1.2x*(1/2x^2-1)-3x(1/3x^2+2/3)2.3(y-z)^2-(2y+z)(-z+2y)
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1.2x*(1/2x^2-1)-3x(1/3x^2+2/3)2.3(y-z)^2-(2y+z)(-z+2y)
1.2x*(1/2x^2-1)-3x(1/3x^2+2/3)
2.3(y-z)^2-(2y+z)(-z+2y)
1.2x*(1/2x^2-1)-3x(1/3x^2+2/3)2.3(y-z)^2-(2y+z)(-z+2y)
2x*(1/2x^2-1)-3x(1/3x^2+2/3)
=x(x²-2)-x(x²+2)
=x(x²-2-x²-2)
=-4x
3(y-z)^2-(2y+z)(-z+2y)
=3y²-6yz+3z²-4y²+z²
=4z²-6yz-y²
没写错吗?
2x*(1/2x^2-1)-3x(1/3x^2+2/3)
=2x*1/2(x^2-2)-3x*1/3*(x^2+2)
=x(x^2-2)-x(x^2+2)
=x[(x^2-2)-(x^2+2)]
=x(x^2-2-x^2-2)
=x(-4)
=-4x
3(y-z)^2-(2y+z)(-z+2y)
=3(y-z)^2-(2y+z)(2y-z)
=3(y-z)^2-(4y^2-z^2)
=3y^2-6yz+3z^2-4y^2+z^2
=4z^2-6yz-y^2
2x*(1/2x^2-1)-3x(1/3x^2+2/3)
=2x*1/2(x^2-2)-3x*1/3*(x^2+2)
=x(x^2-2)-x(x^2+2)
=x[(x^2-2)-(x^2+2)]
=x(x^2-2-x^2-2)
=x(-4)
=-4x
3(y-z)^2-(2y+z)(-z+2y)
=3(y-z)^2-(2y+z)(2...
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2x*(1/2x^2-1)-3x(1/3x^2+2/3)
=2x*1/2(x^2-2)-3x*1/3*(x^2+2)
=x(x^2-2)-x(x^2+2)
=x[(x^2-2)-(x^2+2)]
=x(x^2-2-x^2-2)
=x(-4)
=-4x
3(y-z)^2-(2y+z)(-z+2y)
=3(y-z)^2-(2y+z)(2y-z)
=3(y-z)^2-(4y^2-z^2)
=3y^2-6yz+3z^2-4y^2+z^2
=4z^2-6yz-y^2
此题主要应用平方差公式，完全平方和公式(a+b)(a-b)=a^2-b^2 (a+b)^2=a^2+2ab+b^2 还有提取公因式，因式分解的最终形式应该是几个整式相乘的形式，所以第二题有问题哦！
收起
3(y-z)^2-(2y+z)(-z+2y)
=3(y-z)^2-(2y+z)(2y-z)
=3(y-z)^2-(4y^2-z^2)
=3y^2-6yz+3z^2-4y^2+z^2
=4z^2-6yz-y^2 ( 次题是符号错了，不能分解呀