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a. x2 + 6x + 9 = (x + 3)2
b. 25 + 10x + x2 = (5 + x)2
c. x2 + 8x + 16 = (x + 4)2
d. x2 + 14x + 49 = (x + 7)2
e. 4x2 + 12x + 9 = (2x + 3)2
f. 9x2 + 12x + 4 = (3x + 2)2
h. 16x2 + 8 + 1 = (4x + 1)2
i. 4x2 + 12xy + 9y2 = (2x + 3y)2
k. 25x2 + 20xy + 4y2 = (5x + 2y)2
a) \(=\left(x+3\right)^2\)
b) \(=\left(x+5\right)^2\)
c) \(=\left(x+4\right)^2\)
d) \(=\left(x+7\right)^2\)
e) \(=\left(2x+3\right)^2\)
f) \(=\left(3x+2\right)^2\)
h) \(=\left(4x+1\right)^2\)
i) \(=\left(2x+3y\right)^2\)
k) \(=\left(5x+2y\right)^2\)
a) \(x^2+2x+1=\left(x+1\right)^2\)
b) \(x^2+8x+16=\left(x+4\right)^2\)
c) \(x^2+6x+9=\left(x+3\right)^2\)
d) \(4x^2+4x+1=\left(2x+1\right)^2\)
e) \(36+x^2-12x=x^2-12x+36=\left(x-6\right)^2\)
f) \(4x^2+12x+9=\left(2x+3\right)^2\)
g) \(x^4+81+18x^2=x^4+18x^2+81=\left(x^2+9\right)^2\)
h) \(9x^2+30xy+25y^2=\left(3x+5y\right)^2\)
a, \(x^2\) + 2\(x\) + 1 = (\(x\) + 1)2
b, \(x^2\) + 8\(x\) + 16 = (\(x\) + 4)2
c, \(x^2\) + 6\(x\) + 9 = (\(x\) + 3)2
d, 4\(x^2\) + 4\(x\) + 1 = (2\(x\) + 1)2
1, \(\frac{x^2+2x+1}{2x^2-2}=\frac{\left(x+1\right)^2}{2\left(x^2-1\right)}=\frac{\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}=\frac{x+1}{2\left(x-1\right)}\)= \(\frac{x+1}{2x-2}\)
2 \(\frac{x^2-6x+9}{5x^2-45}=\frac{\left(x-3\right)^2}{5\left(x^2-9\right)}=\frac{\left(x-3\right)^2}{5\left(x-3\right)\left(x+3\right)}=\frac{x-3}{5x+15}\)
3 \(\frac{x^2-12x+36}{2x^2-4x}=\frac{\left(x-6\right)^2}{2x\left(x-2\right)}\)
4 \(\frac{x^2-10x+25}{2x^2-50}=\frac{\left(x-5\right)^2}{2\left(x^2-25\right)}=\frac{\left(x-5\right)^2}{2\left(x-5\right)\left(x+5\right)}=\frac{x-5}{2x+10}\)
Bài làm:
a) \(x^6-6x^4+12x^2-8\)
\(=\left(x^2-2\right)^3\)
b) \(x^2+16-8x=\left(x-4\right)^2\)
c) \(10x-x^2-25=-\left(x-5\right)^2\)
d) \(9\left(a-b\right)^2-4\left(x-y\right)^2\)
\(=\left[3\left(a-b\right)\right]^2-\left[2\left(x-y\right)\right]^2\)
\(=\left(3a-3b-2x+2y\right)\left(3a-3b+2x-2y\right)\)
e) \(\left(x+y\right)^2-2xy+1\)
\(=x^2+2xy+y^2-2xy+1\)
\(=x^2+y^2+1\)
sai sai
a. \(x^6-6x^4+12x^2-8=\left(x^2\right)^3-3\left(x^2\right)^2.2+3x^22-2^3=\left(x^2-2\right)^3\)
b. \(x^2+16-8x=x^2-8x+4^2=\left(x-4\right)^2\)
c. \(10x-x^2-25=10x-x^2-5^2=-\left(x-5\right)^2\)
d. \(9\left(a-b\right)^2-4\left(x-y\right)^2=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\)
\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)=\left(x-5y\right)\left(5x-y\right)\)
e. \(\left(x+y\right)^2-2xy+1=x^2+2xy+y^2-2xy+1=x\left(x+2y\right)-y\left(y+2x\right)+2y^2+1\)
\(=x\left(x+y\right)-y\left(y+x\right)+xy-yx+2y^2+x=\left(x-y\right)\left(x+y\right)+2y^2+x\)
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
Câu trên mình trả lời rồi nha
đề là gì bạn có phải như mình làm ko
\(x^2+6x+9=\left(x+3\right)^2\)
\(x^2+8x+16=\left(x+4\right)^2\)
\(x^2+10x+25=\left(x+5\right)^2\)
\(x^2-12+36=\left(x-6\right)^2\)
\(x^2-14x+49=\left(x-7\right)^2\)