(25x+4x)=21x(45x-12x)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2x2 - 98 = 0
2x2 = 0 + 98
2x2 = 98
x2 = 98 : 2
x2 = 49
x = \(\sqrt{49}\)
=> x = 7
Ta có : 2x2 - 98 = 0
=> 2(x2 - 49) = 0
Mà : 2 > 0
Nên x2 - 49 = 0
=> x2 = 49
=> x2 = -7;7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4x^2+4x+1\)
\(=\left(2x\right)^2+2.2x.1+1\)
\(=\left(2x+1\right)^2\)
\(1+12x+36x^2\)
\(=1+2.6x+\left(6x\right)^2\)
\(=\left(1+6x\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4x^2+12x+9=\left(2x+3\right)^2\)
\(25x^2+10x+5=\left(5x+1\right)^2+4\)( bạn xem lại đề )
\(x^2+6x+9x^2=10x^2+6x=2x\left(5x+3\right)\)
\(25x^2+10xy+y^2=\left(5x+y\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B: rút gọn
a) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-6x^2+12x\)
\(=x^3-6x^2+12x-8\)
\(=\left(x-2\right)^3\)
b) Ta có: \(\left(2x+5\right)\left(5-2x\right)+\left(x-5\right)\left(4x+5\right)\)
\(=25-4x^2+4x^2+5x-20x-25\)
=-15x
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(=\left(5x+2y\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(36-12x+x^2\) \(=6^2-2.6.x+x^2\)
\(=\left(6-x\right)^2\)
b) \(4x^2+12x+9=\left(2x\right)^2+2.2x.3+3^2\)
\(=\left(2x+3\right)^2\)
c) \(-25x^6-y^8+10x^3y^4=-\left[25x^6-10x^3y^4+y^8\right]\)
\(=-\left[\left(5x^3\right)^2-2.5x^3.y^4+\left(y^4\right)^2\right]\)
\(=-\left(5x^3-y^4\right)^2\)
d) \(\dfrac{1}{4}x^2-5xy+25y^2=\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.5y+\left(5y\right)^2\)
\(=\left(\dfrac{1}{2}x-5y\right)^2\)
Học tốt~~~