xin lỗi, bn cóa thể bấm ∑ cái nài để lm lại đề đc hăm :v?

\(\dfrac{2x-3}{4-x}+\dfrac{1}{3}>\dfrac{1}{2}-\dfrac{3-x}{5}\) 

đúng ko ???

2x + 13/6 =8/27

2x            = 8/27 - 13/6

2x            = - 101/54

x            = - 101/54 : 2

x              = - 101/108

1, Hoành độ giao điểm 2 đường thẳng đó là:

\(2x-3=x+1\Leftrightarrow x=4\)

Tung độ giao điểm 2 đường thẳng đó là:

\(y=2x-3=2.1-3=-1\)

Vậy tọa độ giao điểm 2 đường thẳng đó là:\(\left(4;-1\right)\)

2, Để đường thẳng (d₁) đi qua A(1;-2) thì:

\(-2=\left(2m-1\right).1+n+2\\ \Leftrightarrow2m-1+n+2+2=0\\ \Leftrightarrow2m+n+3=0\left(1\right)\)

Để đường thẳng (d₂) đi qua A(1;-2) thì:

\(-2=2n.1+2m-3\\ \Leftrightarrow2n+2m-3+2=0\\ \Leftrightarrow2n+2m-1=0\left(2\right)\)

Từ (1), (2) ta có hệ: \(\left\{{}\begin{matrix}2m+n+3=0\\2n+2m-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=-\dfrac{7}{2}\\n=4\end{matrix}\right.\)

1) Xét phương trình hoành độ giao điểm của 2 đường thẳng trên ta có:

\(2x-3=x+1.\\ \Leftrightarrow2x-x=1+3.\\ \Leftrightarrow x=4.\\ \Rightarrow y=5.\)

Tọa độ giao điểm của 2 đường thẳng trên là \(\left(4;5\right).\)

2. Thay tọa độ điểm \(A\left(1;-2\right)\) vào 2 phương trình đường trên ta có:

\(\left\{{}\begin{matrix}\left(2m-1\right)+n+2=-2.\\2n+2m-3=-2.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m+n=-3.\\2m+2n=1.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m=-\dfrac{7}{2}.\\m=4.\end{matrix}\right.\)

\(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)

\(\Leftrightarrow\frac{6\left(x+5\right)}{24}-\frac{8\left(2x-3\right)}{24}=\frac{3\left(6x-1\right)}{24}+\frac{2\left(2x-1\right)}{24}\)

\(\Leftrightarrow6x+30-16x+24=18x-3+4x-2\)

\(\Leftrightarrow6x-16x-18x-4x=-2-3-24-30\)

\(\Leftrightarrow-32x=-59\)

\(\Leftrightarrow x=\frac{59}{32}\)

1.

Phương trình hoành độ giao điểm:

\(2x-3=x+1\Rightarrow x=4\)

\(\Rightarrow y=5\)

Vậy tọa độ giao điểm là \(\left(4;5\right)\)

2.

Hai đường thẳng cắt nhau tại A khi chúng không song song nhau và cùng đi qua A

\(\Rightarrow\left\{{}\begin{matrix}2m-1\ne2n\\\left(2m-1\right).1+n+2=-2\\2n.1+2m-3=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2m-1\ne2n\\2m+n=-3\\2m+2n=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n=4\\m=-\dfrac{7}{2}\end{matrix}\right.\)

đáp số:5/2

cậu làm bài nhanh đi chúc bạn học tốt nha

(2/3-1-2)x=5/12

1/6x=5/12

x=5/2

dap an la 5/12

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

b) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

\(\Rightarrow\frac{3}{5}x=\left(-\frac{1}{7}\right)+\frac{1}{2}\)

\(\Rightarrow\frac{3}{5}x=\frac{5}{14}\)

\(\Rightarrow x=\frac{5}{14}:\frac{3}{5}\)

\(\Rightarrow x=\frac{25}{42}\)

Vậy \(x=\frac{25}{42}.\)

c) \(5-\left|3x-1\right|=3\)

\(\Rightarrow\left|3x-1\right|=5-3\)

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

\(\Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3:3\\x=\left(-1\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{1;-\frac{1}{3}\right\}.\)

d) \(\left(1-2x\right)^2=9\)

\(\Rightarrow\left(1-2x\right)^2=\left(\pm3\right)^2\)

\(\Rightarrow1-2x=\pm3.\)

\(\Rightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-2\right):2\\x=4:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{-1;2\right\}.\)

Chúc bạn học tốt!