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AH
Akai Haruma
Giáo viên
20 tháng 6 2021

1. a

$(3x+5)^2=(3x)^2+2.3x.5+5^2$

$=9x^2+30x+25$

1.b 

$(6x^2+\frac{1}{3})^2=(6x^2)^2+2.6x^2.\frac{1}{3}+(\frac{1}{3})^2$

$=36x^4+4x^2+\frac{1}{9}$

1.c

$(5x-4y)^2=(5x)^2-2.5x.4y+(4y)^2$

$=25x^2-40xy+16y^2$

1.d

(2x^2y-3y^3x)^2=(2x^2y)^2-2.2x^2y.3y^3x+(3y^3x)^2$

$=4x^4y^2-12x^3y^4+9x^2y^6$

 

AH
Akai Haruma
Giáo viên
20 tháng 6 2021

1.e

$(5x-3)(5x+3)=(5x)^2-3^2=25x^2-9$

1.f

$(6x+5y)(6x-5y)=(6x)^2-(5y)^2=36x^2-25y^2$

1.g

$(-4xy-5)(5-4xy)=(-4xy-5)(-4xy+5)$

$=(-4xy)^2-5^2=16x^2y^2-25$

1.h

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

$=(ab^2)^2-(a^2b)^2=a^2b^4-a^4b^2$

 

DD
6 tháng 10 2021

Bài 3: 

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

\(=\left(-1-2x\right)\left(5-4x\right)\)

b) \(49\left(x-3\right)^2-9\left(x+2\right)^2\)

\(=\left[7\left(x-3\right)\right]^2-\left[3\left(x+2\right)\right]^2\)

\(=\left[\left(7x-21\right)-\left(3x+6\right)\right]\left[\left(7x-21\right)+\left(3x+6\right)\right]\)

\(=\left(4x-27\right)\left(10x-15\right)\)

c) \(2xy-x^2-y^2+16=16-\left(x-y\right)^2=\left(16-x+y\right)\left(16+x-y\right)\)

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

\(=\left(x-3\right)\left(3x+11\right)\)

e) \(16x^2-\left(x^2+4\right)^2=\left(4x-x^2-4\right)\left(4x+x^2+4\right)\)

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

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

\(=\left(x-1-y+z\right)\left(x-1+y-z\right)\)

DD
6 tháng 10 2021

Bài 5: 

a) \(x^2+4x-5=x^2-x+5x-5=x\left(x-1\right)+5\left(x-1\right)=\left(x+5\right)\left(x-1\right)\)

b) \(2x^2-14x+20=2x^2-4x-10x+20=2x\left(x-2\right)-10x\left(x-2\right)=2\left(x-5\right)\left(x-2\right)\)

c) \(3x^2+8x+5=3x^2+3x+5x+5=3x\left(x+1\right)+5\left(x+1\right)=\left(3x+5\right)\left(x+1\right)\)

d) \(6x^2-xy-7y^2=6x^2+6xy-7xy-7y^2=6x\left(x+y\right)-7y\left(x+y\right)\)

\(=\left(6x-7y\right)\left(x+y\right)\)

DD
6 tháng 10 2021

Bài 4: 

a) \(x^3-6x^2+12x-8=x^3-2.3.x^2+3.2^2.x-2^3=\left(x-2\right)^3\)

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

\(=2\left(x^2-2x+1+x^2-4x+3+x^2-6x+9\right)\)

\(=2\left(3x^2-12x+13\right)\)

c) \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

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

\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3xy-3yz-3zx\right]\)

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

25 tháng 10 2021

ai giải giúp em đi ạ em đang cần gấp lắm ạ 

25 tháng 8 2021

18, \(\frac{x}{2}+\frac{x^2}{8}=0\Leftrightarrow4x+x^2=0\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow x=-4;x=0\)

19, \(4-x=2\left(x-4\right)^2\Leftrightarrow\left(4-x\right)-2\left(4-x\right)^2=0\)

\(\Leftrightarrow\left(4-x\right)\left[1-2\left(4-x\right)\right]=0\Leftrightarrow\left(4-x\right)\left(-7+2x\right)=0\Leftrightarrow x=4;x=\frac{7}{2}\)

20, \(\left(x^2+1\right)\left(x-2\right)+2x-4=0\Leftrightarrow\left(x^2+1\right)\left(x-2\right)+2\left(x-2\right)=0\)

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

25 tháng 8 2021

21, \(x^4-16x^2=0\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\Leftrightarrow x=0;x=\pm4\)

22, \(\left(x-5\right)^3-x+5=0\Leftrightarrow\left(x-5\right)^3-\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\Leftrightarrow\left(x-5\right)\left(x-6\right)\left(x-4\right)=0\Leftrightarrow x=4;x=5;x=6\)

23, \(5\left(x-2\right)-x^2+4=0\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\Leftrightarrow x=2;x=3\)

28 tháng 8 2021

a)\(\left(-a+\frac{2}{3}\right)\left(a+\frac{2}{3}\right)=\left(\frac{2}{3}-a\right)\left(\frac{2}{3}+a\right)=\left(\frac{2}{3}\right)^2-a^2=\frac{4}{9}-a^2\)

b)\(\left(x+5\right)\left(x^2-5x+25\right)=x^3+5^3=x^3+125\)

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

d)\(\left(a^2-2a+3\right)\left(a^2+2a+3\right)=\left(a^2+3\right)^2-\left(2a\right)^2=\left(a^2+3\right)^2-4a^2\)

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

f)\(2\left(x-\frac{1}{2}\right)\left(4x^2+2x+1\right)=\left(2x-1\right)\left(4x^2+2x+1\right)=\left(2x\right)^3-1=8x^3-1\)

28 tháng 8 2021

Trả lời:

Bài 1:

a, \(9x^2-4=\left(3x\right)^2-2^2=\left(3x-2\right)\left(3x+2\right)\)

b, \(x^3+27=x^3+3^3=\left(x+3\right)\left(x^2-3x+9\right)\)

c, \(8-y^3=2^3-y^3=\left(2-y\right)\left(4+2y+y^2\right)\)

d, \(x^4-81=\left(x^2\right)^2-9^2=\left(x^2-9\right)\left(x^2+9\right)\)\(=\left(x^2-3^2\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)

e, \(64x^3-1=\left(4x\right)^3-1^3=\left(4x-1\right)\left(16x^2+4x+1\right)\)

f, \(x^6+8y^3=\left(x^2\right)^3+\left(2y\right)^3=\left(x^2+2y\right)\left(x^4-2x^2y+4y^2\right)\)