Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)
b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)
c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Xem lại đề câu d
TL:
\(4x^2-y^2+4x+1\)
\(=\left(2x-1\right)^2-y^2\)
\(=\left(2x-1+y\right)\left(2x-1-y\right)\)
\(x^3-x+y^3-y\)
\(=\left(x+y\right)\left(x^2-xy+x^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+x^2-1\right)\)
Giải pt à bạn:P?
\(\left(x+4\right)\left(x^2-4x+16\right)-\left(x-2\right)^3=0\)
\(\Leftrightarrow x^3+4^3-\left(x^3-8-6x^2+12x\right)=0\)
\(\Leftrightarrow x^3+4^3-x^3+8+6x^2-12x=0\)
\(\Leftrightarrow72+6x^2-12x=0\Leftrightarrow6\left(x^2-2x+12\right)=0\Leftrightarrow x^2-2x+12=0\)
Ta lại có: \(x^2-2x+12=x^2-2x+1+11=\left(x-1\right)^2+11\ge11>0\ne0\)
=> Pt vô nghiệm.
a: ta có: \(\left(x+3\right)\left(x-1\right)-x\left(x-5\right)=11\)
\(\Leftrightarrow x^2+2x-3-x^2+5x=11\)
\(\Leftrightarrow x=2\)
b: Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x+1\right)\left(x+2\right)+3x^2=0\)
\(\Leftrightarrow x^3+64-x^3-3x^2-2x+3x^2=0\)
\(\Leftrightarrow2x=64\)
hay x=32
a)\(12x^3+4x^2-27x-9\)
\(=12x^3-27x+4x^2-9\)
\(=3x\left(4x^2-9\right)+\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
b)\(7x\left(y-4\right)^2-\left(4-y\right)^3\)
\(=7x\left(y-4\right)^2+\left(y-4\right)^3\)
\(=\left(y-4\right)^2\left(7x+y-4\right)\)
\(x^3+1-x^2-x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
\(x+y-x^3-y^3\)
\(=\left(x+y\right)-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left(1-x^2+xy-y^2\right)\)
\(4x^2-28=0\)
\(4x^2=28\)
\(x^2=7\)
\(\)
\(4x^2-28=0\)
\(\Leftrightarrow4\left(x^2-7\right)=0\)
\(\Leftrightarrow x^2-7=0\)
\(\Leftrightarrow x^2=7\)
\(\Leftrightarrow x=\pm\sqrt{7}\)