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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) \(25x^4-10x^2y+y^2=\left(5x^2\right)^2-2.5x^2.y+y^2=\left(5x^2-y\right)^2\)
1) \(x^6+1\)
\(=x^6+x^4-x^4+x^2-x^2+1\)
\(=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)
\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
2) \(x^6-y^6\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
3)(9a)2-(5a-3b)2
= (9a-5a+3b)(9a+5a-3b)
= (4a+3b)(14a-3b)
4/ a/ Ta có \(x^2-2xy+y^2+a^2=\left(x-y\right)^2+a^2\)
Mà \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\a^2\ge0\end{cases}}\)=> \(\left(x-y\right)^2+a^2\ge0\)
=> \(x^2-2xy+y^2+a^2\ge0\)
Vậy \(x^2-2xy+y^2\)chỉ nhận những giá trị không âm.
b/ Ta có \(x^2+2xy+2y^2+2y+1=\left(x^2+2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x+y\right)^2+\left(y+1\right)^2\)
Mà \(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}}\)=> \(\left(x+y\right)^2+\left(y+1\right)^2\ge0\)
=> \(x^2+2xy+2y^2+2y+1\ge0\)
Vậy \(x^2+2xy+2y^2+2y+1\)chỉ nhận những giá trị không âm.
c/ Ta có \(9b^2-6b+4c^2+1=\left(3b-1\right)^2+4c^2\)
Mà \(\hept{\begin{cases}\left(3b-1\right)^2\ge0\\4c^2\ge0\end{cases}}\)=> \(\left(3b-1\right)^2+4c^2\ge0\)
=> \(9b^2-6b+4c^2+1\ge0\)
Vậy \(9b^2-6b+4c^2+1\)chỉ nhận những giá trị không âm.
d/ Ta có \(x^2+y^2+2x+6y+10=\left(x+1\right)^2+\left(y+3\right)^2\)
Mà \(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)=> \(\left(x+1\right)^2+\left(y+3\right)^2\ge0\)
=> \(x^2+y^2+2x+6y+10\ge0\)
Vậy \(x^2+y^2+2x+6y+10\)chỉ nhận những giá trị không âm.
1/
a/ \(x^4-y^4=\left(x^2-y^2\right)\)
b/ \(\left(a+b\right)^3-\left(a-b\right)^3=\left(a+b-a+b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left[a^2+2ab+b^2-\left(a^2-b^2\right)+\left(a^2-2ab+b^2\right)\right]\)
\(=2b\left(a^2+b^2\right)\)
c/ \(\left(a^2+2ab+b^2\right)+\left(a+b\right)\)
= \(\left(a+b\right)^2+\left(a+b\right)\)
= \(\left(a+b\right)\left(a+b+1\right)\)
1)\(144a^2-81=\left(12a\right)^2-9^2=\left(12a-9\right)\left(12a+9\right)\)
2)\(a^4-4b^2=a^2-\left(2b\right)^2=\left(a-2b\right)\left(a+2b\right)\)
3)Ko hỉu
4)\(\left(a-5b\right)^2-16b^2=\left(a-5b\right)^2-\left(4b\right)^2=\left(a-9b\right)\left(a-b\right)\)
5)\(9\left(a+b\right)^2-4\left(a-b\right)^2=\left[3\left(a+b\right)+2\left(a-b\right)\right]\left[3\left(a+b\right)-2\left(a-b\right)\right]\)
\(=\left(5a+b\right)\left(a+5b\right)\)
6)\(x^2+12x+36=\left(x+6\right)^2\)
7)Đề sai
8)\(9x^4+24x^2+16=\left(3x^2+4\right)^2\)
9)Chịu
CẢM ƠN NHÌU
1. 8 - 12x + 6x2 - x3
= 23 - 3.22.x + 3.x2.2 - x3
=(2-x)3
2. 125x3 - 75x2 +15x - 1
=(5x)3 - 3.(5x)2.1 + 3.5x.12 - 13
=(5x - 1)3
3, 4 (sai đề)
5. x3 + 2x2 - 6x - 27
=(x3 - 27) + (2x2 - 6x)
=(x3 - 33) + (2x2 - 6x)
=(x -3)(x2 + 3x + 9) + 2x(x-3)
=(x-3)(x2 + 3x +9 +2x)
=(x-3)(x2 + 5x +9)
6. 12x3 + 4x2- 27x -9
=(12x3 + 4x2) - (27x + 9)
=4x2(3x + 1) - 9(3x +1)
=(3x -1)(4x2 -9)
=(3x-1)(2x-3)(2x+3)
a)25a2-49b4
=(5a)2-(7b2)2
=(5a-7b2)(5a+7b2)
b)100a2-9b4
=(10a)2-(3b2)2
=(10a-3b2)(10a+3b2)
c)a4-4b2
=(a2)2-(2b)2
=(a2-2b)(a2+2b)
a) 25a2 - 49b2
= (5a + 7b)(5a - 7b)
b) 100a2 - 9b2
= (10a - 3b)(10a + 3b)
c) a4 - 4b2
= (a2 - 2b)(a2 + 2b)
d) \(\dfrac{4}{9}a^{4^{ }}-\dfrac{25}{4}\)
= \(\left(\dfrac{2}{3}a^2+\dfrac{5}{2}\right)\left(\dfrac{2}{3}a^2-\dfrac{5}{2}\right)\)
e) \(\dfrac{1}{4}a^2-b^2\)
=\(\left(\dfrac{1}{2}a-b\right)\left(\dfrac{1}{2}a+b\right)\)
f) \(\dfrac{1}{4}a^2-\dfrac{1}{9}b^2\)
= \(\left(\dfrac{1}{2}a-\dfrac{1}{3}b\right)\left(\dfrac{1}{2}a+\dfrac{1}{3}b\right)\)
g) \(\dfrac{1}{25}-36x^2\)
= \(\left(\dfrac{1}{5}-6x\right)\left(\dfrac{1}{5}+6x\right)\)
h) \(25a^2-\dfrac{1}{4}b^2\)
= \(\left(5a-\dfrac{1}{2}b\right)\left(5a+\dfrac{1}{2}b\right)\)
a. \(9a^2-1=\left(3a-1\right)\left(3a+1\right)\)
\(b.196a^2-4b^2=\left(14a+2b\right)\left(14a-2b\right)\)
\(c.\dfrac{4}{9}a^4-\dfrac{25}{4}=\left(\dfrac{2}{3}a^2-\dfrac{5}{2}\right)\left(\dfrac{2}{3}a^2+\dfrac{5}{2}\right)\)
\(d.\left(a+3b\right)^2-9b^2=\left(a+3b+3b\right)\left(a+3b-3b\right)\\ =\left(a+6b\right)a\)
\(e.81a^2-\left(5a-3b\right)^2=\left(9a-5a+3b\right)\left(9a+5a-3b\right)\\ =\left(4a+3b\right)\left(14a-3b\right)\)
\(f.4\left(2a-b\right)^2-16\left(a-b\right)^2\\ =\left[2\left(2a-b\right)-4\left(a-b\right)\right]\left[2\left(2a-b\right)+4\left(a-b\right)\right]\\ =\left(4a-2b-4a+4b\right)\left(4a-2b+4a-4b\right)\\ =4b\left(5a-3b\right)\)
\(g.x^4-4x^2y+4y^2=\left(x^2-2y\right)^2\)
\(h.9x^6-12x^7+4x^8=x^6\left(9-12x+4x^2\right)\\ =x^6\left(3-2x\right)^2\)