1. ( 3x + 2)² - 4

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

= 3x(3x+4)

2. 4x² - 25y²

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

3. 4x²- 49

=(2x-7)(2x+7)

4. 8z³ + 27

=(2z+3)(4x²-6z+9)

5. \(\dfrac{9}{25}x^4-\dfrac{1}{4}\)

= \((\dfrac{3}{5}x^2-\dfrac{1}{2})(\dfrac{3}{5}x^2+\dfrac{1}{2})\)

6. x³²  - 1

=(x¹⁶-1)(x¹⁶+1)

7. 4x² + 4x + 1

=(2x+1)²

8. x² - 20x + 100

=(x-10)²

9. y⁴ -14y² + 49

=(y²-7)²

10.  125x³ - 64y³

= (5x-4y)(25x²+20xy+16y²)

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

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

3) \(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)

4) \(8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)

5) \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)

6) \(x^{32}-1=\left(x^{16}-1\right)\left(x^{16}+1\right)\)

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

\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

7) \(4x^2+4x+1=\left(2x+1\right)^2\)

8) \(x^2-20x+100=\left(x-10\right)^2\)

9) \(y^4-14y^2+49=\left(y^2-7\right)^2\)

lam kieu gi

1: \(4x^2+4x+1=\left(2x+1\right)^2\)

2: \(x^2-20x+100=\left(x-10\right)^2\)

3: \(y^4-14y^2+49=\left(y^2-7\right)^2\)

4: \(125x^3-64y^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)

a, \(x^4+2x^2+1-x^2\)

= \(\left(x^2+1\right)^2-x^2\)

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

b, \(x^4+x^2+1\)

= \(x^4+2x^2+1-x^2\)

= .. ( như phần a )

c, \(y^4+64\)

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

d, \(4xy+3z-12y-xz\)

\(=4y\left(x-3\right)-z\left(x-3\right)\)

\(=\left(x-3\right)\left(4y-z\right)\)

e, \(x^2-4xy+4y^2-z^2+6z-9\)

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

g, \(x^2-4xy+5x+4y^2-10y\)

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

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

\(=\left(x-2y\right)\left(x-2y+5\right)\)

h, \(x^2-7x+6\)

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

\(=x\left(x-6\right)-\left(x-6\right)\)

\(=\left(x-6\right)\left(x-1\right)\)

i, \(x^3+5x^2+6x+2\)

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

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

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

phần b là 6^4 nhé

nhầm câu c là 6^4

Lời giải:

$x^2+55=4y^2$

$4y^2-x^2=55$
$(2y-x)(2y+x)=55$

Vì $x,y$ là số tự nhiên nên $2y+x, 2y-x$ là số nguyên và $2y+x>0$.

Mà $(2y-x)(2y+x)=55>0$ nên $2y-x>0$

Kết hợp với $2y+x\geq 2y-x$ ta có các TH sau:

TH1: $2y-x=1; 2y+x=55\Rightarrow y=14; x=27$

TH2: $2y-x=5; 2y+x=11\Rightarrow y=4; x=3$

Lời giải:

$x^2+55=4y^2$

$\Leftrightarrow 55=4y^2-x^2=(2y-x)(2y+x)$

Do $x,y$ là stn nên $2y+x$ là stn. 

$\Rightarrow 2y+x>0$. Mà $(2y+x)(2y-x)=55>0$ nên $2y-x>0$.

Vậy $2y+x> 2y-x>0$.

Khi đó ta có các TH sau:

TH1: $2y-x=1, 2y+x=55\Rightarrow y=14; x=27$ (tm) 

TH2: $2y-x=5; 2y+x=11\Rightarrow y=4; x=3$ (tm)

x=9,16515139?

4 /x=y/21=28/49

vì 21.4/12=7 nên x=7

    28.21/49 =12 nên y=12 (công thức nhân chéo)

vậy ,x=7 ;y=12

\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}\\ \Rightarrow\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{4}{7}\)

\(\dfrac{4}{x}=\dfrac{4}{7}\Rightarrow x=7\)

\(\dfrac{y}{21}=\dfrac{4}{7}\Rightarrow\dfrac{y}{21}=\dfrac{12}{21}\Rightarrow y=12\)

vì 4/x=28/49 nên x.28=4.49=196 ( dấu . là dấu nhân , vì sao x.28=4.49 cấp 2 học )

x=196:28=7

vì y/21=28/49 nên y.49=21.28=588

y=588:49=12

\(\frac{4}{x}=\frac{y}{21}=\frac{28}{49}\)

\(\Rightarrow\frac{4}{x}=\frac{12}{21}=\frac{28}{49}\)

\(\Rightarrow\frac{4}{7}=\frac{12}{21}=\frac{28}{49}\)

Vậy x = 7 , y = 12

Cách làm:

y = \(\frac{28\times21}{49}=12\)

x = \(\frac{21\times4}{12}=7\)

Công thức : nhân chéo

#Hok tốt !

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