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Ta có
( a 2 + 9 ) 2 – 36 a 2 = ( a 2 + 9 ) 2 – ( 6 a ) 2 = ( a 2 + 9 + 6 a ) ( a 2 + 9 – 6 a ) = ( a + 3 ) 2 ( a – 3 ) 2
Đáp án cần chọn là: A
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\(a,Sửa:25x^2-20xy+4y^2=\left(5x-2y\right)^2\\ b,=\dfrac{1}{4}\left(\dfrac{1}{9}a^2-b^2\right)=\dfrac{1}{4}\left(\dfrac{1}{3}a-b\right)\left(\dfrac{1}{3}a+b\right)\\ c,=\dfrac{1}{8}\left(a+2\right)^3-1=\left[\dfrac{1}{2}\left(a+2\right)\right]^3-1=\left[\dfrac{1}{2}a+1\right]^3-1\\ =\left(\dfrac{1}{2}a+1-1\right)\left(\dfrac{1}{4}a^2+a+1+\dfrac{1}{2}a+1+1\right)\\ =\dfrac{1}{2}a\left(\dfrac{1}{4}a^2+\dfrac{3}{2}a+3\right)\\ d,=\left(x^3-1\right)\left(x^3+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(50x^5-8x^3\)
\(=2x^3\left(25x^2-4\right)\)
\(=2x^3\left(5x-2\right)\left(5x+2\right)\)
b: \(x^4-5x^2-4y^2+10y\)
\(=\left(x^2-2y\right)\left(x^2+2y\right)-5\left(x^2-2y\right)\)
\(=\left(x^2-2y\right)\left(x^2+2y-5\right)\)
c: \(36a^2+12a+1-b^2\)
\(=\left(6a+1\right)^2-b^2\)
\(=\left(6a+1-b\right)\left(6a+1+b\right)\)
d: \(x^3+y^3-xy^2-x^2y\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+y\right)\cdot\left(x-y\right)^2\)
e: Ta có: \(4x^2+4x-3\)
\(=4x^2+6x-2x-3\)
\(=2x\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(2x-1\right)\)
f: Ta có: \(9x^4+16x^2-4\)
\(=9x^4+18x^2-2x^2-4\)
\(=9x^2\left(x^2+2\right)-2\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(9x^2-2\right)\)
g: Ta có: \(-6x^2+5xy+4y^2\)
\(=-6x^2+8xy-3xy+4y^2\)
\(=-2x\left(3x-4y\right)-y\left(3x-4y\right)\)
\(=\left(3x-4y\right)\left(-2x-y\right)\)
h: Ta có: \(\left(x^2+4x\right)^2+8\left(x^2+4x\right)+15\)
\(=\left(x^2+4x\right)^2+3\left(x^2+4x\right)+5\left(x^2+4x\right)+15\)
\(=\left(x^2+4x+3\right)\cdot\left(x^2+4x+5\right)\)
\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Phân tích a 2 – 6ab + 9 b 2 = ( a – 3 b ) 2 ; thực hiện phép chia được kết quả a – 3b.
b) Phân tích a 3 + 9 a 2 b + 27a b 2 – 27 b 3 = ( a – 3 b ) 3 ; thực hiện phép chia được kết quả a – 3b.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(25a^2-1=\left(5a-1\right)\left(5a+1\right)\)
b) \(a^2-9=\left(a-3\right)\left(a+3\right)\)
c) \(\dfrac{1}{4}a^2-\dfrac{9}{25}=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)\)
d) \(\dfrac{9}{4}a^4-\dfrac{16}{25}=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\)
e) \(\left(2a+b\right)^2-a^2=\left(2a+b-a\right)\left(2a+b+a\right)=\left(a+b\right)\left(3a+b\right)\)
f) \(16\left(x-1\right)^2-25\left(x+y\right)^2=\left(4x-4-5x-5y\right)\left(4x-4+5x+5y\right)=\left(-x-4-5y\right)\left(9x+5y-4\right)\)
a/ $25x^2-1\\=(5x)^2-1^2\\=(5x-1)(5x+1)$
b/ $a^2-9\\=a^2-3^2\\=(a-3)(a+3)$
c/ $\dfrac{1}{4}a^2-\dfrac{9}{25}\\=\left(\dfrac{1}{2}a\right)^2-\left(\dfrac{3}{5}\right)^2\\=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)$
d/ $\dfrac{9}{4}a^4-\dfrac{16}{25}\\=\left(\dfrac{3}{2}a^2\right)^2-\left(\dfrac{4}{5}\right)^2\\=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left[\left(\sqrt{\dfrac 3 2}a\right)^2-\left(\dfrac{2\sqrt 5}{5}\right)^2\right]\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left(\sqrt{\dfrac 3 2}a-\dfrac{2\sqrt 5}{5}\right)\left(\sqrt{\dfrac 3 2}a+\dfrac{2\sqrt 5}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)$
e/ $(2a+b)^2-a^2\\=(2a+b-a)(2a+b+a)\\=(a+b)(3a+b)$
f/ $16(x-1)^2-25(x+y)^2\\=[4(x-1)]^2-[5(x-y)]^2\\=[4(x-1)-5(x-y)][4(x-1)+5(x-y)]\\=[4x-4-5x+5y][4x-4+5x-5y]\\=(-x+5y-4)(9x-5y-4)$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (a - 2b)x(a + 2b)
b) x2-(y-3)2
=> (x-y+3)(x+y-3)
c) (2a + b - a)(2a + b + a)
=> (a+b)(3a+b)
d) (4(x - 1))2 - (5(x + y))2
⇔ (4x - 4 - 5x - 5y)(4x - 4 + 5x + 5y)
⇔ -(x + 5y + 4)(9x + 5y + -4)
e) (x + 5)2
f) (5x - 2y)2
h) (x - 5)(x2 + 5x + 25)
k) (x + 5)3
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\(a,=\left(3x-5\right)\left(3x+3\right)=3\left(x+1\right)\left(3x-5\right)\\ b,=\left(5x-4-7x\right)\left(5x-4+7x\right)=\left(-2x-4\right)\left(12x-4\right)\\ =-8\left(x+2\right)\left(x-3\right)\\ c,=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\\ =\left(x+14\right)\left(3x-4\right)\\ d,=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\\ =\left(x+5\right)\left(5x-3\right)\\ e,=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\\ =\left(4x+7\right)\left(8x+11\right)\\ f,=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\\ =\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\\ =\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\\ g,=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\\ =\left(a-b\right)\left(x-y\right)\left(a+b\right)\left(x+y\right)\)
\(h,=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\\ =\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\\ =\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
a: \(\left(3x-1\right)^2-16\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x+3\right)\left(3x-5\right)\)
\(=3\left(x+1\right)\left(3x-5\right)\)
b: \(\left(5x-4\right)^2-49x^2\)
\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)
\(=\left(-2x-4\right)\left(12x-4\right)\)
\(=-8\left(x+2\right)\left(3x-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(a^2-1\right)^3-\left(a^4+a^2+1\right)\left(a^2-1\right)\)
\(=a^6-3a^4+3a^2-1-\left(a^6-1\right)\)
\(=-3a^4+3a^2\)
b: Ta có: \(\left(a^4-3a^2+9\right)\left(a^2+3\right)-\left(a^2+3\right)^3\)
\(=a^6+27-a^6-9a^4-27a^2-27\)
\(=-9a^4-27a^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,a^2+b^2=\left(a+b\right)^2-2ab=9^2-2\cdot20=41\\ b,a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=41^2-2\left(ab\right)^2\\ =1681-2\cdot400=881\\ c,\left(a-b\right)^2=a^2+b^2-2ab=41-2\cdot20=1\\ \Rightarrow a-b=1\\ \Rightarrow C=a^2-b^2=\left(a-b\right)\left(a+b\right)=9\cdot1=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^2-9+6x-x^2=a^2-\left(x^2-6x+9\right)=a^2-\left(x-3\right)^2=\left(a-x+3\right)\left(a+x-3\right)\)
a 2 − 9 + 6 x − x 2
= a 2 − ( x 2 − 6 x + 9 )
= a 2 − ( x − 3 ) 2
= ( a − x + 3 ) ( a + x − 3 )
\(\left(a^2+9\right)^2-36a^2\)
\(=\left(a^2+9+6a\right)\left(a^2+9-6a\right)\)
\(=\left(a+3\right)^2.\left(a-3\right)^2\)
p/s:chúc bạn hk tốt
\(\left(a^2+9\right)^2-36a^2\)
\(=\left(a^2+9+36a^2\right)\cdot\left(a^2+9-36a^2\right)\)
\(=\left(37a^2+9\right)\cdot\left(-35a^2+9\right)\)