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Ta có ; x2 - 11x + 24
= x2 - 3x - 8x + 24
= x(x - 3) - (8x - 24)
= x(x - 3) - 8(x - 3)
= (x - 3)(x - 8)
a,(x-y)^2-2(x+y)+1 b, x^2-y^2+4x+4 c, 4x^2-y^2+8(y-2)
=(x-y-1)^2 =(x^2+4x+4)-y^2 =4x^2-y^2+8y-16
=(x+2)^2-y^2 =4x^2-(y^2-8y+16)
=(x+2-y)(x+2+y) =4x^2-(y-4)^2
a) (x+y)2-2(x+y)+1=(x+y-1)2
b) x2-y2+4x+4 = (x2+4x+4)-y2=(x+2)2-y2=(x+y+2)(x-y+2)
c)4x2-y2+8(y-2) = 4x2-(y2-8y+16) = (2x)2-(y-4)2=(2x+y-4)(2x-y+4)
d)x3-2x2+2x-4 = x2(x-2)+2(x-2) = (x-2)(x2+2)
e)xy-4+2x-2y=x(y+2) - 2(y+2) = (x-2)(y+2)
a) x2 + 4x + 3 - y2 -2y
= x2 +4x + 4 - y2 -2y-1
= (x+2)2 - (y+1)2
= (x+2-y-1).(x+2+y+1)
= (x-y+1).(x+y+3)
b) 2a2 -5ab + 2b2
= 2a2 -4ab + 2b2 - ab
= 2.(a2 - 2ab+b2) - ab
= 2.(a-b)2 -ab
...
c) (x+y)2 - 2x - 2y + 1
= (x+y)2 - 1 - 2x -2y +2
= (x+y-1).(x+y+1) - 2.(x+y-1)
= (x+y-1)2
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
Bài 1 :
a, \(\left(a-2\right)^2-b^2=\left(a-2-b\right)\left(a-2+b\right)\)
b, \(2a^3-54b^3=2\left(a^3-27b^3\right)=2\left(a-3b\right)\left(a^2+3ab+9b\right)\)
Bài 2 : tự kết luận nhé, ngại mà lười :(
a, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
\(\Leftrightarrow\frac{4x-3}{5}-\frac{5x-4}{3}=\frac{6x-2}{7}+3\)
\(\Leftrightarrow\frac{12x-9-25x+20}{15}=\frac{6x-2+21}{7}\)
\(\Leftrightarrow\frac{-13x-29}{15}=\frac{6x+19}{7}\Rightarrow-91x-203=90x+285\)
\(\Leftrightarrow181x=-488\Leftrightarrow x=-\frac{488}{181}\)
b, \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)
\(\Leftrightarrow\frac{4x+8+9\left(2x-1\right)}{12}-\frac{10x-6}{12}=\frac{12x+5}{12}\)
\(\Rightarrow4x+8+18x-9-10x+6=12x+5\)
\(\Leftrightarrow12x+5=12x+5\Leftrightarrow0x=0\)
Vậy phương trình có vô số nghiệm
c, \(\left|2x-3\right|=4\)
Với \(x\ge\frac{3}{2}\)pt có dạng : \(2x-3=4\Leftrightarrow x=\frac{7}{2}\)
Với \(x< \frac{3}{2}\)pt có dạng : \(2x-3=-4\Leftrightarrow x=-\frac{1}{2}\)
d, \(\left|3x-1\right|-x=2\Leftrightarrow\left|3x-1\right|=x+2\)
Với \(x\ge\frac{1}{3}\)pt có dạng : \(3x-1=x+2\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Với \(x< \frac{1}{3}\)pt có dạng : \(3x-1=-x-2\Leftrightarrow4x=-1\Leftrightarrow x=-\frac{1}{4}\)
a) \(4x^3\left(x^2+x\right)-\left(x^2+x\right)=\left(x^2+x\right)\left(4x^3-1\right)\)
b)\(\left(1-2a+a^2\right)-\left(b^2-2bc+c^2\right)=\left(1-a\right)^2-\left(b-c\right)^2=\)\(\left(1-a+b-c\right)\left(1-a-b+c\right)\)
lm tiếp câu c
c) \(C=\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-72\)
\(=\left[\left(x-7\right)\left(x-2\right)\right]\left[\left(x-5\right)\left(x-4\right)\right]-72\)
\(=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\)
Đặt \(x^2-9x+17=a\) ta có:
\(C=\left(a-3\right)\left(a+3\right)-72\)
\(=a^2-9-72\)
\(=a^2-81=\left(a-9\right)\left(a+9\right)\)
Thay trở lại ta được: \(C=\left(x^2-9x++8\right)\left(x^2-9x+26\right)\)
\(e,-5x+x^2-14\)
\(=x^2+2x-7x-14\)
\(=x\left(x+2\right)-7\left(x+2\right)\)
\(=\left(x+2\right)\left(x-7\right)\)
\(f,x^3+8+6x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+2x+4\right)+6x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+8x+4\right)\)
\(g,15x^2-7xy-2y^2\)
\(=15x^2+3xy-10xy-2y^2\)
\(=3\left(5x+y\right)-2y\left(5x+y\right)\)
\(=\left(5x+y\right)\left(3-2y\right)\)
\(h,3x^2-16x+5\)
\(=3x^2-x-15x+5\)
\(=x\left(3x-1\right)+5\left(3x-1\right)\)
\(=\left(3x-1\right)\left(x+5\right)\)
\(a,x^3+2x^2y+xy^2=x\left(x^2+2xy+y^2\right)\)
\(=x\left(x+y\right)^2\)
\(b,4x^2-9y^2+4x-6y\)
\(=4x^2+4x+1-\left(9y^2+6y+1\right)\)
\(=\left(2x+1\right)^2-\left(3y+1\right)^2\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
\(c,-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left(y-x+5\right)\)
\(d,x^2+4x-12\)
\(=x^2-2x+6x-12\)
\(=x\left(x-2\right)+6\left(x-2\right)\)
\(=\left(x-2\right)\left(x+6\right)\)
a) a2−b2−4a+4
=(a2-4a+4)-b2
=(a-2)2-b2
=(a-2-b)(a-2+b)a2−b2−4a+4
b) x2+2x−3
=x2-x+3x-3
=x(x-1)+3(x-1)
=(x+3)(x-1)x2+2x−3
c) 4x2y2−(x2+y2)2
=(2xy-x2-y2)(2xy+x2+y2)
=-(x-y)2(x+y)2
d) 2a3−54b3
=2(a3-27b3)
=2(a-3b)(a2+3ab+9b2)