\(2x\left(x^2-25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-25=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

\(2x\left(3x-5\right)+\left(3x-5\right)=0\)

\(\left(2x+1\right)\left(3x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x-5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{3}\end{cases}}\)

\(9\left(3x-2\right)-x\left(2-3x\right)=0\)

\(9\left(3x-2\right)+x\left(3x-2\right)=0\)

\(\left(9+x\right)\left(3x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}9+x=0\\3x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-9\\x=\frac{2}{3}\end{cases}}\)

\(\left(2x-1\right)^2=25\)

\(\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

(16 - 4x)(x + 3) - x(1 - 4x) = 39

16x + 48 - 4x² - 12x - x + 4x² = 39

3x = 39 - 48

3x = -9

x = -3

Thank you Lộc nhìu nha!  ^ - ^

(Ko chép lại đề)

\(a.\Rightarrow\orbr{\begin{cases}3x+5=0\\4-3x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{4}{3}\end{cases}}\)

\(b.\left(3x-2\right)\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\x-7=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=7\end{cases}}\)

\(c.7x^2=28\)

\(x^2=4\)

\(x^2=2^2\)

\(x=\pm2\)

\(d.\left(2x+1\right)\left(1+x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\1+x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-1\end{cases}}\)

a)\(\left(3x+5\right)\left(4-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+5=0\\4-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{4}{3}\end{cases}}}\)

b)\(3x\left(x-7\right)-2\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\3x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=\frac{2}{3}\end{cases}}}\)

c) \(7x^2-28=0\)

\(\Leftrightarrow7x^2=28\)

\(\Leftrightarrow7x^2=7.4\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

d)\(\left(2x+1\right)+x\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(1+x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\1+x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-1\end{cases}}}\)

#H

Giúp mình câu này với

Số mol \(0,6.10^{23}\text{ phân tử }O_2=0,6.16.2.10^{23}=19,2.10^{23}\)

Suy ra \(M_A=\frac{m_A}{n_A}=\frac{8}{19,2.10^{23}}\left(\text{g/mol}\right)\)

P/s: Ko chắc

a, điều kiện xác định là \(x\ne1;x\ne-1\)

\(\frac{3x+3}{x^2-1}\)

\(=\frac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{3}{x-1}\)

b, để \(\frac{3x+3}{x^2-1}=-2\Rightarrow\frac{3}{x-1}=-2\)

\(\Rightarrow-2x+2=3\)

\(\Rightarrow-2x=1\)

\(\Rightarrow x=-\frac{1}{2}\)

a. ĐKXĐ: x² - 1\(\ne\)0 (=) x \(\ne\)\(\pm\)1

b. \(\frac{3x+3}{x^2-1}\)

\(=\frac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{3}{x+1}\)với x \(\pm\)1

c. \(\frac{3}{x+1}=-2\)

\(\Rightarrow\)\(\left(x+1\right).\left(-2\right)=3\)

\(-2x-2=3\)

\(-2x=5\)

\(x=-\frac{5}{2}\)(t/m đk)