giải giúp mình bài 2 thôi ạ

dạ thôi mình ko cần nữa ạ. Cảm ơn pạn nào có ý định giúp mik nhe

\(\dfrac{2-x}{4}=\dfrac{3x-1}{-3}\\ \Rightarrow-3\left(2-x\right)=4\left(3x-1\right)\\ \Rightarrow3x-6=12x-4\\ \Rightarrow12x-4-3x+6=0\\ \Rightarrow9x+2=0\\ \Rightarrow9x=-2\\ \Rightarrow x=-\dfrac{2}{9}\)

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

⇒\(\dfrac{3.\left(2-x\right)}{12}\)=\(\dfrac{4.\left(3x-1\right)}{12}\)

⇒\(6x-3=12x-4\)

⇒1=6x

⇒x=\(\dfrac{1}{6}\)

Chúc bạn học tốt!

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

\(\Rightarrow-3\left(2-x\right)=4\left(3x-1\right)\)

     \(-6+3x=12x-4\)

      \(3x-12x=6-2\)

        \(-9x=4\)

             \(x=\dfrac{4}{-9}\)

Đáp án:x+2/2x

Ta có:2x+6=2(x+3);x²+3x=x(x+3)

➞MTC:2x(x+3)

Ta co:(x+1/2x+6)+(2x+3/x²+3x)={[(x+1)x]+[(2x+3)2]}/2x(x+3)=x²+5x+6/2x(x+3)=(x+2)(x+3)/2x(x+3)=x+2/2x

\(\Leftrightarrow\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{x\left(x+1\right)}=-1\left(đkxđ:x\ne\pm1;0;2;3\right)\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x}-\dfrac{1}{x+1}=-1\)

\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x+1}=-1\)

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

\(\Leftrightarrow x^2-2x-3=-4\)

\(\Leftrightarrow x^2-2x+1=0\)

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

\(\Leftrightarrow x=1\left(loai\right)\)

Vậy không có giá trị x thỏa mãn

Giải:

Ta có:

|x+1/3|=2/3

⇒x+1/3=2/3 hoặc x+1/3=-2/3

          x=1/3 hoặc x=-1

+)TH1: (nếu như có ngoặc)

Khi x=1/3:

A=(1/3)²-3.(1/3)+1

A=1/9

Khi x=-1

A=(-1)²-3.(-1)+1

A=5

+)TH2: (nếu x ko có ngoặc)

Khi x=-1

A=-1²-3.-1+1

A=3

Trường hợp này chỉ có -1 vì 1/3 ² =1/9 ; còn ko có ngoặc hay có ngoặc còn tùy thuộc vào đề bài và cách suy nghĩ của bạn nhé!

Chúc bạn học tốt!

\(\dfrac{3-3x}{5}=\dfrac{x-1}{2}\\ \Leftrightarrow2\left(3-3x\right)=5x-5\\ \Leftrightarrow6-6x=5x-5\\ \Leftrightarrow-6x+5x=-5-6\\ \Leftrightarrow-x=-11\\ \Rightarrow x=11\)

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt{\dfrac{19}{16}}-\sqrt{\left(-0,75\right)^2}\\ \Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt{19}}{4}-\dfrac{3}{4}\\ \Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt{19}-3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt{19}}{4}\)

\(TH_1:x\le\dfrac{2}{3}\\ 2-3x=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{3+\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{3+\sqrt{19}}{12}\left(tm\right)\)

\(TH_2:x>\dfrac{2}{3}\\ 3x-2=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{13-\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{13-\sqrt{19}}{12}\left(tm\right)\)

Vậy \(x\in\left\{\dfrac{3+\sqrt{19}}{12};\dfrac{13-\sqrt{19}}{12}\right\}\)

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt[]{\dfrac{19}{16}}-\sqrt[]{\left(-0,75\right)^2}\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-0,75\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt[]{19}}{4}+\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt[]{19}}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2-3x=\dfrac{5-\sqrt[]{19}}{4}\\2-3x=\dfrac{-5+\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=2-\dfrac{5-\sqrt[]{19}}{4}\\3x=2-\dfrac{\sqrt[]{19}-5}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{3+\sqrt[]{19}}{4}\\3x=\dfrac{13-\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt[]{19}}{12}\\x=\dfrac{13-\sqrt[]{19}}{12}\end{matrix}\right.\)

=\(^{\dfrac{-x^2-xy}{5\left(x^2-y^2\right)}}\).\(\dfrac{3\left(x^3-y^3\right)}{x^2-xy}\)

=\(\dfrac{-3\left(x-y\right)}{5}\)