1/ \(\left(2x-1\right)^2-3\left(2x-1\right)^2=0\)
\(\left(2x-1\right)^2\left(1-3\right)=0\)
\(\left(2x-1\right)^2\cdot\left(-2\right)=0\)
\(\Rightarrow\text{ }\left(2x-1\right)^2=0\)
\(2x-1=0\)
\(2x=0+1=1\)
\(x=\frac{1}{2}\)
1) \(\left(2x-1\right)^2-3\left(2x-1\right)^2=0\)
=> \(\left(2x-1\right)^2\left(1-3\right)=0\)
=> \(\left(2x-1\right)^2.\left(-2\right)=0\)
=> \(\left(2x-1\right)^2=0\)
=> \(2x-1=0\)
=> \(2x=1\)
=> \(x=1:2=\frac{1}{2}\)
2/ \(\left(x-1\right)^2\left(x+1\right)=x+1\)
\(\left(x-1\right)^2\left(x+1\right)\text{ : }\left(x+1\right)=1\)
\(\left(x-1\right)^2\left(x+1\right)\cdot\frac{1}{\left(x+1\right)}=1\)
\(\left(x-1\right)^2\left[\left(x+1\right)\cdot\frac{1}{\left(x+1\right)}\right]=1\)
\(\left(x-1\right)^2\cdot1=1\)
\(\Rightarrow\text{ }\left(x-1\right)^2=1\)
\(x-1=\pm1\)
\(\Rightarrow\orbr{\begin{cases}x=-1+1\\x=1+1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{0\text{ ; }2\right\}\)
huhu, e ghi thiếu đề câu 1, mọi người thông cảm cho e với ạ
1/ \((2x-1)^2x-3(2x-1)^2=0\)
2) \(\left(x-1\right)^2\left(x+1\right)=\left(x+1\right)\)
=> \(\left(x-1\right)^2\left(x+1\right)-\left(x+1\right)=0\)
=> \(\left[\left(x-1\right)^2-1\right]\left(x+1\right)=0\)
=> \(\orbr{\begin{cases}\left(x-1\right)^2-1=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}\left(x-1\right)^2=1\\x=-1\end{cases}}\)
=> x - 1 = 1 hoặc x - 1 = -1
hoặc x = -1
=> x = 2 hoặc x= 0
hoặc x = -1
3) \(x^4-3x^2=x^2\)
=> \(x^4-3x^2-x^2=0\)
=> \(x^4-4x^2=0\)
=> \(x^2\left(x^2-4\right)=0\)
=> \(\orbr{\begin{cases}x^2=0\\x^2-4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x^2=2^2\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
3/ \(x^4-3x^2=x^2\)
\(x^4-3x^2-x^2=0\)
\(x^4-\left[3x^2+x^2\right]=0\)
\(x^4-\left[x^2\left(3+1\right)\right]=0\)
\(x^4-4x^2=0\)
\(x^2\left[x^2-4\right]=0\)
Do \(x^2-4\ne0\) ( \(x^2\ge0\) )
\(\Rightarrow\text{ }x^2=0\)
\(\Rightarrow\text{ }x=0\)
3/ \(x^4-3x^2=x^2\)
\(x^4-3x^2-x^2=0\)
\(x^4-\left[3x^2+x^2\right]=0\)
\(x^4-\left[x^2\left(3+1\right)\right]=0\)
\(x^4-4x^2=0\)
\(x^2\left[x^2-4\right]=0\)
Do \(x^2-4\ne0\) ( \(x^2\ge0\) )
\(\Rightarrow\text{ }x^2=0\)
\(\Rightarrow\text{ }x=0\)
@ hang tran @ Đùa à ? Giải lại này ! Mà đây là toán lớp 6
1/ \(\left(2x-1\right)^2x-3\left(2x-1\right)^2=0\)
\(\left(2x-1\right)^2\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-1\right)^2=0\\x-3=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x-1=0\\x=0+3=3\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x=0+1=1\\x=3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=3\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{1}{2}\text{ ; }3\right\}\)
