\(x^8+3x^4+4\)

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2 tháng 9 2018

\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)

\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)

\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)

\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)

\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)

13 tháng 8 2018

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AH
Akai Haruma
Giáo viên
20 tháng 10 2020

Lời giải:

a)

$yz(y+z)+xz(z-x)-xy(x+y)=yz(y+z)+xz^2-x^2z-x^2y-xy^2$

$=yz(y+z)+x(z^2-y^2)-x^2(z+y)$

$=yz(y+z)+x(z-y)(z+y)-x^2(z+y)$

$=(y+z)(yz+xz-xy-x^2)$

$=(y+z)[z(x+y)-x(x+y)]=(y+z)(x+y)(z-x)$

b)

$2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4abc$

$=(2a^2b+4ab^2)-(a^2c+2abc)+(ac^2+2bc^2)-(4b^2c+2abc)$

$=2ab(a+2b)-ac(a+2b)+c^2(a+2b)-2bc(a+2b)$

$=(a+2b)(2ab-ac+c^2-2bc)$

$=(a+2b)[2b(a-c)-c(a-c)]$

$=(a+2b)(2b-c)(a-c)$

c)

$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$

$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$

17 tháng 8 2020

Lời giải:

a)

$yz(y+z)+xz(z-x)-xy(x+y)=yz(y+z)+xz^2-x^2z-x^2y-xy^2$

$=yz(y+z)+x(z^2-y^2)-x^2(z+y)$

$=yz(y+z)+x(z-y)(z+y)-x^2(z+y)$

$=(y+z)(yz+xz-xy-x^2)$

$=(y+z)[z(x+y)-x(x+y)]=(y+z)(x+y)(z-x)$

b)

$2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4abc$

$=(2a^2b+4ab^2)-(a^2c+2abc)+(ac^2+2bc^2)-(4b^2c+2abc)$

$=2ab(a+2b)-ac(a+2b)+c^2(a+2b)-2bc(a+2b)$

$=(a+2b)(2ab-ac+c^2-2bc)$

$=(a+2b)[2b(a-c)-c(a-c)]$

$=(a+2b)(2b-c)(a-c)$

c)

$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$

$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$

Bài 1: Phân tích đa thức thành nhân tử: a) \(2x\left(x+1\right)+2\left(x+1\right)\) b) \(y^2\left(x^2+y\right)-zx^2-zy\) c) \(4x\left(x-2y\right)+8y\left(2y-x\right)\) d) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\) e) \(x^2-6xy+9y^2\) f) \(x^3+6x^2y+12xy^2+8y^3\) g) \(x^3-64\) h) \(125x^3+y^6\) k) \(0,125\left(a+1\right)^3-1\) t) \(x^2-2xy+y^2-xz+yz\) q) \(x^2-y^2-x+y\) p) \(a^3x-ab+b-x\) đ)...
Đọc tiếp

Bài 1: Phân tích đa thức thành nhân tử:

a) \(2x\left(x+1\right)+2\left(x+1\right)\)

b) \(y^2\left(x^2+y\right)-zx^2-zy\)

c) \(4x\left(x-2y\right)+8y\left(2y-x\right)\)

d) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)

e) \(x^2-6xy+9y^2\)

f) \(x^3+6x^2y+12xy^2+8y^3\)

g) \(x^3-64\)

h) \(125x^3+y^6\)

k) \(0,125\left(a+1\right)^3-1\)

t) \(x^2-2xy+y^2-xz+yz\)

q) \(x^2-y^2-x+y\)

p) \(a^3x-ab+b-x\)

đ) \(3x^2\left(a+b+c\right)+36xy\left(a+b+c\right)+108y^2\left(a+b+c\right)\)

l) \(x^2-x-6\)

i) \(x^4+4x^2-5\)

m) \(x^3-19x-30\)

j) \(x^4+x+1\)

y) \(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

o) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

ê) \(4a^2b^2-\left(a^2+b^2+c^2\right)^2\)

w) \(\left(1+x^2\right)^2-4x\left(1-x^2\right)\)

z) \(\left(x^2-8\right)^2+36\)

u) \(81x^4+4\)

Bài 2 : Tìm x

a)\(\left(2x-1\right)^2-25=0\)

b) \(8x^3-50x=0\)

c) \(\left(x-2\right)\left(x^2+2+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)

d) \(3x\left(x-1\right)+x-1=0\)

e) \(2\left(x+3\right)-x^2-3x\) =0

f) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

5
12 tháng 10 2017

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 ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)

12 tháng 10 2017

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 ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)

20 tháng 3 2016

b) a3 + b3 + c3 - 3abc

= ( a + b)3 - 3ab - 3ba + c - 3abc

= (a3 + 3a2b + 3ab2 + b3) + c3 - (3a2b + 3ab2 + 3ab) 

= (a + b)3 + c2 - 3ab(a + b + c)

= (a + b + c) [ (a  + b)2 - ( a + b )c + c^2 ]  - 3ab(a + b + c)

=  ( a + b + c ) ( a2 + b2 + 2ab - ac - bc + c2 -3ab )

=  ( a + b + c ) ( a2 + b2 + c2 - ab - ac - bc 

Phân tích các đa thức sau thành nhân tử: * \(x^3-7x+6\) * \(x^3-9x^2+6x+16\) * \(x^3-6x^2-x+30\) * \(2x^3-x^2+5x+3\) * \(27x^3-27x^2+18x-4\) * \(x^2+2xy+y^2-x-y-12\) * \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\) * \(4x^4-32x^2+1\) * \(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\) * \(64x^4+y^4\) * \(a^6+a^4+a^2b^2+b^4-b^6\) * \(x^3+3xy+y^3-1\) * \(4x^4+4x^3+5x^2+2x+1\) * \(x^8+x+1\) * \(x^8+3x^4+4\) * \(3x^2+22xy+11x+37y+7y^2+10\) *...
Đọc tiếp

Phân tích các đa thức sau thành nhân tử:

* \(x^3-7x+6\)

* \(x^3-9x^2+6x+16\)

* \(x^3-6x^2-x+30\)

* \(2x^3-x^2+5x+3\)

* \(27x^3-27x^2+18x-4\)

* \(x^2+2xy+y^2-x-y-12\)

* \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

* \(4x^4-32x^2+1\)

* \(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)

* \(64x^4+y^4\)

* \(a^6+a^4+a^2b^2+b^4-b^6\)

* \(x^3+3xy+y^3-1\)

* \(4x^4+4x^3+5x^2+2x+1\)

* \(x^8+x+1\)

* \(x^8+3x^4+4\)

* \(3x^2+22xy+11x+37y+7y^2+10\)

* \(x^4-8x+63\)

* \(\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)

* \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)

* \(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)

* \(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)

* \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab^2+c^3-3abc\)

* \(\left(a+b+c\right)^3-a^3-b^3-c^3=[\left(a+b\right)c]^3-a^3-b^3-c^3\)

* \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

\([\) Các bạn làm được bài nài thì làm giúp mk với nha,làm vài câu cũng được\(]\)

Mk mệt quá rồi làm giúp mk với nha

3
4 tháng 12 2017

\(1,x^3-7x+6\)

\(=x^3+3x^2-3x^2-9x+2x+6\)

\(=x^2\left(x+3\right)-3x\left(x+3\right)+2\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-3x+2\right)\)

\(=\left(x+3\right)\left(x^2-2x-x+2\right)\)

\(=\left(x+3\right)\left(x-2\right)\left(x-1\right)\)

\(2,x^3-9x^2+6x+16\)

\(=x^3+x^2-10x^2-10x+16x+16\)

\(=x^2\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-10x+16\right)\)

\(=\left(x+1\right)\left(x^2-2x-8x+16\right)\)

\(=\left(x+1\right)\left(x-8\right)\left(x-2\right)\)

4 tháng 12 2017

mk ms lm hai câu thôi mà đã mệt r , bh mk lm bt mai đi học ,lúc khác lm đ cko bn