Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a.\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\\ =\left(2x\right)^3-3^3-\left[\left(2x\right)^3+3^3\right]\\ =8x^3-9-\left(8x^3+9\right)\\ =8x^3-9-8x^3-9=-18\)
\(b.\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\\ =x^3+1-\left(x^3-1\right)\\ =x^3+1-x^3+1=2\)
\(c.\left(3x-1\right)\left(3x+1\right)-\left(3x-2\right)^2\\ =9x^2-1-\left(9x^2-12x+4\right)\\ =9x^2-1-9x^2+12x-4\\ =12x-5\)
\(d.\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\\ =\left(2x-3\right)\cdot\left[\left(2x-3\right)-\left(2x+3\right)\right]\\ =\left(2x-3\right)\cdot\left(2x-3-2x-3\right)\\ =\left(2x-3\right)\cdot\left(-6\right)\\ =-12x\cdot18\)
\(e.\left(3x-4\right)^2-\left(2x+4\right)^2\\ =9x^2-24x+16-\left(4x^2+16x+16\right)\\ =9x^2-24x+16-4x^2-16x-16\\ =5x^2-40x\)
\(f.\left(3x-5\right)^3-\left(3x+5\right)^3\\ =27x^3-135x^2+225x-125-\left(27x^3+135x^2+225x+125\right)\\ =27x^3-135x^2+225x-125-27x^3-135x^2-225x-125\\ =-270x^2-250\)
\(g.\left(2x-1\right)^2-\left(3x-1\right)^2\\ =4x^2-4x+1-\left(9x^2-6x+1\right)\\ =4x^2-4x+1-9x^2+6x-1\\ =-5x^2+2x\)
\(h.\left(x-2y\right)\left(x^2+2xy+4y^2\right)+\left(x^3-6y^3\right)\\ =x^3-8y^3+x^3-6y^3\\ =2x^3-14y^3\)
\(a.x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Leftrightarrow x\left(x^2-5^2\right)-\left(x^3+2^3\right)=3\)
\(\Leftrightarrow x^3-25x-x^3-8=3\)
\(\Leftrightarrow x^3-x^3-25x=8+3\)
\(\Leftrightarrow x=\frac{11}{-25}\)
Vậy x có nghiệm là \(\frac{-11}{25}.\)
\(\)
\(a,-2x\left(2-3x\right)+3\left(-5+7x-6x^2\right)=-4\)
\(\Rightarrow-4x+6x^2-15+21x-18x^2=-4\)
\(\Rightarrow-12x^2+17x-11=0\)
\(\Rightarrow12x^2-17x+11=0\)
\(\Rightarrow9x^2-2.3.\frac{17}{6}x+\left(\frac{17}{6}\right)^2-\left(\frac{17}{6}\right)^2+11=0\)
\(\Rightarrow\left(3x-\frac{17}{6}\right)^2+\frac{107}{36}=0VN\)
Không có gt x thỏa mãn
\(b,-3x\left(-1+3x-4x^2\right)+6x^2\left(-2x+3\right)=0\)
\(\Rightarrow3x-9x^2+12x^3-12x^3+18x^2=0\)
\(\Rightarrow9x^2+3x=0\)
\(\Rightarrow3x\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\3x=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-\frac{1}{3}\end{cases}}}\)
\(\left(x^2+3\right)\left(3-x^2\right)\)
\(\left(x^2+3\right)\left(-x^2+3\right)\)
\(\left(-x^2+3\right).x^2+3\left(-x^2+3\right)\)
\(-x^2.x^2+3x^2+3\left(-x^2+3\right)\)
\(-x^2.x^2+3x^2-3x^2+9\)
\(-x^2.x^2+9\)
a,\(\text{Để }\left(4x+3\right)^3-\left(2x-5\right)^3=\left(2x+8\right)^3\) thì
\(3\left(4x+3\right)\left(2x-5\right)\left(2x+8\right)=0\)
\(\Leftrightarrow\left(4x+3\right)\left(2x-5\right)\left(2x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\2x-5=0\\2x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)
Vậy..
b,\(Để\left(3x+2016\right)^3+\left(3x-2019\right)^3=\left(6x-3\right)^3\) thì
\(3\left(3x+2016\right)\left(3x-2019\right)\left(6x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2016=0\\3x-2019=0\\6x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2016}{3}\\x=\dfrac{2019}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy...