Bài 11: 1) \(\left(7+3\right)\left(49-21+9\right)=7^3+3^3\)2) \(\left(1+x\right)\left(1-x+x^2\right)=1+x^3\)4) \(\left(3x+2\right)\left(9x^2-6x+4\right)=27x^3+8\)6) \(\left(3y+2x\right)\left(9y^2-6xy+4x^2\right)=27y^3+8x^3\)10) \(\left(x+5\right)\left(x^2-5x+25\right)=x^3+125\)11) \(\left(5x-25-x^2\right)\left(x+5\right)=-\left(x^2-5x+25\right)\left(x+5\right)=-\left(x^3+125\right)=-x^3-125\)12) \(\left(2x+5y\right)\left(4x^2-10xy+25y^2\right)=8x^3+125y^3\)14) \(\left(x+\dfrac{1}{2}y\right)\left(x^2-\dfrac{1}{2}xy+\dfrac{1}{4}y^4\right)=x^3+\dfrac{1}{8}y^3\)15) \(\left(3x+\dfrac{1}{2}y\right)\left(9x^2-\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)=27x^3+\dfrac{1}{8}y^3\)16) \(\left(\dfrac{1}{2}a+\dfrac{5}{3}b\right)\left(\dfrac{1}{4}a^2-\dfrac{5}{6}ab+\dfrac{25}{9}b^2\right)=\dfrac{1}{8}a^3+\dfrac{125}{27}b^3\)Câu trả lời: Bài 11: 1) \(\left(7+3\right)\left(49-21+9\right)=7^3+3^3\)2) \(\left(1+x\right)\left(1-x+x^2\right)=1+x^3\)4) \(\left(3x+2\right)\left(9x^2-6x+4\right)=27x^3+8\)6) \(\left(3y+2x\right)\left(9y^2-6xy+4x^2\right)=27y^3+8x^3\)10) \(\left(x+5\right)\left(x^2-5x+25\right)=x^3+125\)11) \(\left(5x-25-x^2\right)\left(x+5\right)=-\left(x^2-5x+25\right)\left(x+5\right)=-\left(x^3+125\right)=-x^3-125\)12) \(\left(2x+5y\right)\left(4x^2-10xy+25y^2\right)=8x^3+125y^3\)14) \(\left(x+\dfrac{1}{2}y\right)\left(x^2-\dfrac{1}{2}xy+\dfrac{1}{4}y^4\right)=x^3+\dfrac{1}{8}y^3\)15) \(\left(3x+\dfrac{1}{2}y\right)\left(9x^2-\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)=27x^3+\dfrac{1}{8}y^3\)16) \(\left(\dfrac{1}{2}a+\dfrac{5}{3}b\right)\left(\dfrac{1}{4}a^2-\dfrac{5}{6}ab+\dfrac{25}{9}b^2\right)=\dfrac{1}{8}a^3+\dfrac{125}{27}b^3\)