a, \(40x^2y-8xy-16x^2y^2=8xy\left(5x-1-2xy\right)\)

b, \(\left(x-y\right)\left(x+y\right)-3\left(x-y\right)=\left(x-y\right)\left(x+y-3\right)\)

c, \(\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)

a) 40x²y - 8xy - 16x²y²

= 8xy( 5x - 1 - 2xy)

b) (x-y)(x+y)-3(x-y)    

= (x - y)( x + y - 3)

c)(x-y)² -5(x-y)

=(x - y)(x - y - 5)

a: =(16x+20)^2-(10x+10)^2

=(16x+20-10x-10)(16x+20+10x+10)

=(26x+30)(6x+10)

=4(13x+15)(3x+5)

b: =(x-y+4-2x-3y+1)(x-y+4+2x+3y-1)

=(-x-4y+5)(3x+2y+3)

c: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]

=(x^2+2x+1-x^2+2x-1)(x^2+2x+1+x^2-2x+1)

=2(x^2+1)*4x

=8x(x^2+1)

$16(4x+5)^2-25(2x+2)^2$

$=[4(4x+5)]^2-[5(2x+2)]^2$

$=[4(4x+5)-5(2x+2)][4(4x+5)+5(2x+2)]$

$=(16x+20-10x-10)(16x+20+10x+10)$

$=(6x+10)(26x+30)$

$=4(3x+5)(13x+15)$

$(x-y+4)^2-(2x+3y-1)^2$

$=[(x-y+4)-(2x+3y-1)][(x-y+4)+(2x+3y-1)]$

$=(-x-4y+5)(3x+2y+3)$

Vậy:

$=-(x+4y-5)(3x+2y+3)$

$(x+1)^4-(x-1)^4$

$=[(x+1)^2]^2-[(x-1)^2]^2$

$=[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]$

$=[x^2+2x+1-(x^2-2x+1)][x^2+2x+1+x^2-2x+1]$

$=(4x)(2x^2+2)$

$=8x(x^2+1)$

d)

Câu này không phân tích được thành nhân tử với hệ số nguyên/rational theo các hằng đẳng thức thông thường: $x^6-y^5$

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

\(=x^2\left(x-1\right)-2x\cdot\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-2x\right)=x\left(x-1\right)\left(x-2\right)\)

5: \(y^2\left(x^2+y\right)-z\cdot x^2-zy\)

\(=y^2\left(x^2+y\right)-z\left(x^2+y\right)\)

\(=\left(x^2+y\right)\left(y^2-z\right)\)

7: \(5\left(x+y\right)^2+15\left(x+y\right)\)

\(=5\left(x+y\right)\cdot\left(x+y\right)+5\left(x+y\right)\cdot3\)

=5(x+y)(x+y+3)

9: \(7x\left(y-4\right)^2-\left(4-y\right)^3\)

\(=7x\left(y-4\right)^2+\left(y-4\right)^3\)

\(=\left(y-4\right)^2\left(7x+y-4\right)\)

11: \(\left(x+1\right)\left(y-2\right)-\left(2-y\right)^2\)

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

=(y-2)(x+1-y+2)

=(y-2)(x+y+3)

2: \(5x\left(x-2\right)-3x^2\left(x-2\right)\)

\(=\left(x-2\right)\left(5x-3x^2\right)=x\left(5-3x\right)\left(x-2\right)\)

4: 3x(x-5y)-2y(5y-x)

=3x(x-5y)+2y(x-5y)

=(x-5y)(3x+2y)

6: b(a-c)+5c-5a

=b(a-c)-5(a-c)

=(a-c)(b-5)

8: 9x(x-y)-10(y-x)^2

=9x(x-y)-10(x-y)^2

=(x-y)(9x-10x+10y)

=(x-y)(-x+10y)

10: \(\left(a-b\right)^2-\left(a+b\right)\left(b-a\right)\)

\(=\left(a-b\right)^2+\left(a+b\right)\left(a-b\right)\)

=(a-b)(a-b+a+b)

=2a(a-b)

12: 2x(x-3)+y(x-3)+(3-x)

=(x-3)(2x+y)-(x-3)

=(x-3)(2x+y-1)

1) \(x^2+2xy+y^2-x-y-12\)

= \(\left(x+y\right)^2-\left(x+y\right)-12\)

Đặt \(x+y=z\) (đặt ẩn phụ)

\(\Rightarrow z^2-z-12\)

\(=z^2+3z-4z-12\)

\(=z\left(z+3\right)-4\left(z+3\right)\)

\(=\left(z+3\right)\left(z-4\right)\)

Khi đó: \(\left(x+y+3\right)\left(x+y-4\right)\)

#HuyenAnh

vại

fdvfdverberrgtrgrgg

a: \(x^2-9-x^2\left(x^2-9\right)\)

\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)

\(=\left(x^2-9\right)\left(1-x^2\right)\)

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

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

\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2-y^2\right)\)

\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)

c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)

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

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

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

d: \(x^2+5x+6\)

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

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

e: \(3x^2-4x-4\)

\(=3x^2-6x+2x-4\)

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

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

g: \(x^4+64y^4\)

\(=x^4+16x^2y^2+64y^4-16x^2y^2\)

\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)

\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)

h: \(a^2+b^2+2a-2b-2ab\)

\(=a^2-2ab+b^2+2a-2b\)

\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)

i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)

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

\(=\left(x-y+4\right)^2\)

k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)

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

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

a) \(x^6-x^4+2x^3+2x^2\)

\(=x^2\left(x^4-x^2+2x+2\right)\)

\(=x^2\left[x^2\left(x^2-1\right)+2\left(x+1\right)\right]\)

\(=x^2\left(x+1\right)\left[x^2\left(x-1\right)+2\right]\)

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

b) Ta có: \(4x^4+y^4\)

\(=4x^4+y^4+4x^2y^2-4x^2y^2\)

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

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

a, \(x^6-x^4+2x^3+2x^2\)

\(=x^2\left(x^4-x^2+2x+2\right)=x^2\left[x^2\left(x^2-1\right)+2\left(x+1\right)\right]\)

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

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

b, \(4x^4+y^4=\left(2x^2\right)^2+2.2x^2.y^2+\left(y^2\right)^2-4x^2y^2\)

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

1;

a: \(\frac{x-20}{y+4}=13\)

=>x-20=13(y+4)

=>x-20=13y+52

=>x-13y=72

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{20}=\frac{y}{9}=\frac{z}{6}=\frac{x-13y}{20-13\cdot9}=\frac{72}{20-117}=\frac{72}{-97}\)

=>\(\begin{cases}x=\frac{-72}{97}\cdot20=-\frac{1440}{97}\\ y=-\frac{72}{97}\cdot9=-\frac{648}{97}\\ z=-\frac{72}{97}\cdot6=-\frac{432}{97}\end{cases}\)

b: \(\frac{x}{3}=\frac{y}{4}\)

=>\(\frac{x}{15}=\frac{y}{20}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{7}\)

=>\(\frac{y}{20}=\frac{z}{28}\) (2)

Từ (1),(2) suy ra \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

mà x-y-z=46

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{x-y-z}{15-20-28}=\frac{46}{-33}=\frac{-46}{33}\)

=>\(\begin{cases}x=-\frac{46}{33}\cdot15=-46\cdot\frac{5}{11}=-\frac{230}{11}\\ y=-\frac{46}{33}\cdot20=-\frac{920}{33}\\ z=-\frac{46}{33}\cdot28=-\frac{1288}{33}\end{cases}\)

d: \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\)

=>\(\frac{x}{2}\cdot12=\frac{2y}{3}\cdot12=\frac{3z}{4}\cdot12\)

=>6x=8y=9z

=>\(\frac{6x}{72}=\frac{8y}{72}=\frac{9z}{72}\)

=>\(\frac{x}{12}=\frac{y}{9}=\frac{z}{8}\)

Đặt \(\frac{x}{12}=\frac{y}{9}=\frac{z}{8}=k\)

=>x=12k; y=9k; z=8k

xyz=108

=>\(12k\cdot9k\cdot8k=108\)

=>\(8k^3=1\)

=>\(k^3=\frac18\)

=>\(k=\frac12\)

=>\(\begin{cases}x=12\cdot\frac12=6\\ y=9\cdot\frac12=4,5\\ z=8\cdot\frac12=4\end{cases}\)

Bài 2:

b: \(\frac{a}{7}=\frac{b}{6}\)

=>\(\frac{a}{35}=\frac{b}{30}\)

\(\frac{b}{5}=\frac{c}{8}\)

=>\(\frac{b}{30}=\frac{c}{48}\)

=>\(\frac{a}{35}=\frac{b}{30}=\frac{c}{48}\)

mà a-2b+c=46

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{a}{35}=\frac{b}{30}=\frac{c}{48}=\frac{a-2b+c}{35-2\cdot30+48}=\frac{46}{23}=2\)

=>\(\begin{cases}a=2\cdot35=70\\ b=2\cdot30=60\\ c=48\cdot2=96\end{cases}\)

c: 5a=8b=3c

=>\(\frac{5a}{120}=\frac{8b}{120}=\frac{3c}{120}\)

=>\(\frac{a}{24}=\frac{b}{15}=\frac{c}{40}\)

mà a-2b+c=24

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{a}{24}=\frac{b}{15}=\frac{c}{40}=\frac{a-2b+c}{24-2\cdot15+40}=\frac{24}{34}=\frac{12}{17}\)

=>\(\begin{cases}a=\frac{12}{17}\cdot24=\frac{288}{17}\\ b=\frac{12}{17}\cdot15=\frac{180}{17}\\ c=\frac{12}{17}\cdot40=\frac{480}{17}\end{cases}\)

d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{a+3}{5}=\frac{b-2}{3}=\frac{c-1}{7}=\frac{a+b+c+3-2-1}{5+3+7}=\frac{24}{15}=1,6\)

=>\(\begin{cases}a+3=5\cdot1,6=8\\ b-2=3\cdot1,6=4,8\\ c-1=7\cdot1,6=11,2\end{cases}\Rightarrow\begin{cases}a=5\\ b=6,8\\ c=12,2\end{cases}\)

mk ghi đáp án, ko phân tích đc thì IB mk

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

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

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

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

e)  \(25x^4-10x^2y^2+y^4=\left(5x^2-y^2\right)^2\)

f) \(-a^2-2a-1=-\left(a+1\right)^2\)

g)  \(27b^3-8a^3=\left(3b-2a\right)\left(9b^2+6ab+4a^2\right)\)

h)  \(x^3+9x^2y+27xy^2+27y^3=\left(x+3y\right)^3\)

i) \(16x^2-9\left(x+y\right)^2=\left(x-3y\right)\left(7x+3y\right)\)