\(\dfrac{2x}{x-1}+\dfrac{3\left(x+1\right)}{x}=5\left(\text{đ}k\text{x}\text{đ}:x\ne1\right)\\ \Leftrightarrow\dfrac{2x^2}{x\left(x-1\right)}+\dfrac{3\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}=\dfrac{5x\left(x-1\right)}{x\left(x-1\right)}\\ \Rightarrow2x^2+\left(3x+3\right)\left(x-1\right)=5x^2-5x\\ \Leftrightarrow2x^2+3x^2-3x+3x-3=5x^2-5x\\ \Leftrightarrow5x^2-3-5x^2+5x=0\\ \Leftrightarrow5x-3=0\\ \Leftrightarrow5x=3\\ \Leftrightarrow x=\dfrac{3}{5}\) \(b,\left|1-2x\right|=2x-1\) `(1)` Nếu `1-2x ≥0<=> 2x≥1<=>x≥` \(\dfrac{1}{2}\) thì biểu thức `(1)` trở thành `1-2x=2x-1` `<=> 1+1=2x+2x` `<=> 2=4x` `<=> -4x=-2` `<=>x=` \(\dfrac{-2}{-4}=\dfrac{1}{2}\) ( thoả mãn đk ) Nếu `1-2x <0<=> 2x<1<=>x<` \(\dfrac{1}{2}\) thì biểu thức `(1)` trở thành `-(1-2x)=2x-1` `<=>-1+2x=2x-1` `<=> 2x-2x=-1+1` `<=>0=0` ( luôn đúng ) `c,` \(\dfrac{2\left(x+1\right)}{3}-2\ge\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{4\left(x+1\right)}{6}-\dfrac{2}{6}\ge\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow4x+4-2\ge3x-6\\ \Leftrightarrow4x+2\ge3x-6\\ \Leftrightarrow4x-3x\ge-6-2\\ \Leftrightarrow x\ge-8\) Câu trả lời: \(\dfrac{2x}{x-1}+\dfrac{3\left(x+1\right)}{x}=5\left(\text{đ}k\text{x}\text{đ}:x\ne1\right)\\ \Leftrightarrow\dfrac{2x^2}{x\left(x-1\right)}+\dfrac{3\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}=\dfrac{5x\left(x-1\right)}{x\left(x-1\right)}\\ \Rightarrow2x^2+\left(3x+3\right)\left(x-1\right)=5x^2-5x\\ \Leftrightarrow2x^2+3x^2-3x+3x-3=5x^2-5x\\ \Leftrightarrow5x^2-3-5x^2+5x=0\\ \Leftrightarrow5x-3=0\\ \Leftrightarrow5x=3\\ \Leftrightarrow x=\dfrac{3}{5}\) \(b,\left|1-2x\right|=2x-1\) `(1)` Nếu `1-2x ≥0<=> 2x≥1<=>x≥` \(\dfrac{1}{2}\) thì biểu thức `(1)` trở thành `1-2x=2x-1` `<=> 1+1=2x+2x` `<=> 2=4x` `<=> -4x=-2` `<=>x=` \(\dfrac{-2}{-4}=\dfrac{1}{2}\) ( thoả mãn đk ) Nếu `1-2x <0<=> 2x<1<=>x<` \(\dfrac{1}{2}\) thì biểu thức `(1)` trở thành `-(1-2x)=2x-1` `<=>-1+2x=2x-1` `<=> 2x-2x=-1+1` `<=>0=0` ( luôn đúng ) `c,` \(\dfrac{2\left(x+1\right)}{3}-2\ge\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{4\left(x+1\right)}{6}-\dfrac{2}{6}\ge\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow4x+4-2\ge3x-6\\ \Leftrightarrow4x+2\ge3x-6\\ \Leftrightarrow4x-3x\ge-6-2\\ \Leftrightarrow x\ge-8\)

a) \(\dfrac{2x}{x-1}+\dfrac{3\left(x+1\right)}{x}=5\) \(\dfrac{x\cdot2x}{x\left(x-1\right)}+\dfrac{3\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}=5\) \(\dfrac{2x^2}{x^2-x}+\dfrac{3\left(x^2-1^2\right)}{x^2-x}=5\) \(\dfrac{2x^2}{x^2-x}+\dfrac{3x^2-3}{x^2-x}=5\) \(\dfrac{2x^2+3x^2-3}{x^2-x}=\dfrac{5x^2-3}{x^2-x}=5\) \(\Rightarrow5x^2-3=5\left(x^2-x\right)=5x^2-5x\) \(\Rightarrow3=5x\) \(x=\dfrac{3}{5}\) b) \(\left|1-2x\right|=2x-1\) TH1: \(1>2x\) \(\Rightarrow\left[{}\begin{matrix}1-2x>0\\2x-1< 0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left|1-2x\right|>0\\2x-1< 0\end{matrix}\right.\) => Vô lí TH2: \(1\le2x\) \(\Rightarrow\left[{}\begin{matrix}1-2x\le0\Rightarrow\left|1-2x\right|\ge0\\2x-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|1-2x\right|=2x-1\ge0\) \(\Leftrightarrow2x-1\ge0\Rightarrow2x-1+1=2x\ge0+1=1\) \(\Leftrightarrow\dfrac{2x}{2}=x\ge\dfrac{1}{2}\) Câu trả lời: a) \(\dfrac{2x}{x-1}+\dfrac{3\left(x+1\right)}{x}=5\) \(\dfrac{x\cdot2x}{x\left(x-1\right)}+\dfrac{3\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}=5\) \(\dfrac{2x^2}{x^2-x}+\dfrac{3\left(x^2-1^2\right)}{x^2-x}=5\) \(\dfrac{2x^2}{x^2-x}+\dfrac{3x^2-3}{x^2-x}=5\) \(\dfrac{2x^2+3x^2-3}{x^2-x}=\dfrac{5x^2-3}{x^2-x}=5\) \(\Rightarrow5x^2-3=5\left(x^2-x\right)=5x^2-5x\) \(\Rightarrow3=5x\) \(x=\dfrac{3}{5}\) b) \(\left|1-2x\right|=2x-1\) TH1: \(1>2x\) \(\Rightarrow\left[{}\begin{matrix}1-2x>0\\2x-1< 0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left|1-2x\right|>0\\2x-1< 0\end{matrix}\right.\) => Vô lí TH2: \(1\le2x\) \(\Rightarrow\left[{}\begin{matrix}1-2x\le0\Rightarrow\left|1-2x\right|\ge0\\2x-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|1-2x\right|=2x-1\ge0\) \(\Leftrightarrow2x-1\ge0\Rightarrow2x-1+1=2x\ge0+1=1\) \(\Leftrightarrow\dfrac{2x}{2}=x\ge\dfrac{1}{2}\)

Phùng Phạm Quỳnh Trang

13 tháng 7 2023 - olm

#Hỏi cộng đồng OLM #Toán lớp 8

⭐Hannie⭐

13 tháng 7 2023

\(\dfrac{2x}{x-1}+\dfrac{3\left(x+1\right)}{x}=5\left(\text{đ}k\text{x}\text{đ}:x\ne1\right)\\ \Leftrightarrow\dfrac{2x^2}{x\left(x-1\right)}+\dfrac{3\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}=\dfrac{5x\left(x-1\right)}{x\left(x-1\right)}\\ \Rightarrow2x^2+\left(3x+3\right)\left(x-1\right)=5x^2-5x\\ \Leftrightarrow2x^2+3x^2-3x+3x-3=5x^2-5x\\ \Leftrightarrow5x^2-3-5x^2+5x=0\\ \Leftrightarrow5x-3=0\\ \Leftrightarrow5x=3\\ \Leftrightarrow x=\dfrac{3}{5}\)

\(b,\left|1-2x\right|=2x-1\) `(1)`

Nếu `1-2x ≥0<=> 2x≥1<=>x≥`\(\dfrac{1}{2}\) thì biểu thức `(1)` trở thành

`1-2x=2x-1`

`<=> 1+1=2x+2x`

`<=> 2=4x`

`<=> -4x=-2`

`<=>x=` \(\dfrac{-2}{-4}=\dfrac{1}{2}\) ( thoả mãn đk )

Nếu `1-2x <0<=> 2x<1<=>x<`\(\dfrac{1}{2}\) thì biểu thức `(1)` trở thành

`-(1-2x)=2x-1`

`<=>-1+2x=2x-1`

`<=> 2x-2x=-1+1`

`<=>0=0` ( luôn đúng )

`c,`

\(\dfrac{2\left(x+1\right)}{3}-2\ge\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{4\left(x+1\right)}{6}-\dfrac{2}{6}\ge\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow4x+4-2\ge3x-6\\ \Leftrightarrow4x+2\ge3x-6\\ \Leftrightarrow4x-3x\ge-6-2\\ \Leftrightarrow x\ge-8\)

Đúng(1)

Duc Nguyen

13 tháng 7 2023

a)\(\dfrac{2x}{x-1}+\dfrac{3\left(x+1\right)}{x}=5\)

\(\dfrac{x\cdot2x}{x\left(x-1\right)}+\dfrac{3\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}=5\)

\(\dfrac{2x^2}{x^2-x}+\dfrac{3\left(x^2-1^2\right)}{x^2-x}=5\)

\(\dfrac{2x^2}{x^2-x}+\dfrac{3x^2-3}{x^2-x}=5\)

\(\dfrac{2x^2+3x^2-3}{x^2-x}=\dfrac{5x^2-3}{x^2-x}=5\)

\(\Rightarrow5x^2-3=5\left(x^2-x\right)=5x^2-5x\)

\(\Rightarrow3=5x\)

\(x=\dfrac{3}{5}\)

b) \(\left|1-2x\right|=2x-1\)

TH1: \(1>2x\)

\(\Rightarrow\left[{}\begin{matrix}1-2x>0\\2x-1< 0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left|1-2x\right|>0\\2x-1< 0\end{matrix}\right.\) => Vô lí

TH2: \(1\le2x\)

\(\Rightarrow\left[{}\begin{matrix}1-2x\le0\Rightarrow\left|1-2x\right|\ge0\\2x-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left|1-2x\right|=2x-1\ge0\)

\(\Leftrightarrow2x-1\ge0\Rightarrow2x-1+1=2x\ge0+1=1\)

\(\Leftrightarrow\dfrac{2x}{2}=x\ge\dfrac{1}{2}\)

Đúng(0)