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a) \(4x^2-12x+9=\left(2x\right)^2-2.2x.3+3^2=\left(2x-3\right)^2\)
b) \(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
c) \(1+12x+36x^2=1^2+2.6x.1+\left(6x\right)^2=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2=\left(3x-4y\right)^2\)
f) \(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
g) \(-16a^4b^6-24a^5b^5-9a^6b^4=-\left(16a^4b^6+24a^5b^5+9a^6b^4\right)\)
\(=-\left[\left(4a^2b^3\right)^2+2.4a^2b^3.3a^3b^2+\left(3a^3b^2\right)^2\right]\)
\(=-\left(4a^2b^3+3a^3b^2\right)^2\)
h) \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\) \(=\left(5x-2y\right)^2\)
i) \(25x^4-10x^2y+y^2=\left(5x^2\right)^2-2.5x^2.y+y^2=\left(5x^2-y\right)^2\)
a) Ta có: \(\left(4x^2-12x+9\right)-1\)
\(=\left(2x-3\right)^2-1^2\)
\(=\left(2x-3-1\right)\left(2x-3+1\right)\)
\(=\left(2x-4\right)\left(2x-2\right)\)
\(=4\left(x-2\right)\left(x-1\right)\)
b) Ta có: \(\left(\frac{x^2}{4}+2xy+4y^2\right)-25\)
\(=\left[\left(\frac{x}{2}\right)^2+2\cdot\frac{x}{2}\cdot2y+\left(2y\right)^2\right]-5^2\)
\(=\left(\frac{x}{2}+2y\right)^2-5^2\)
\(=\left(\frac{x}{2}+2y-5\right)\left(\frac{x}{2}+2y+5\right)\)
c) Ta có: \(1+12x+35x^2\)
\(=35x^2+12x+1\)
\(=35x^2+5x+7x+1\)
\(=5x\left(7x+1\right)+\left(7x+1\right)\)
\(=\left(7x+1\right)\left(5x+1\right)\)
d) Ta có: \(9x^2-24xy+15y^2\)
\(=9x^2-9xy-15xy+15y^2\)
\(=9x\left(x-y\right)-15y\left(x-y\right)\)
\(=\left(x-y\right)\left(9x-15y\right)\)
\(=3\left(x-y\right)\left(3x-5y\right)\)
e) Ta có: \(25x^2-20xy+3y^2\)
\(=25x^2-15xy-5xy+3y^2\)
\(=5x\left(5x-3y\right)-y\left(5x-3y\right)\)
\(=\left(5x-3y\right)\left(5x-y\right)\)
f) Ta có: \(24x^4-10x^2y+y^2\)
\(=24x^4-4x^2y-6x^2y+y^2\)
\(=4x^2\left(6x^2-y\right)-y\left(6x^2-y\right)\)
\(=\left(6x^2-y\right)\left(4x^2-y\right)\)
câu đầu mình nghĩ là 4x2 mới đúng chắc bạn viết thiếu
4x2-12x+9=(2x-3)2
4x2+4x+1=(2x+1)2
1+12x+36x2=(6x+1)2
9x2-24xy+16y2=(3x-4y)2
x2/4+2xy+4y2=(x/2 + 2y)2
-x2+10x2y+y2? câu này bạn có úp sai đề không ta ?
a)Ta có : 9(a + b)2 - 4(a - 2b)2
= [3(a + b) - 2(a - 2b)].[3(a + b) + 2(a - 2b)]
= (3a + 3b - 2a + 4b)(3a + 3b + 2a - 4b)
= (a + 7b)(5a - b)
1.
\(x^2-22x+12\) : biểu thức không phân tích được thành nhân tử nữa.
2.
\(9x^2+6x+1=(3x)^2+2.3x.1+1^2=(3x+1)^2\)
3.
\(x^2-10x+2\): không p. tích được thành nhân tử.
4.
\(x^3+1=x^3+1^3=(x+1)(x^2-x+1)\)
5.
\(8x^3-27y^3=(2x)^3-(3y)^3=(2x-3y)[(2x)^2+(2x)(3y)+(3y)^2]\)
\(=(2x-3y)(4x^2+6xy+9y^2)\)
6.
\((x+3y)^2-(3y+1)^2=[(x+3y)-(3y+1)][(x+3y)+(3y+1)]\)
\(=(x-1)(x+6y+1)\)
7.
\(4y^2-36x^2=(2y)^2-(6x)^2=(2y-6x)(2y+6x)=4(y-3x)(y+3x)\)
8.
\(27-(x+4)^3=3^3-(x+4)^3=[3-(x+4)][3^2+3(x+4)+(x+4)^2]\)
\(=-(x+1)(37+x^2+11x)\)
9.
\(25x^2-10xy+y^2=(5x)^2-2.5x.y+y^2=(5x-y)^2\)
10.
\(9x^6-12x^7+4x^8=x^6(9-12x+4x^2)=x^6[3^2-2.3.2x+(2x)^2]\)
\(=x^6(3-2x)^2\)
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
toàn hằng đẳng thức (1) và (2) thôi mà bạn, đọc SGK 8 tập 1 là hiểu ngay. Có gì khó hiểu hỏi nhé!
a, x2-6x +9 = (x-3)2
b, 4x2+4x +1 = (2x)2+2.2x.1 +12=(2x+1)2
c, 9x2 -12x +4 = (3x-2)2
d, 25x2 -10x +1= (5x -1)2
e, x4-4x2+4 = (x2 -2)2
f, x2 +8x +16 = (x+4)2
Bài 1:
a) 25\(x^2\) - 0,09
= \(\left(5x\right)^2-0,3^2\)
= (5x - 0,3) (5x +0,3)
Bài 5:
a: \(=\left(2x-3\right)^2\)
b: \(=\left(2x+1\right)^2\)
c: \(=\left(6x+1\right)^2\)
d: \(=\left(3x-4y\right)^2\)
e: \(=\left(\dfrac{1}{2}x-2y\right)^2\)
f: \(=-\left(x-5\right)^2\)
1. \(\left(5x-y\right)^2\)
2. \(\left(x^2-2y\right)^2\)
3. \(\left(3x^2-2y\right)^2\)
4. \(\left(9^2-2x^3\right)^2\)
1/ \(25x^2-10xy+y^2=\left(5x\right)^2-2.5xy+y^2=\left(5x-y\right)^2\)
2/\(x^4-4x^2y+4y^2=\left(x^2\right)^2-2.x^2.2y+\left(2y\right)^2=\left(x^2-2y\right)^2\)
3/\(9x^4-12x^2y+4y^2=\left(3x^2\right)^2-2.3x^2.2y+4y^2=\left(3x^2-2y\right)^2\)
4/\(9x^4-12x^5+4x^6=\left(3x^2\right)^2-2.3x^2.2x^3+\left(2x^3\right)^2=\left(3x^2-2x^3\right)^2\)
\(4x^2+4x+1\)
\(=\left(2x\right)^2+2.2x.1+1\)
\(=\left(2x+1\right)^2\)
\(1+12x+36x^2\)
\(=1+2.6x+\left(6x\right)^2\)
\(=\left(1+6x\right)^2\)