Question

【急~！高一数学~！高分追加】~！

计算(x^2-√2x+1/3)^2

【我想要简单的解题过程~谢谢、】
就是化简、

Answer 1

(x^2-√2x+1/3)^2

=(x^2-√2x)^2+1/9+(2/3)(x^2-√2x)

=x^4+2x^2-2√2x^3+1/9+(2/3)(x^2)-(2/3)(√2x)

=x^4-2√2x^3+(8/3)x^2-(2/3)(√2x)+1/9

Answer 2

(x^2-√2x+1/3)^2

x等于多少啊

Answer 3

原式=【（x^2-√2x）+1/3】^2
=(x^2- √2x)^2+2/3(x^2- √2x)+1/9
=x^4-2 √2x^3+2x^2+2/3*x^2-2/3* √2x+1/9
=x^4-2 √2x^3+8/3*x^2-2/3* √2x+1/9

Answer 4

(x^2-√2x+1/3)^2

=(x^2- √2x)^2+2/3(x^2- √2x)+1/9
=x^2(x- √2)^2+2/3(x^2- √2x)+1/9
=x^2(x^2-2√2x-2)+2/3(x^2- √2x)+1/9
=x^4-2 √2x^3+8/3*x^2-2/3* √2x+1/9