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Câu 1:
\(\text{ a) }x^3-x^2z+x^2y-xyz\\ \\=x\left(x^2-xz+xy-yz\right)\\ \\=x\left[\left(x^2-xz\right)+\left(xy-yz\right)\right]\\ =x\left[x\left(x-z\right)+y\left(x-z\right)\right]\\ \\=x\left(x+y\right)\left(x-z\right)\)
\(\text{b) }\left(x^2+1\right)^2-4x^2\\ \\=\left(x^2+1\right)^2-\left(2x\right)^2\\ \\=\left(x^2+1+2x\right)\left(x^2+1-2x\right)\\ \\=\left(x+1\right)^2\left(x-1\right)^2\)
\(\text{c) }x^2-10x-9y^2+25\\ \\=\left(x^2-10x+25\right)-9y^2\\ \\=\left(x-5\right)^2-\left(3y\right)^2\\ \\=\left(x-5-3y\right)\left(x-5+3y\right)\)
\(\text{d) }4x^2-36x+56\\ \\ =4\left(x^2-9x+14\right)\\ \\ =4\left(x^2-7x-2x+14\right)\\ \\ =4\left[\left(x^2-7x\right)-\left(2x-14\right)\right]\\ \\ =4\left[x\left(x-7\right)-2\left(x-7\right)\right]\\ \\ =4\left(x-2\right)\left(x-7\right)\\ \)
Câu 2:
\(\text{a) }\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\\ \Leftrightarrow9x^2+24x+16-9x^2+1=49\\ \Leftrightarrow4x+17=49\\ \Leftrightarrow4x=32\\ \Leftrightarrow x=8\\ \text{Vậy }x=8\)
\(\text{b) }x^2-25=3x-15\\ \Leftrightarrow x^2-25-3x+15=0\\ \Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow x^2+2x-5x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\text{Vậy }x=5\text{ hoặc }x=-2\)
\(\text{d) }\left(x-1\right)^3+3\left(x+1\right)^2=\left(x-2x+4\right)\left(x+2\right)\\ \Leftrightarrow\left(x-1\right)^3+3\left(x+1\right)^2-\left(x-2x+4\right)\left(x+2\right)=0\\ \Leftrightarrow x^3-3x^2+3x-1+3\left(x^2+2x+1\right)-\left(x^3+8\right)=0\\ \Leftrightarrow x^3-3x^2+3x-1+3x^2+6x+3-x^3-8=0\\ \Leftrightarrow9x-6=0\\ \Leftrightarrow3\left(3x-2\right)=0\\ \Leftrightarrow3x-2=0\\ \Leftrightarrow3x=2\\ \Leftrightarrow x=\dfrac{2}{3}\\ \text{Vậy }x=\dfrac{2}{3}\)
Bài 1
a) x3 - x2z + x2y - xyz
= (x3 - x2z) + (x2y - xyz)
= x2(x - z) + xy(x - z)
= (x2 + xy)(x - z)
= x(x + y)(x - z)
b) (x2 + 1)2 - 4x2
= (x2 + 1 + 2x)(x2 + 1 - 2x)
= (x2 + x + x + 1)(x2 - x - x + 1)
= [(x2 + x) + (x + 1)].[(x2 - x) - (x - 1)]
= [x(x + 1) + (x + 1)].[x(x - 1) - (x - 1)]
= (x + 1)(x + 1)(x - 1)(x - 1)
= (x + 1)2(x - 1)2
c) x2 - 10x - 9y2 + 25
= (x2 - 10x + 25) - 9y2
= (x - 5)2 - (3y)2
= (x - 5 - 3y)(x - 5 + 3y)
= (x - 3y - 5)(x + 3y - 5)
d) 4x2 - 36x + 56
= 4x2 - 36x + 81 - 25
= (4x2 - 2.2x.9 + 92) - 52
= (2x - 9)2 - 52
= (2x - 9 - 5)(2x - 9 + 5)
= (2x - 14)(2x - 4)
Bài 2.
a) (3x + 4)2 - (3x + 1)(3x - 1) = 0
9x2 + 12x + 16 - 9x2 + 1 = 0
12x + 17 = 0
\(\Rightarrow\) 12x = -17
\(\Rightarrow\) x = \(\dfrac{-17}{12}\)
b) x2 - 4x + 4 = 9(x - 2)
x2 - 4x + 4 - 9(x - 2) = 0
x2 - 4x + 4 - 9x + 18 = 0
x2 - 13x + 22 = 0
x2 - 2x - 11x + 22 = 0
(x2 - 2x) - (11x - 22) = 0
x(x - 2) - 11(x - 2) = 0
(x - 11)(x - 2) = 0
\(\Rightarrow\left\{{}\begin{matrix}x-11=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)
c) x2 - 25 = 3x - 15
\(\Rightarrow\) x2 - 25 - 3x + 15 = 0
x2 - 3x - 10 = 0
x2 + 2x - 5x - 10 = 0
(x2 + 2x) - (5x + 10) = 0
x(x + 2) - 5(x + 2) = 0
(x - 5)(x + 2) = 0
\(\Rightarrow\left\{{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
d) (x - 1)3 + 3(x + 1)2 = (x2 - 2x + 4)(x + 2)
x3 - 3x2 + 3x - 1 + 3(x2 + 2x + 1) = x3 + 8
x3 - 3x2 + 3x - 1 + 3x2 + 6x + 3 - x3 - 8 = 0
9x - 6 = 0
\(\Rightarrow\) 9x = 6
\(\Rightarrow\) x = \(\dfrac{2}{3}\)
Bài 3
a) (10x3y - 5x2y2 - 25x4y3) : (-5xy)
=(-5xy)(-2x2 + xy + 5x3y2) : (-5xy)
= -2x2 + xy + 5x3y2
b) [15(x - y)5 - 9(x - y)4 + 12(y - x)2] : (y - x)2
= [15(x - y)5 - 9(x - y)4 + 12(x - y)2] : (x-y)2
= (x - y)2[15(x - y)3 - 9(x - y)2 + 12] : (x - y)2
= 15(x - y)3 - 9(x - y)2 + 12
c) (27x3 - y3) : (3x - y)
= (3x - y)(9x2 + 3xy + y2) : (3x - y)
= 9x2 + 3xy + y2
d) (15x4 + 4x3 + 11x2 + 14x - 8) : (5x2 + 3x - 2)
= (15x4 + 15x3 - 11x3 - 11x2 + 22x2 + 22x - 8x - 8) : (5x2 + 3x - 2)
= [ (15x4 + 15x3) - (11x3 + 11x2) + (22x2 + 22x) - (8x + 8)] : (5x2 + 3x - 2)
= [15x3(x + 1) - 11x2(x + 1) + 22x(x + 1) - 8(x + 1)] : (5x2 + 3x - 2)
= (15x3 - 11x2 + 22x - 8)(x + 1) : (5x2 + 3x - 2)
= (15x3 - 6x2 - 5x2 + 2x + 20x - 8)(x + 1) : (5x2 + 3x - 2)
= [(15x3 - 6x2) - (5x2 - 2x) + (20x - 8)](x + 1) : (5x2 + 3x - 2)
= [3x2(5x - 2) - x(5x - 2) + 4(5x - 2)](x + 1) : (5x2 + 3x - 2)
= (3x2 - x + 4)(5x - 2)(x + 1) : (5x2 + 3x - 2)
= (3x2 - x + 4)(5x2 + 5x - 2x - 2) : (5x2 + 3x - 2)
= (3x2 - x + 4)(5x2 + 3x - 2) : (5x2 + 3x - 2)
= 3x2 - x + 4
NHỚ TIK MK NHÉ 
Câu 3: Bài này thuộc dạng cơ bản. Bạn ráng làm nha.
giúp mình nhanh với mình sắp học rồi.Camon các bạn rất nhìu :>>
