Tìm các số tự nhiên: (x2 + 4y2 + 28) = 17(x4 + y4 + 14y2 + 49)
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![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(y^4-14y^2+49=\left(y^2-7\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(125x^3-64y^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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, \(x^3+5x^2+6x+2\)
\(=x^3+x^2+4x^2+4x+2x+2\)
\(=x^2\left(x+1\right)+4x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+4x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: $2y-x=5; 2y+x=11\Rightarrow y=4; x=3$
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
Tìm các số tự nhiên x và y biết : 4/x = y/21=28/49
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}\\ \Rightarrow\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{4}{7}\)
\(\dfrac{4}{x}=\dfrac{4}{7}\Rightarrow x=7\)
\(\dfrac{y}{21}=\dfrac{4}{7}\Rightarrow\dfrac{y}{21}=\dfrac{12}{21}\Rightarrow y=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 !