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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) \(x^6-y^6\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
a, bằng cách tìm nhân tử chung
1,\(x^2-3x\)
=x.(\(\left(x-3\right)\)
2,\(15x^2-6x\)
=3x.(5x-2)
3,\(4x\left(x-y\right)\)\(+2y\left(x-y\right)\)
=(x-y).(4x+2y)
=2(x-y).(x+y)
=2(\(x^2-y^2\left(\right)\)
b, dùng hằng đẳng thức
1,\(64x^2-25y^2\)
=\(\left(8x\right)^2-\left(5y\right)^2\)
=(8x-5y)(8x+5y)
2,\(9x^2-30x-25\)
=\(\left(3x-5\right)^2\)
3,
\(\dfrac{1}{4}x^2+2x+4\)
=\(\left(\dfrac{1}{2}x+2\right)^2\)
4,\(25a^2-2a+\dfrac{1}{25}\)
=(\(\left(5a-\dfrac{1}{5}\right)^2\)
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) \(25a^2-\dfrac{1}{4}b^2\)
= \(\left(5a-\dfrac{1}{2}b\right)\left(5a+\dfrac{1}{2}b\right)\)
1) 196a2 - 4b2 = (14a)2 - (2b)2 = (14a + 2b)(14a - 2b)
2) 25a2 - 49b2 = (5a)2 - (7b)2 = (5a + 7b)(5a - 7b)
3) x2 + 10x + 25 = (x + 5)2
1) \(196a^2-4b^2=4\left(49a^2-b^2\right)=4\left(7a-b\right)\left(7a+b\right)\)
2) \(25a^2-49b^4=\left(5a\right)^2-\left(49b^2\right)^2=\left(5a-49b^2\right)\left(5a+49b^2\right)\)
3) \(x^2+10x+25=x^2+5x+5x+25=x\left(x+5\right)+5\left(x+5\right)=\left(x+5\right)^2\)
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
\(1,a^2-4b^2=\left(a-2b\right)\left(a+2b\right)\)
\(2,4a^2-b^2=\left(2a-b\right)\left(2a+b\right)\)
\(3,a^2-25=\left(a-5\right)\left(a+5\right)\)
\(4,25a^2-\dfrac{1}{2}=\left(5a\right)^2-\left(\dfrac{1}{\sqrt{2}}\right)^2=\left(5a-\dfrac{1}{\sqrt{2}}\right)\left(5a+\dfrac{1}{\sqrt{2}}\right)\)
\(5,x^2+10x+25=x^2+2.x.5+5^2=\left(x+5\right)^2\)
1,\(a^2-4b^2=\left(a-2b\right)\left(a+2b\right)\)
2, \(4a^2-b^2=\left(2a-b\right)\left(2a+b\right)\)
3, \(a^2-25=a^2-5^2=\left(a-5\right)\left(a+5\right)\)
4, \(25a^2-\dfrac{1}{2}=\left(5a\right)^2-\sqrt{\left(\dfrac{1}{2}\right)^2}=\left(5a-\sqrt{\dfrac{1}{2}}\right)\left(5a+\sqrt{\dfrac{1}{2}}\right)\)
5, \(x^2+10x+25=x^2+2.x.5+5^2=\left(x+5\right)^2\)