Thực hiện phép tính:
a. (5x – 2y)(x2 – xy + 1)
b. (x – 1)(x + 1)(x + 2)
c. 1/2.x2y2 (2x + y)(2x – y)
Giúp vs, cô mk chuẩn bị thu bài.
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2,5 + 1,5 + 3 + 2 = 9x
Một hình có bốn cạnh có tổng tất cả mọi góc là 360 độ
Vậy 1x= 360 : 9= 40
Minh cung khong biet nua vi minh moi hoc lop 4 thoi.
a, \(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
b, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)
c, \(\frac{x^2}{2}\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)=\frac{6x^3}{2}-\frac{3x^2}{2}-x^3-\frac{1}{2}x+\frac{1}{2}x+2\)
\(=3x^3-\frac{3x^2}{2}-x^3+2=2x^3-\frac{3}{2}x^2+2\)
a. (\(\frac{1}{2}\) x – 1) (2x – 3)
= x2 - \(\frac{3}{2}\) x – 2x + 3
= x2 - \(\frac{7}{2}\) x + 3
b. (x –7)(x –5)
= x2 – 5x – 7x + 35
= \(x^2\) – 12x + 35
c. (x - \(\frac{1}{2}\) )(x + \(\frac{1}{2}\) )(4x - 1)
= (\(x^2\) + \(\frac{1}{2}\) x - \(\frac{1}{2}\) x - \(\frac{1}{4}\) )(4x - 1)
= (\(x^2\) - \(\frac{1}{4}\) )(4x - 1)
= 4\(x^3\) – \(x^2\) – x + \(\frac{1}{4}\)
a, \(\left(\frac{1}{2}x-1\right)\left(2x-3\right)=x^2-\frac{3}{2}x-2x+3=x^2-\frac{7}{2}x+3\)
b, \(\left(x-7\right)\left(x-5\right)=x^2-12x+35\)
c, \(\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\left(4x-1\right)=\left(x^2-\frac{1}{4}\right)\left(4x-1\right)=4x^3-x^2-x+\frac{1}{4}\)
\(\left(\frac{a}{b^2}+\frac{1}{a}-\frac{1}{b}\right):\left(\frac{b}{a}+\frac{a^2}{b^2}\right)=\left(\frac{a^2+b^2-ab}{ab^2}\right):\left(\frac{b^3+a^3}{ab^2}\right)=\frac{a^2+b^2-ab}{\left(a^3+b^3\right)}\)
\(=\frac{a^2+b^2-ab}{\left(a+b\right)\left(a^2+b^2-ab\right)}=\frac{1}{a+b}\)
\(\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4-x^3y+x^3y-x^2y^2+x^2y^2-xy^3+xy^3-y^4=x^4-y^4\)
3x3 - 5x2 + 5x - 2
= 3x3 - 2x2 - 3x2 +2x + 3x-2
= x2(3x -2) - x(3x -2) + (3x-2)
= (3x-2)(x2 -x +1)
a. (5\(x\) – 2\(y\))(\(x^2\) –\(x\) \(y\) + 1)
= 5\(x\).(\(x^2\) –\(x\) \(y\) + 1) – 2\(y\)(\(x^2\) – \(x\) \(y\)+ 1)
= (5 \(x^3\) – 5\(x^2\)\(y\) + 5\(x\)) – (2\(x\)2\(y\) – 2\(x\)\(y\)2 + 2\(y\))
= 5\(x\)3 – 5\(x\)2\(y\) + 5\(x\) – 2\(x\)2\(y\) + 2\(x\)\(y\)2 – 2\(y\)
= 5\(x\)3 – 7\(x\)2\(y\) + 5\(x\) + 2\(x\)\(y\)2 – 2\(y\)
b. (\(x\) – 1)(\(x\) + 1)(\(x\) + 2)
= (\(x^2\) + \(x\) – \(x\) – 1)(\(x\) + 2)
= (\(x^2\) – 1)(\(x\) + 2)
= \(x^2\)( \(x\) + 2) – 1.(\(x\) +2)
= \(x^3\) + 2\(x\) – \(x\) – 2
c. \(\frac{1}{2}\).\(x\) 2\(y\)2(2\(x\) + \(y\))(2\(x\) – \(y\))
= \(\frac{1}{2}\).\(x\)2\(y\)2 (4\(x\)2 – 2\(x\) \(y\)+ 2\(x\) \(y\)– \(y\)2)
= \(\frac{1}{2}\).\(x\)2\(y\)2 (4\(x\)2 – \(y\)2)
= \(\frac{1}{2}\).\(x\)2.\(y\)2.4\(x\)2 + \(\frac{1}{2}\).\(x\)2\(y\)2. (-\(y\)2)
= 2\(x\)4\(y\)2 - \(\frac{1}{2}\).\(x\)2\(y\)4