đi mà xin ny

gõ văn bản gì ?

\(\left(4x-3\right)\left(x-2\right)=\left(2-x\right)^2\\ \Leftrightarrow\left(4x-3\right)\left(x-2\right)-\left(x-2\right)^2=0\\ \Leftrightarrow\left(x-2\right)\left(4x-3-x+2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

a, 

b, 

c,

d, 

Khi đó điểm A và điểm C trùng nhau

hong bé ơi

ĐKXĐ:\(x\ne\pm2\)

\(\dfrac{x-4}{x+2}+\dfrac{x+1}{2-x}=\dfrac{24}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{24}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2-6x+8-x^2-3x-2-24}{\left(x+2\right)\left(x-2\right)}=0\\ \Rightarrow-9x-18=0\\ \Leftrightarrow x=-2\left(ktm\right)\)

1, ĐKXĐ:\(x\ne2,y\ne1\)

Đặt `1/(x-2)` = a, `1/(y-1)` = b

\(Hệ.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\2a-3b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{7}{5}\\b=\dfrac{3}{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{7}{5}\\\dfrac{1}{y-1}=\dfrac{3}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x-14=5\\3y-3=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\\y=\dfrac{8}{3}\end{matrix}\right.\)\(2,\Delta'=\left[-\left(m+1\right)\right]^2-4m=m^2+2m+1-4m=m^2-2m+1=\left(m-1\right)^2\ge0\)

Để pt có 2 nghiệm phân biệt thì \(\Delta'>0\Leftrightarrow\left(m-1\right)^2>0\Leftrightarrow m-1\ne0\Leftrightarrow m\ne1\)

b, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=4m\end{matrix}\right.\)

\(\left(x_1-x_2\right)^2-x_1x_2=3\\ \Leftrightarrow\left(x_1+x_2\right)^2-5x_1x_2=3\\ \Leftrightarrow\left(2m+2\right)^2-5.4m-3=0\\ \Leftrightarrow4m^2+8m+4-20m-3=0\\ \Leftrightarrow4m^2-12m+1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+2\sqrt{2}}{2}\\x=\dfrac{3-2\sqrt{2}}{2}\end{matrix}\right.\)

\(e,\dfrac{7}{12}+\dfrac{x}{15}=\dfrac{1}{20}\\ \Rightarrow\dfrac{x}{15}=\dfrac{1}{20}-\dfrac{7}{12}\\ \Rightarrow\dfrac{x}{15}=-\dfrac{8}{15}\\ \Rightarrow x=-8\\ f,\dfrac{-7}{x}+\dfrac{8}{15}=\dfrac{-1}{20}\\ \Rightarrow\dfrac{-7}{x}=\dfrac{-1}{20}-\dfrac{8}{15}\\ \Rightarrow\dfrac{-7}{x}=\dfrac{-7}{12}\\ \Rightarrow x=12\)

\(g,\dfrac{-3}{31}+\dfrac{-6}{17}+\dfrac{1}{25}+\dfrac{-28}{31}+\dfrac{-11}{17}+\dfrac{-1}{5}\\ =\left(\dfrac{-3}{31}+\dfrac{-28}{31}\right)+\left(\dfrac{-6}{17}+\dfrac{-11}{17}\right)+\left(\dfrac{1}{25}+\dfrac{-5}{25}\right)\\ =\dfrac{-31}{31}+\dfrac{-17}{17}+\dfrac{-1}{25}\\ =\left(-1\right)+\left(-1\right)+\dfrac{-4}{25}\\ =-\dfrac{54}{25}\\ h,\dfrac{-4}{12}+\dfrac{18}{45}+\dfrac{-6}{9}+\dfrac{-21}{35}+\dfrac{6}{30}\\ =\dfrac{-1}{2}+\dfrac{2}{5}+\dfrac{-2}{3}+\dfrac{-3}{5}+\dfrac{1}{5}\\ =\dfrac{-1}{2}+\dfrac{-2}{3}+\left(\dfrac{2}{5}+\dfrac{-3}{5}+\dfrac{1}{5}\right)\\ =\dfrac{-3}{6}+\dfrac{-4}{6}+\dfrac{0}{5}\\ =\dfrac{-7}{6}+0\\ =-\dfrac{7}{6}\)

hết tuần này 4 coin, lâu lắm r ko trả