Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. \(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)\left(x+1\right)\left(2x-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x+5=0\\x+1=0\\2x-9=0\end{matrix}\right.\) \(\Rightarrow x=\)
b. \(\Leftrightarrow x^3+x+3x^2+3=0\)
\(\Leftrightarrow x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+1=0\left(vn\right)\end{matrix}\right.\)
c. \(\Leftrightarrow2x\left(3x-1\right)^2-\left(9x^2-1\right)=0\)
\(\Leftrightarrow\left(6x^2-2x\right)\left(3x-1\right)-\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(6x^2-5x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-1\right)\left(6x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-1=0\\6x+1=0\end{matrix}\right.\)
d.
\(\Leftrightarrow x^3-3x^2+2x-3x^2+9x-6=0\)
\(\Leftrightarrow x\left(x^2-3x+2\right)-3\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\\x-2=0\end{matrix}\right.\)
e.
\(\Leftrightarrow x^3+2x^2+x+3x^2+6x+3=0\)
\(\Leftrightarrow x\left(x^2+2x+1\right)+3\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+1=0\end{matrix}\right.\)
Mk năm nay lên lớp 9 nên chỉ làm bài 1 đc thôi
Câu 1:
a)\(\left(2x+3\right)^2-\left(x+1\right)^2=0\)
\(\left(2x+3+x+1\right)\left(2x+3-x-1\right)=0\)
\(\left(3x+4\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+4=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{4}{3}\\x=-2\end{cases}}\)
b)\(x^2-6x+5=0\)
\(x^2-5x-x+5=0\)
\(\left(x-5\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
c)\(3x^2-5x+2=0\)
\(3x^2-3x-2x+2=0\)
\(\left(3x-2\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\x-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=1\end{cases}}\)
\(\left(4x-5\right)\left(2x-3\right)\left(x-1\right)=9\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-5=9\\2x-3=9\\x-1=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=6\\x=10\end{matrix}\right.\)
Vậy \(x=\left\{3,5;6;10\right\}\)
d: Sửa đề: \(\left(4x-5\right)^2\cdot\left(2x-3\right)\left(x-1\right)=9\)
a: \(\Leftrightarrow\left(2x^2+x\right)^2-3\left(2x^2+x\right)-\left(2x^2+x\right)+3=0\)
\(\Leftrightarrow\left(2x^2+x\right)\left(2x^2+x-3\right)-\left(2x^2+x-3\right)=0\)
\(\Leftrightarrow\left(2x^2+x-3\right)\left(2x^2+x-1\right)=0\)
\(\Leftrightarrow\left(2x^2+3x-2x-3\right)\left(2x^2+2x-x-1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)\left(x+1\right)\left(2x-1\right)=0\)
hay \(x\in\left\{-\dfrac{3}{2};1;-1;\dfrac{1}{2}\right\}\)