac{1}{7}\)
Vậy nghiệm của phương trình trên là \(\frac{1}{7}\)
\(d.3,6-0,5\left(2x+1\right)=x-0,25\left(2-4x\right)\\ \Leftrightarrow3,6-x-0,5=x-0,5+x\\\Leftrightarrow -x-x-x=-3,6-0,5+0,5\\ \Leftrightarrow-3x=-3,6\\\Leftrightarrow x=1,2\)
Vậy nghiệm của phương trình trên là \(1,2\)
\(e.\left(x-3\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\\\Leftrightarrow x^2-x^2+4x-3x-6x+8x=12-4+16\\ \Leftrightarrow3x=24\\ \Leftrightarrow x=8\)
Vậy nghiệm của phương trình trên là \(8\)
7 tháng 3 2020
\(b,\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(2x^2-2x=x+3-x^2-3x\)
\(2x^2-2x=-2x+3-x^2\)
\(2x^2=3-x^2\)
\(2x^2+x^2=3\)
\(3x^2=3\Leftrightarrow x^2=1\Leftrightarrow x=\pm\sqrt{1}\)
tớ n g u nên cần tg suy nghĩ thêm :v
7 tháng 3 2020
câu a tìm ra r nè , vất vả :v ( kiên trì lắm đấy )
\(a,\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2+1\right)\)
\(9x^3+9x^2-4x-4-3x^2-3x-2x^2-2=0\)
\(6x^3+7x^2-7x-6=0\)
\(\left(6x^2+13x+6\right)\left(x-1\right)=0\)
\(Th1:6x^2+9x+4x+6=0\)
\(\Leftrightarrow\left[3x\left(2x+3\right)+2\left(2x+3\right)\right]=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\3x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\3x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{2}{3}\end{cases}}}\)
\(Th2:x-1=0\Leftrightarrow x=1\)
PN
9 tháng 5 2021
a,\(2x+5=2-x\)
\(< =>2x+x+5-2=0\)
\(< =>3x+3=0\)
\(< =>x=-1\)
b, \(/x-7/=2x+3\)
Với \(x\ge7\)thì \(PT< =>x-7=2x+3\)
\(< =>2x-x+3+7=0\)
\(< =>x+10=0< =>x=-10\)( lọai )
Với \(x< 7\)thì \(PT< =>7-x=2x+3\)
\(< =>2x+x+3-7=0\)
\(< =>3x-4=0< =>x=\frac{4}{3}\) ( loại )
PN
9 tháng 5 2021
c,\(\frac{4}{x+2}-\frac{4x-6}{4x-x^3}=\frac{x-3}{x\left(x-2\right)}\left(đk:x\ne-2;0;2\right)\)
\(< =>\frac{4x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{4x-6}{x\left(x-2\right)\left(2+x\right)}=\frac{\left(x-3\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
\(< =>4x^2-8x+4x-6=x^2-x-6\)
\(< =>4x^2-x^2-4x+x-6+6=0\)
\(< =>3x^2-3x=0< =>3x\left(x-1\right)=0< =>\orbr{\begin{cases}x=0\left(loai\right)\\x=1\left(tm\right)\end{cases}}\)
TN
18 tháng 3 2020
\(a.\frac{4x-3}{x-5}=\frac{29}{3}\\ \Leftrightarrow\frac{3\left(4x-3\right)}{3\left(x-5\right)}=\frac{29\left(x-5\right)}{3\left(x-5\right)}\\ \Leftrightarrow3\left(4x-3\right)=29\left(x-5\right)\\ \Leftrightarrow3\left(4x-3\right)-29\left(x-5\right)=0\\ \Leftrightarrow12x-9-29x+145=0\\ \Leftrightarrow-17x+136=0\\ \Leftrightarrow-17x=-136\\ \Leftrightarrow x=\frac{-136}{-17}=8\)
\(b.\frac{2x-1}{5-3x}=2\\ \Leftrightarrow\frac{2x-1}{5-3x}=\frac{4}{2}\\ \Leftrightarrow\frac{2\left(2x-1\right)}{2\left(5-3x\right)}=\frac{4\left(5-3x\right)}{2\left(5-3x\right)}\\ \Leftrightarrow2\left(2x-1\right)=4\left(5-3x\right)\\ \Leftrightarrow2\left(2x-1\right)-4\left(5-3x\right)=0\\ \Leftrightarrow4x-2-20+12x=0\\ \Leftrightarrow16x-22=0\\ \Leftrightarrow16x=22\\ \Leftrightarrow x=\frac{22}{16}=\frac{11}{8}\)
\(c.\frac{4x-5}{x-1}=\frac{2+x}{x-1}\\ \Leftrightarrow4x-5=2+x\\ \Leftrightarrow4x-5-2-x=0\\ \Leftrightarrow3x-7=0\\ \Leftrightarrow3x=7\\ \Leftrightarrow x=\frac{7}{3}\)
TN
18 tháng 3 2020
\(d.\frac{7}{x+2}=\frac{3}{x-5}\\ \Leftrightarrow\frac{7\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}=\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-5\right)}\\ \Leftrightarrow7\left(x-5\right)=3\left(x+2\right)\\ \Leftrightarrow7\left(x-5\right)-3\left(x+2\right)=0\\ \Leftrightarrow7x-35-3x-6=0\\ \Leftrightarrow4x-41=0\\ \Leftrightarrow4x=41\\ \Leftrightarrow x=\frac{41}{4}\)
\(e.\frac{2x+5}{2x}-\frac{x}{x+5}=0\\ \Leftrightarrow\frac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\frac{x.2x}{2x\left(x+5\right)}=0\\ \Leftrightarrow\left(2x+5\right)\left(x+5\right)-2x^2=0\\ \Leftrightarrow2x^2+10x+5x+25-2x^2=0\\ \Leftrightarrow15x+25=0\\ \Leftrightarrow15x=-25\\ \Leftrightarrow x=\frac{-25}{15}=\frac{-5}{3}\)
\(f.\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\\\Leftrightarrow\frac{18\left(12x+1\right)}{18\left(11x-4\right)}+\frac{\left(10x-4\right).2\left(11x-4\right)}{9.2\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\\ \Leftrightarrow18\left(12x+1\right)+\left(10x-4\right).2\left(11x-4\right)=\left(20x+17\right)\left(11x-4\right)\\ \Leftrightarrow220x^2+48x+50=220x^2+107x-68\\ \Leftrightarrow48x+50=107x-68\\ \Leftrightarrow48x-107x=-68-50\\ \Leftrightarrow59x=-118\\ \Leftrightarrow x=-2\)
e: \(\dfrac{1}{6}+\dfrac{x}{4}=\dfrac{2-x}{3}\)
=>\(\dfrac{2+3x}{12}=\dfrac{4\left(2-x\right)}{12}\)
=>4(2-x)=3x+2
=>8-4x=3x+2
=>-7x=-6
=>\(x=\dfrac{6}{7}\)
f: \(\dfrac{x+2}{3}-\dfrac{3}{4}=\dfrac{x-1}{2}\)
=>\(\dfrac{4\left(x+2\right)-9}{12}=\dfrac{6\left(x-1\right)}{12}\)
=>4x+8-9=6x-6
=>4x-1=6x-6
=>-2x=-5
=>x=2,5
Đố thịnh giải đc