tìm x
(2x-1)2= \(\frac{1}{9}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
MA
7
6 tháng 4 2017
\(\left(X+5\right)-\left(X-9\right)=X+2\)\(2\)
\(=>X+5-X+9=X+2\)
\(=>\left(X-X\right)+\left(5+9\right)=X+2\)
\(=>0+14=X+2\)
\(=>14=X+2\)
\(=>X=12\)
6 tháng 4 2017
(x+5)-(x-9)=x+2
x+5-x+9=x+2
x-x+14=x+2
12=x(cũng bớt mới về đi 2 đơn vị)
hoặc x=14-2suy ra x=12
1 tháng 6 2019
\(1,\)\(\frac{x+2}{x+3}+\frac{x-1}{x+1}=\frac{2}{x^2+4x+3}+1\)
\(\Rightarrow\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}\)
\(\Rightarrow\)\(x^2+3x+2+x^2-2x-3=2+x^2+4x+3\)
\(\Rightarrow x^2-3x-6=0\)
.....
1 tháng 6 2019
\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)
\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)
.....
PP
1
6 tháng 11 2018
\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{z+4}{9}=\frac{2x+3y-1}{6x}\)(1)
Áp dụng tính chất dãy tỉ sổ bằng nhau, ta được
\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{z+4}{9}=\frac{2x+3y-1}{6x}=\frac{\left(2x+1\right)+\left(3y-2\right)}{5+7}=\frac{2x+3y-1}{12}\)
\(\Rightarrow\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)
\(\Rightarrow\frac{12}{6x}=\frac{2x+3y-1}{2x+3y-1}=1\)
\(\Rightarrow\frac{2}{x}=1\)
\(\Rightarrow x=2\)
Thay x=2 vào (1), ta được
\(\frac{3y-2}{7}=\frac{z+4}{9}=\frac{2\cdot2+1}{5}=1\)
\(\Rightarrow\hept{\begin{cases}3y-2=7\\z+4=9\end{cases}}\Rightarrow\hept{\begin{cases}3y=9\\z=5\end{cases}}\Rightarrow\hept{\begin{cases}y=3\\z=5\end{cases}}\)
Vậy...hok tốt
PA
30 tháng 11 2016
\(MTC:\left(x-3\right)^2\left(x^2+3x+9\right)\)
\(\frac{x}{x^3-27}=\frac{x}{\left(x-3\right)\left(x^2+3x+9\right)}=\frac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(\frac{2x}{x^2-6x+9}=\frac{2x}{\left(x-3\right)^2}=\frac{2x\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(\frac{1}{x^2+3x+9}=\frac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(MTC:2\left(x-1\right)\left(x+1\right)\)
\(\frac{x-1}{2x+2}=\frac{x-1}{2\left(x+1\right)}=\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(\frac{x+1}{2x-2}=\frac{x+1}{2\left(x-1\right)}=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(\frac{1}{1-x^2}=-\frac{1}{\left(x-1\right)\left(x+1\right)}=-\frac{2}{2\left(x-1\right)\left(x+1\right)}\)
\(MTC:2\left(x+1\right)\left(x^2-x+1\right)\)
\(\frac{1}{x^3+1}=\frac{1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{2}{2\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{3}{2x+2}=\frac{3}{2\left(x+1\right)}=\frac{3\left(x^2-x+1\right)}{2\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2}{x^2-x+1}=\frac{4\left(x+1\right)}{2\left(x+1\right)\left(x^2-x+1\right)}\)
S
26 tháng 6 2018
a, \(\frac{1}{\left(2x-3\right)^2}=9\Leftrightarrow\left(\frac{1}{2x-3}\right)^2=3^2\Leftrightarrow\frac{1}{2x-3}=3\Leftrightarrow1=6x-9\Leftrightarrow x=\frac{5}{3}\)
b, \(\frac{1}{\left(x-1\right)^3}=-\frac{1}{27}\Leftrightarrow\left(\frac{1}{x-1}\right)^3=\left(\frac{-1}{3}\right)^3\Leftrightarrow\frac{1}{x-1}=\frac{1}{-3}\Leftrightarrow x-1=-3\Leftrightarrow x=-2\)
c, \(\left(x-1\right)^2=\left(2x-5\right)^2\Leftrightarrow\orbr{\begin{cases}x-1=2x-5\\x-1=-2x+5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}}\)
C
24 tháng 9 2019
a) Đặt \(x-1=a\)
\(pt\Leftrightarrow\frac{13}{a}+\frac{5}{2a}=\frac{6}{3a}\)
\(\Leftrightarrow\frac{31}{2a}=\frac{6}{3a}\)
\(\Leftrightarrow\frac{31}{2}=2\)(vô lí)
Vậy pt vô nghiệm
24 tháng 9 2019
a) \(\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}\)
\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{6}{3\left(x-1\right)}\)
\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{2}{x-1}\)
\(\frac{31}{2\left(x-1\right)}=\frac{2}{x-1}\)
\(\frac{31}{2}=2\)
=> không có x thỏa mãn đề bài.
b) \(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}\)
\(10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)\)
\(7-4x-3x^2=25x-25\)
\(7-4x-3x^2-25x+25=0\)
\(32-29x-3x^2=0\)
\(3x^2+29x-30=0\)
\(3x^2+32x-3x-32=0\)
\(x\left(3x+32\right)-\left(3x+32\right)=0\)
\(\left(3x+32\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}\)
\(\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}\)
1 tháng 7 2017
a ) MTC : \(2x\left(x+3\right)\left(x-3\right)\)
\(\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{3-2x}{x^2-9}=\frac{3-2x}{\left(x-3\right)\left(x+3\right)}=\frac{2x\left(3-2x\right)}{2x\left(x+3\right)\left(x-3\right)}\)
b ) MTC : \(2\left(-x\right)\left(x-1\right)^2\)
\(\frac{2x-1}{x-x^2}=\frac{2x-1}{-x\left(x-1\right)}=\frac{2\left(2x-1\right)\left(x-1\right)}{2\left(-x\right)\left(x-1\right)^2}\)
\(\frac{x+1}{2-4x+2x^2}=\frac{x+1}{2\left(x^2-2x+1\right)}=\frac{-x\left(x+1\right)}{2\left(-x\right)\left(x-1\right)^2}\)
