(3x-5)/18=-2/(3x-5)
/ là phần nhé giúp mik với
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\(\frac{1}{3}x+\frac{2}{5}=\frac{6}{5}+\frac{1}{2}x\)
\(\frac{1}{3}x-\frac{1}{2}x=\frac{6}{5}-\frac{2}{5}\)
\(-\frac{1}{6}x=\frac{4}{5}\)
\(x=\frac{4}{5}:\left(-\frac{1}{6}\right)\)
\(x=-\frac{24}{5}\)
Ta có: \(\frac{1}{3}x+\frac{2}{5}=\frac{6}{5}+\frac{1}{2}x\)
\(\Leftrightarrow\frac{1}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{6}{5}\)
\(\Leftrightarrow\frac{1}{6}x=-\frac{4}{5}\)
\(\Rightarrow x=-\frac{24}{5}\)
a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
a) \(7x^2=28\Leftrightarrow x^2=7\Leftrightarrow x=\sqrt{7}\)
c) \(\left(x-1\right)\left(x+\dfrac{5}{2}\right)=0\Leftrightarrow x\in\left\{1;\dfrac{-5}{2}\right\}\)
\(\left(-3x+2\right)-\left(5-3x\right)=-3\)
\(\Rightarrow-3x+2-5+3x=-3\)
\(\Rightarrow-3x+3x=-3+5-2\)
\(\Rightarrow0x=0\Rightarrow x\in Z\)
\(3+x-\left(3x-1\right)=6-2x\)
\(\Rightarrow3+x-3x+1=6-2x\)
\(\Rightarrow x-3x+2x=6-1-3\)
\(\Rightarrow0x=2\left(loại\right)\)
\(\left(x-5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{4}{3}\end{cases}}}\)
\(7x\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
\(\left(3x-1\right)2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}}\)
\(\Leftrightarrow2\left(x^2-\dfrac{3}{2}x+\dfrac{5}{2}\right)=0\\ \Leftrightarrow2\left(x-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{31}{16}\right)=0\\ \Leftrightarrow2\left(x-\dfrac{3}{4}\right)^2+\dfrac{31}{8}=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-\dfrac{3}{4}\right)^2+\dfrac{31}{8}\ge\dfrac{31}{8}>0\right]\)
a)
\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)
\(=16\)
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
ta gọi
ab=0,5 (a+b)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} ax+bx=67 kết quả =67\)
a) A= x^2 - 6x + 5
A=x^2-6x+9-4
A=(x-3)^2-4>hoặc= -4
Pmin =-4 <=> x-3=0 <=> x=3
P/s máy mình lag nên ko sủ dụng được cồn thức
à à mình nhân nhầm đấy, do công thức bị lỗi
dòng thứ 2 phải là : \(\left(3x-5\right)^2=-36\)( vô lí )
vì \(\left(3x-5\right)^2\ge0\forall x;-36< 0\)
\(\frac{3x-5}{18}=-\frac{2}{3x-5}\)
\(\Rightarrow\left(3x-5\right)^2=-16\)( vô lí )
vì \(\left(3x-5\right)^2\ge0\forall x;-16< 0\)