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\(16x-5x^2-3\)
\(=-5x^2+16x-3\)
\(=-5x^2+15x+x-3\)
\(=\left(-5x^2+15x\right)+\left(x-3\right)\)
\(=-5x.\left(x-3\right)+\left(x-3\right)\)
\(=\left(-5x+1\right).\left(x-3\right)\)
\(2x^2+7x+5\)
\(=2x^2+2x+5x+5\)
\(=\left(2x^2+2x\right)+\left(5x+5\right)\)
\(=2x.\left(x+1\right)+5.\left(x+1\right)\)
\(=\left(2x+5\right).\left(x+1\right)\)
\(2x^2+3x+5\) (Bạn xem lại đề nhé.)
\(x^3-3x^2+1-3x\)
\(=\left(x^3+1\right)-\left(3x^2+3x\right)\)
\(=\left(x+1\right).\left(x^2-x+1\right)-3x.\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right).\left(x^2-4x+1\right)\)
\(x^2-4x-5\)
\(=x^2-5x+x-5\)
\(=\left(x^2-5x\right)+\left(x-5\right)\)
\(=x.\left(x-5\right)+\left(x-5\right)\)
\(=\left(x+1\right).\left(x-5\right)\)
\(\left(a^2+1\right)^2-4a^2\)
\(=\left(a^2+1\right)^2-\left(2a\right)^2\)
\(=\left(a^2-2a+1\right).\left(a^2+2a+1\right)\)
\(=\left(a-1\right)^2.\left(a+1\right)^2\)
\(x^3-0,25x=0\)
\(\Rightarrow x.\left(x^2-0,25\right)=0\)
\(\Rightarrow x.\left(x-0,5\right).\left(x+0,5\right)=0\)
Trường hợp 1: \(x=0\)
Trường hợp 2: \(x-0,5=0\Rightarrow x=0,5\)
Trường hợp 3: \(x+0,5=0\Rightarrow x=-0,5\)
\(2x.\left(3x-5\right)-\left(5-3x\right)=0\)
\(\Rightarrow2x.\left(3x-5\right)+\left(3x-5\right)=0\)
\(\Rightarrow\left(3x-5\right).\left(2x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-5=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{-1}{2}\end{cases}}\)
\(49x^2+14x+1=0\)
\(\Rightarrow\left(7x+1\right)=0\)
\(\Rightarrow7x=-1\)
\(\Rightarrow x=\frac{-1}{7}\)
Ai giúp mik với ạ
3. Tìm x biết
\(9x^2-4=0\)
\(\Rightarrow\left(3x\right)^2-2^2=0\)
\(\Rightarrow\left(3x-2\right).\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\3x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=2\\3x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-2}{3}\end{cases}}\)
\(\left(3x+1\right)^2=\left(5x-2\right)^2\)
\(\Rightarrow\left(3x+1\right)^2-\left(5x-2\right)^2=0\)
\(\Rightarrow\left(3x+1-5x+2\right).\left(3x+1+5x-2\right)=0\)
\(\Rightarrow\left(-2x+3\right).\left(8x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}-2x+3=0\\8x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}-2x=-3\\8x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{8}\end{cases}}\)
\(4x^2-25+\left(2x+5\right).\left(2x+7\right)=0\)
\(\Rightarrow\left(4x^2-25\right)+\left(2x+5\right).\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right).\left(2x+5\right)+\left(2x+5\right).\left(2x+7\right)=0\)
\(\Rightarrow\left(2x+5\right).\left(2x-5+2x+7\right)=0\)
\(\Rightarrow\left(2x+5\right).\left(4x+2\right)=0\)
\(\Rightarrow2.\left(2x+5\right).\left(2x+1\right)=0\)
\(\Rightarrow\left(2x+5\right).\left(2x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+5=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=-5\\2x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=\frac{-1}{2}\end{cases}}\)
\(x^3+4x^2+4x=0\)
\(\Rightarrow x.\left(x^2+4x+4\right)=0\)
\(\Rightarrow x.\left(x+2\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
\(x^2-10x+9=0\)
\(\Rightarrow x^2-x-9x+9=0\)
\(\Rightarrow x.\left(x-1\right)-9.\left(x-1\right)=0\)
\(\Rightarrow\left(x-9\right).\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-9=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
\(-3x^2+2x+1=0\)
\(\Rightarrow-3x^2+3x-x+1=0\)
\(\Rightarrow-3x.\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(-3x-1\right).\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}-3x-1=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=1\end{cases}}\)
nếu ko có vào tcn nhé
\(A=-2x^2+6x+9\)
\(A=-\left(2x^2-6x-9\right)\)
\(A=-[\left(\sqrt{2}x\right)^2-2.\sqrt{2}.\frac{3\sqrt{2}}{2}+\frac{9}{2}-\frac{27}{2}]\)
\(A=-\left(x\sqrt{2}-\frac{9}{2}\right)^2+\frac{27}{2}\le\frac{27}{2}\)
Dấu bằng xảy ra \(\Leftrightarrow x\sqrt{2}-\frac{9}{2}=0\)
\(\Leftrightarrow x\sqrt{2}=\frac{9}{2}\)
\(\Leftrightarrow x=\frac{9\sqrt{2}}{4}\)