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bài 3
a) (xy+1)2-(x-y)2
=[(xy+1)-(x-y)][(xy+1)+(x-y)]
=(xy+1-x+y)(xy+1+x-y)
b) x2-4y4+x+2y2
=(x2-4y4)+(x+2y2)
=(x-2y2)(x+2y2)+(x+2y2)
=(x+2y2)(x-2y2+1)
c) (x2+2x)2+9x2+18x
=(x2+2x)2+(9x2+18x)
=(x2+2x)2+9(x2+2x)
=(x2+2x)(x2+2x+9)
d) (x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8) (x+4)(x+6) +16
=(x2+8x+2x+16)(x2+6x+4x+24)+16
=(x2+10x+16)(x2+10x+24)+16
đặt x2+10x+16=a ta có
a(a+8)+16
=a2+8a+16
=(a+4)2
thay a=(x2+10x+16) ta đc
(x2+10x+16)2
=(x2+8x+2x+16)2
=[x(x+8)+2(x+8)]2
=[ (x+2)(x+8)]2
Bài 2:
a)A= \(6x^2\)\(-11x+3\)
<=>A=\(6x^2\)\(-2x-9x+3\)
<=>A=(\(6x^2\)\(-2x\))-\(\left(9x-3\right)\)
=>A=\(2x\left(3x-1\right)\)\(-3\left(3x+1\right)\)
<=>A=\(2x\left(3x-1\right)+3\left(3x-1\right)\)
=>A=(3x-1)(2x+3)
\(1,x^3-7x+6\)
\(=x^3+3x^2-3x^2-9x+2x+6\)
\(=x^2\left(x+3\right)-3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-3x+2\right)\)
\(=\left(x+3\right)\left(x^2-2x-x+2\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x-1\right)\)
\(2,x^3-9x^2+6x+16\)
\(=x^3+x^2-10x^2-10x+16x+16\)
\(=x^2\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left(x^2-2x-8x+16\right)\)
\(=\left(x+1\right)\left(x-8\right)\left(x-2\right)\)
mk ms lm hai câu thôi mà đã mệt r , bh mk lm bt mai đi học ,lúc khác lm đ cko bn
Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
\(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(=\left(x^4-x^2+1\right)\left(x^4+2x^2-x^2+1\right)\)
\(=\left(x^4-x^2+1\right)[\left(x^2+1\right)^2-x^2]\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
mk ghi đáp án, còn lại bạn tự biến đổi
a) \(2x^3-x^2+5x+3=\left(2x+1\right)\left(x^2-x+3\right)\)
b) \(x^3+5x^2+8x+4=\left(x+1\right)\left(x+2\right)^2\)
c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
d) \(4x^4+1=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)
e) \(x^4-7x^3+14x^2-7x+1=\left(x^2-4x+1\right)\left(x^2-3x+1\right)\)
mk làm chi tiết theo yêu của của người hỏi đề:
a) \(2x^3-x^2+5x+3\)
\(=\left(2x^3-2x^2+6x\right)+\left(x^2-x+3\right)\)
\(=2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
b) \(x^3+5x^2+8x+4\)
\(=\left(x^3+4x^2+4x\right)+\left(x^2+4x+4\right)\)
\(=x\left(x^2+4x+4\right)+\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x+2\right)^2\)
a) \(5x-10x^2\) = \(5x\left(1-2x\right)\)
b) Mạn phép sửa đề:
\(\dfrac{1}{2}x\left(x^2-4\right)+4\left(x+2\right)\) = \(\left(x+2\right)\left[\dfrac{1}{2}x\left(x-2\right)+4\right]\)
= \(\left(x+2\right)\left(\dfrac{1}{2}x^2-x+4\right)\)
c) \(x^4-y^6=\left(x^2-y^3\right)\left(x^2+y^3\right)\)
e) \(x^3-4x^2+4x-1=x^3-x^2-3x^2+3x+x-1\)
= \(x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2-3x+1\right)\)
g) \(x^4+6x^3-12x^2-8x\)
= \(x\left(x^3-2x^2+8x^2-16x+4x-8\right)\)
= \(x\left[x^2\left(x-2\right)+8x\left(x-2\right)+4\left(x-2\right)\right]\)
= \(x\left(x-2\right)\left(x^2+8x+4\right)\)
h) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\) (*)
Đặt \(x^2+4x+8=a\) => (*) trở thành:
\(a^2+3ax+2x^2\) = \(a^2+ãx+2ax+x^2\)
= \(a\left(a+x\right)+2x\left(a+x\right)\)
= \(\left(a+x\right)\left(a+2x\right)\) (1)
Thay \(a=x^2+4x+8\) vào (1) ta được:
\(\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)
=\(\left(x^2+5x+8\right)\left(x^2+2x+4x+8\right)\)
= \(\left(x^2+5x+8\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]\)
= \(\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)
P/s: Còn câu f đang suy nghĩ!
Bài 2:
a: \(=\left(x+y\right)^2-\left(x+y\right)-12\)
\(=\left(x+y-4\right)\left(x+y+3\right)\)
b: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)
c: \(4x^4-32x^2+1\)
\(=4x^4+4x^2+1-36x^2\)
\(=\left(2x^2+1\right)^2-36x^2\)
\(=\left(2x^2-6x+1\right)\left(2x^2+6x+1\right)\)
d: \(=\left(x^2+3\right)\left(x^4-3x^2+9\right)\)