1) Rút gọn phân thức
a) (x-5) (3x+3)- 3x (×-3) +3x+7
b) (x-3) (x^2+3x+9)- (54 +x^3 )
c) (3x +y ) (9x^2-3y +y^2)- (3x-y) (9x^2+3x-y^2)
2) Phân Tích đa thức thành nhân tử
a)14 x^2 y^2- 21xy^2+ 28x^2y
b) (x+y)^2 -4x^2
c) 2x^2- 2xy -5x+ 5y
d)2xy-x^2-y^2+16
3)Tìm x biết
a)x^2 (x+1) + 2x(x+1)=0
b) x(3x-2)-5(2-3x)=0
c)16 -25x^2=o
4) Chứng minh hằng đẳng thức
a) a^3+b^3= (a+b)^3 - 3ab (a+b)
b)a^3-b^3=(a-b)^3 +3ab (a-b)
Đây là đề cương ôn tập mai mình sửa rùi giúp dùm nha
Bài 1:
a) \((x-5)(3x+3)-3x(x-3)+3x+7\)
\(=3(x-5)(x+1)-3x(x-3)+3x+7\)
\(=3(x^2+x-5x-5)-(3x^2-9x)+3x+7\)
\(=3(x^2-4x-5)-(3x^2-9x)+3x+7\)
\(=-8\)
b) \((x-3)(x^2+3x+9)-(54+x^3)\)
\(=(x-3)(x^2-3.x+3^2)-(54+x^3)\)
\(=x^3-3^3-(54+x^3)=-81\)
c) Sửa đề:
\((3x+y)(9x^2-3xy+y^2)-(3x-y)(9x^2+3xy+y^2)\)
\(=(3x+y)[(3x)^2-3x.y+y^2]-(3x-y)[(3x)^2+3x.y+y^2]\)
\(=(3x)^3+y^3-[(3x)^3-y^3]\)
\(=2y^3\)
Câu 2:
\(a)14x^2y^2-21xy^2+28x^2y\)
\(=7xy(2xy-3y+4x)\)
b) \((x+y)^2-4x^2\)\(=(x+y)^2-(2x)^2=(x+y-2x)(x+y+2x)\)
\(=(y-x)(3x+y)\)
c) \(2x^2-2xy-5x+5y\)
\(=(2x^2-2xy)-(5x-5y)\)
\(=2x(x-y)-5(x-y)=(x-y)(2x-5)\)
d) \(2xy-x^2-y^2+16\)
\(=16-(x^2+y^2-2xy)=4^2-(x-y)^2\)
\(=[4-(x-y)][4+(x-y)]=(4-x+y)(4+x-y)\)
Bài 3:
a)
\(x^2(x+1)+2x(x+1)=0\)
\(\Leftrightarrow (x+1)(x^2+2x)=0\)
\(\Leftrightarrow (x+1)x(x+2)=0\)
\(\Rightarrow \left[\begin{matrix} x+1=0\\ x=0\\ x+2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-1\\ x=0\\ x=-2\end{matrix}\right.\)
b)
\(x(3x-2)-5(2-3x)=0\)
\(\Leftrightarrow x(3x-2)+5(3x-2)=0\)
\(\Leftrightarrow (3x-2)(x+5)=0\)
\(\Rightarrow \left[\begin{matrix} 3x-2=0\\ x+5=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{2}{3}\\ x=-5\end{matrix}\right.\)
c)
\(16-25x^2=0\)
\(\Leftrightarrow 4^2-(5x)^2=0\)
\(\Leftrightarrow (4-5x)(4+5x)=0\)
\(\Rightarrow \left[\begin{matrix} 4-5x=0\\ 4+5x=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{4}{5}\\ x=-\frac{4}{5}\end{matrix}\right.\)
Bài 4: Theo hằng đẳng thức đáng nhớ ta có:
a)
\((a+b)^3-3ab(a+b)\)
\(=a^3+3a^2b+3ab^2+b^3-(3a^2b+3ab^2)\)
\(=a^3+b^3\) (đpcm)
b)
\((a-b)^3+3ab(a-b)\)
\(=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2\)
\(=a^3-b^3\) (đpcm)
