\(x^2-x-\frac{3}{4}=0\)

\(\Rightarrow\frac{x^2-x}{1}-\frac{3}{4}=0\)

\(\Rightarrow\frac{\left(x^2-x\right).4}{4}-\frac{3}{4}=0\)

\(\Rightarrow\frac{4x^2-4x-3}{4}=0\)

\(\Rightarrow4x^2-4x-3=0\)

\(\Rightarrow4x^2-6x+2x-3=0\)

\(\Rightarrow2x\left(2x-3\right)+\left(2x-3\right)=0\)

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

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

\(x^2-x-\frac{3}{4}=0\)

<=>  \(x^2-x+\frac{1}{4}-1=0\)

<=>  \(\left(x-\frac{1}{2}\right)^2-1=0\)

<=>  \(\left(x-\frac{1}{2}-1\right)\left(x-\frac{1}{2}+1\right)=0\)

<=> \(\left(x-\frac{3}{2}\right)\left(x+\frac{1}{2}\right)=0\)

làm nốt nhé

không bít

     \(x^2-x-\frac{3}{4}=0\)

\(\Rightarrow x^2+\frac{1}{2}x-\frac{3}{2}x-\frac{3}{4}=0\)

\(\Rightarrow x\left(x+\frac{1}{2}\right)-\frac{3}{2}\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\left(x-\frac{3}{2}\right)\left(x+\frac{1}{2}\right)=0\)

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

Chúc bạn học tốt.

a) \(x+2x+3x+...+100x=-213\)

\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)

\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)

\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)

d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)

                                                                    \(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)

c) \(3\left(x-2\right)+2\left(x-1\right)=10\)

\(\Rightarrow3x-6+2x-2=10\)

\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)

a) \(x+2x+3x+4x+...+100x=-213\)

\(x.\left(1+2+3+4+...+100\right)=-213\)

\(x.5050=-213\)

\(x=-\frac{213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)

\(\frac{1}{2}x=-\frac{43}{12}\)

\(x=\frac{-43}{6}\)

a)    thiếu điều kiện

b)  ta thấy: \(VP\ge0\forall x\)

mà   \(VT\ge0\forall x\)

        | x²+| 6x-2|=x²+4

\(\Rightarrow x^2+\left|6x-2\right|=x^2+4\)

\(\Rightarrow\)\(\left|6x-2\right|=4\)

Bạn tự làm tiếp nhé

Mọi người giúp mk với ạ

đề bài bị lỗi :(