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\(1.\)
\(x^2-2x+1-xy-y=\left(x-1\right)^2-y\left(x-1\right)=\left(x-1\right)\left(x-1-y\right)\)
\(2.\)
\(x^3-4x^2+4x-2x+2=x\left(x^2-4x+4\right)-2\left(x-1\right)=x\left(x-2\right)^2-2\left(x-1\right)\)
\(3.\)
\(10x-25-x^2+4y^2=4y^2-\left(x^2-10x+25\right)=4y^2-\left(x-5\right)^2=\left(2y+x-5\right)\left(2y-x+5\right)\)
\(4.\)
\(4x^2-2x+2xy-y=2x\left(2x-1\right)+y\left(2x-1\right)=\left(2x-1\right)\left(2x+y\right)\)
\(5.\)
\(4x\left(x-3\right)^2-3x^2+9x=4x\left(x-3\right)^2-3x\left(x-3\right)=\left(x-3\right)\left(4x^2-12x-3x\right)\)
a) nhận xét hệ số : 1 + 4 - 29 + 24 = 0
=> x3 + 4x2 - 29x + 24 = x2(x-1) + 5x(x-1) - 24(x-1)
= (x-1)(x2+5x-24) = (x-1)(x-3)(x+8)
b) ...
a) \(x^3+4x^2-29x+24\)=\(\left(x+8\right)\left(x^2-4x+3\right)\)=\(\left(x+8\right)\left(x^2-x-3x+3\right)\)=\(\left(x+8\right)\left(x-1\right)\left(x-3\right)\)
b) \(x^4+6x^3+7x^2-6x+1\)=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)=\(x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)^2\)
24:
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)
\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
\(\Leftrightarrow x+5=0\)
hay x=-5
b) 6x - 9 - x2
= - (x2 - 6x + 9 )
= - ( x2 - 2.x.3 + 32 )
= - ( x - 3 )2
c) x2 - 16
= x2 - 42
= ( x - 4 )( x + 4)
d) 9x2 - 25
= ( 3x )2 - 52
= ( 3x - 5 )( 3x + 5 )
e ) x4 - y4
= ( x2)2 - ( y2 )2
= ( x2 - y2 )( x2 + y2 )
f) x6 -y6
= ( x3 )2 - ( y3)2
=( x3 - y3 )( x3 + y3 )
g) 8x3 - \(\dfrac{1}{27}\)
= ( 2x )3 - ( \(\dfrac{1}{3}\))3
= ( 2x - \(\dfrac{1}{3}\) ) ( 2x + \(\dfrac{2}{3}\)x + \(\dfrac{1}{3}\))
a) \(12x-9-4x^2\)
\(=-\left(4x^2-12x+9\right)\)
\(=-\left(2x-3\right)^2\)
b)\(1-9x+27x^2-27x^3\)
\(=\left(1-3x\right)^{^3}\)
c)\(\frac{x^2}{4}+2xy+4y^2-25\)
\(=\left(\frac{x}{2}+2y\right)^2-5^2\)
\(=\left(\frac{x}{2}+2y-5\right)\left(\frac{x}{2}+2y+5\right)\)
d)\(\left(x^2-4x\right)^2-8\left(x^2-4x\right)+15\)
\(=\left(x^2-4x\right)^2-3\left(x^2-4x\right)-5\left(x^2-4x\right)+15\)
\(=\left(x^2-4x\right)\left(x^2-4x-3\right)-5\left(x^2-4x-3\right)\)
\(=\left(x^2-4x-5\right)\left(x^2-4x-3\right)\)
\(=\left(x^2+x-5x-5\right)\left(x^2-4x-3\right)\)
\(=\left[x\left(x+1\right)-5\left(x+1\right)\right]\left(x^2-4x-3\right)\)
\(=\left(x-5\right)\left(x+1\right)\left(x^2-4x-3\right)\)
Chúc bạn học tốt !
1) \(4x^2+4x+1=\left(2x+1\right)^2\)
2)\(9x^2-24xy+16y^2=\left(3x-4y\right)^2\)
3)\(-x^2+10x-25=-\left(x-5\right)^2\)
4)\(1+12x+36x^2=\left(1+6x\right)^2\)
5) \(\dfrac{x^2}{4}+2xy+4y^2=\left(\dfrac{x}{2}+2y\right)^2\)
6) \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
câu a đề có sai số mũ ko vậy
b) \(\dfrac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)
\(=\dfrac{x^3\left(x+1\right)-\left(x+1\right)}{x^4+x^3+x^2+x^2+x+1}\)
\(=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2+1\right)}=\dfrac{x^2-1}{x^2+1}\)
c) \(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)
\(=\dfrac{\left(x^2+3x\right)^2-1}{x^4+6x^3+9x^2-2x^2-6x+1}\)
\(=\dfrac{\left(x^2+3x-1\right)\left(x^2+3x+1\right)}{\left(x^2+3x\right)^2-2\left(x^2+3x\right)+1}\)
\(=\dfrac{\left(x^2+3x-1\right)\left(x^2+3x+1\right)}{\left(x^2+3x-1\right)^2}=\dfrac{x^2+3x+1}{x^2-3x+1}\)
\(4x^4+4x^2+1=\left(2x^2+1\right)^2\)
\(9x^4-6x^2+1=\left(3x^2-1\right)^2\)
\(\dfrac{x^2}{9}-\dfrac{2}{3}x+1=\left(\dfrac{x}{3}+1\right)^2\)
\(x^2-25=\left(x-5\right)\left(x+5\right)\)