5: \(\left(2x-3\right)^2-16\)
\(=\left(2x-3\right)^2-4^2\)
\(=\left(2x-3-4\right)\left(2x-3+4\right)\)
\(=\left(2x-7\right)\left(2x+1\right)\)
6: \(\left(2x+1\right)^2-\left(\dfrac{1}{2}x-7\right)^2\)
\(=\left(2x+1-\dfrac{1}{2}x+7\right)\left(2x+1+\dfrac{1}{2}x-7\right)\)
\(=\left(\dfrac{3}{2}x+8\right)\left(\dfrac{5}{2}x-6\right)\)
7: \(\left(x+7\right)^2-\left(4x+5\right)^2\)
\(=\left(x+7-4x-5\right)\left(x+7+4x+5\right)\)
\(=\left(-3x+2\right)\left(5x+12\right)\)
8: \(\left(\dfrac{1}{4}x+3\right)^2-\left(4x-\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{4}x+3-4x+\dfrac{1}{4}\right)\left(\dfrac{1}{4}x+3+4x-\dfrac{1}{4}\right)\)
\(=\left(-\dfrac{15}{4}x+\dfrac{13}{4}\right)\left(\dfrac{17}{4}x+\dfrac{11}{4}\right)\)
9: \(\left(\dfrac{1}{3}x+3\right)^2-\left(4x-\dfrac{2}{3}\right)^2\)
\(=\left(\dfrac{1}{3}x+3-4x+\dfrac{2}{3}\right)\left(\dfrac{1}{3}x+3+4x-\dfrac{2}{3}\right)\)
\(=\left(-\dfrac{11}{3}x+\dfrac{11}{3}\right)\left(\dfrac{13}{3}x+\dfrac{7}{3}\right)\)
\(=-\dfrac{11}{3}\left(x-1\right)\left(\dfrac{13}{3}x+\dfrac{7}{3}\right)\)
10: \(9\left(x+5\right)^2-4\left(2x+8\right)^2\)
\(=\left[3\left(x+5\right)\right]^2-\left[2\left(2x+8\right)\right]^2\)
\(=\left(3x+15\right)^2-\left(4x+16\right)^2\)
\(=\left(3x+15-4x-16\right)\left(3x+15+4x+16\right)\)
\(=\left(-x-1\right)\left(7x+31\right)\)
11: \(\left(x+5\right)^2-25\left(2x+8\right)^2\)
\(=\left(x+5\right)^2-\left[5\left(2x+8\right)\right]^2\)
\(=\left(x+5\right)^2-\left(10x+40\right)^2\)
\(=\left(x+5-10x-40\right)\left(x+5+10x+40\right)\)
\(=\left(-9x-35\right)\left(11x+45\right)\)
12: \(16\left(2x+5\right)^2-9\left(2x-1\right)^2\)
\(=\left[4\left(2x+5\right)\right]^2-\left[3\left(2x-1\right)\right]^2\)
\(=\left(8x+20\right)^2-\left(6x-3\right)^2\)
\(=\left(8x+20+6x-3\right)\left(8x+20-6x+3\right)\)
\(=\left(14x+17\right)\left(2x+23\right)\)
1: \(=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x\cdot2y=4xy\)
2: \(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3=2y\left(3x^2+y^2\right)\)
3: \(=x^3+3x^2y+3xy^2+y^3+x^3-3x^2y+3xy^2-y^3\)
\(=2x^3+6xy^2=2x\left(x^2+3y^2\right)\)
4: \(=\left(7-4x+5\right)\left(7+4x-5\right)=\left(12-4x\right)\left(4x+2\right)\)
\(=8\left(3-x\right)\left(2x+1\right)\)
giúp e vớiiii
