a , 4*2 + 4*5 +4 = 4* 2 + 4*5 + 4*1  = 4* ( 2+5+1 ) = 4*8 

b, a*7 +a*2 +a= a*7 + a*2 + a*1 = a* ( 7 + 2+1 ) = a* 10

a) 4 x 8

b) 10a

@Doris (Team ASL) 

bn lm nhanh ghê rứa? :v

a, 4⁸.8⁴

= (2²)⁸.(2³)⁴

= 2¹⁶.2¹²

= 2²⁸

b, 4¹⁵.5¹⁵

= (4.5)¹⁵

= 20¹⁵

c, 2¹⁰.15 + 2¹⁰.85

= 2¹⁰.(15 + 85)

= 2¹⁰.100

=2¹⁰.(2.5)²

= 2¹².5²

d, 3³.9²

= 3³ . (3²)²

= 3³.3⁴

= 3⁷

e, 5¹².7 - 5¹¹.10

= 5¹¹.(5.7 - 10)

= 5¹¹.25

=5¹¹.5²

=5¹³

f, \(x^1\).\(x^2\).\(x^3\)....\(x^{100}\)

= \(x^{1+2+3+...+100}\)

= \(x^{\left(1+100\right).100:2}\)

= \(x^{5050}\)

❤Em cảm ơn cô Nguyễn Thương hoài rất nhiều a!❤

a, 48.84

= (22)8.(23)4

= 216.212

= 228

b, 415.515

= (4.5)15

= 2015

c, 210.15 + 210.85

= 210.(15 + 85)

= 210.100

=210.(2.5)2

= 212.52

d, 33.92

= 33 . (32)2

= 33.34

= 37

e, 512.7 - 511.10

= 511.(5.7 - 10)

= 511.25

=511.52

=513

f, �1x1.�2x2.�3x3....�100x100

= �1+2+3+...+100x1+2+3+...+100

= �(1+100).100:2x(1+100).100:2

= �5050x5050

1a/ z² - 6z + 5 - t² - 4t = z² - 2 . 3z + 3² - 4 - t² - 4t = (z² - 2 . 3z + 3²) - (2² + 2 . 2t + t²) = (z - 3)² - (2 + t)²

b/ x² - 2xy + 2y² + 2y² + 1 = x² - 2xy + y² + y² + 2y + 1 = (x² - 2xy + y²) + (y² + 2y + 1) = (x - y)² + (y + 1)²

c/ 4x² - 12x - y² + 2y + 8 = (2x)² - 12x - y² + 2y + 3² - 1 = [ (2x)² - 2 . 3 . 2x + 3² ] - (y² - 2y + 1) = (2x - 3)² - (y - 1)²

2a/ (x + y + 4)(x + y - 4) = x² + xy - 4x + xy + y² - 4y + 4x + 4y + 16 = x² + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y² + 16

= x² + 2xy + y² + 4² = (x + y)² + 4²

b/ (x - y + 6)(x + y - 6) = x² + xy - 6x - xy - y² + 6y + 6x + 6y - 36 = x² + (xy - xy) + (-6x + 6x) + (6y + 6y) - y² - 36

= x² - y² + 12y - 6² = x² - (y² - 12y + 6²) = x² - (y² - 2 . 6y + 6²) = x² - (y - 6)²

c/ (y + 2z - 3)(y - 2z - 3) = y² -2yz - 3y + 2yz - 4z² - 6z - 3y + 6z + 9 = y² + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z² + 9

= y² - 6y - 4z² + 9 = (y² - 6y + 9) - 4z² = (y - 3)² - (2z)²

d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x² + 4y² + 6yz - 2xy + 6yz + 9z² - 3xz = (2xy - 2xy) + (3xz - 3xz) - x² + (6yz + 6yz) + 9z² + 4y²

= -x² + 4y² + 12yz + 9z² = (4y² + 12yz + 9z²) - x² = [ (2y)² + 2 . 2 . 3yz + (3z)² ] - x² = (2y + 3z)² - x²

:v dễ mà có trong nâng cao mới hc qua :3

a, x2+10x+26+y2+2y

=(x2+2.x.5+52)+(12+2.1.y+y2)

=(x+5)2+(y+1)2

b, x2−2xy+2y2+2y+1

=x2−2xy+y2+y2+2y+1

=(x2−2.x.y+y2)+(y2+2.y.1+12)

=(x−y)2+(y+1)2

c,z2−6z+5−t2−4t

=−(t2+4t−z2+6z−5)

=−(t2+2.t.2+22−z2+2.z.3−32)

=−((t2+2.t.2+22)−(z2−2.z.3+32))

=−((t+2)2−(z−3)2)

=(z−3)2−(t+2)2

Để thu gọn và sắp xếp các hạng tử của mỗi đa thức, ta cần thực hiện các bước sau:
Đối với đa thức P(x): P(x) = (4x + 1 - x^2 + 2x^3) - (x^4 + 3x - x^3 - 2x^2 - 5) = 4x + 1 - x^2 + 2x^3 - x^4 - 3x + x^3 + 2x^2 + 5 = -x^4 + 3x^3 + x^2 + x + 6
Đối với đa thức Q(x): Q(x) = 3x^4 + 2x^5 - 3x - 5x^4 - x^5 + x + 2x^5 - 1 = 2x^5 - x^5 + 3x^4 - 5x^4 + x - 3x - 1 = x^5 - 2x^4 - 2x - 1
Sau khi thu gọn và sắp xếp các hạng tử, ta có: P(x) = -x^4 + 3x^3 + x^2 + x + 6 Q(x) = x^5 - 2x^4 - 2x - 1

a: \(P\left(x\right)=\left(4x+1-x^2+2x^3\right)-\left(x^4+3x-x^3-2x^2-5\right)\)

\(=4x+1-x^2+2x^3-x^4-3x+x^3+2x^2+5\)

\(=-x^4+3x^3+x^2+x+6\)

\(Q\left(x\right)=3x^4+2x^5-3x-5x^4-x^5+x+2x^5-1\)

\(=\left(2x^5-x^5+2x^5\right)+\left(3x^4-5x^4\right)+\left(-3x+x\right)-1\)

\(=-x^5-2x^4-2x-1\)

b: Bạn ghi lại đề đi bạn

\(\begin{array}{l}a)A = 3x - 4{x^4} + {x^3}\\ =  - 4{x^4} + {x^3} + 3x\\b)B =  - 2{x^3} - 5{x^2} + 2{x^3} + 4x + {x^2} - 5\\ = ( - 2{x^3} + 2{x^3}) + \left( { - 5{x^2} + {x^2}} \right) + 4x - 5\\ = 0 + ( - 4{x^2}) + 4x - 5\\ =  - 4{x^2} + 4x - 5\\c)C = {x^5} - \dfrac{1}{2}{x^3} + \dfrac{3}{4}x - {x^5} + 6{x^2} - 2\\ = \left( {{x^5} - {x^5}} \right) - \dfrac{1}{2}{x^3} + 6{x^2} + \dfrac{3}{4}x - 2\\ =  - \dfrac{1}{2}{x^3} + 6{x^2} + \dfrac{3}{4}x - 2\end{array}\)

Bài 1:

a) \(x^2+10x+26+y^2+2y=(x^2+10x+25)+(y^2+2y+1)\)

..................................................= \(\left(x+5\right)^2+\left(y+1\right)^2\)

b) \(z^2-6z+5-t^2-4t=(z^2-6t+9)-(t^2+4t+4)\)

............................................= \(\left(z-3\right)^2-\left(t+2\right)^2\)

c) \(x^2-2xy+2y^2+2y+1=(x^2-2xy+y^2)+(y^2+2y+1)\)

..................................................= \(\left(x-y\right)^2+\left(y+1\right)^2\)

d) \(4x^2-12x-y^2+2y+8=\left(4x^2-12x+9\right)-\left(y^2-2y+1\right)\)

.................................................= \(\left(2x-3\right)^2-\left(y-1\right)^2\)

Bài 2:

a) \(\left(x+y+4\right)\left(x+y-4\right)=\left(x+y\right)^2-16\)

b) \(\left(x-y+6\right)\left(x+y-6\right)=x^2-\left(y-6\right)^2\)

c) \(\left(y+2z-3\right)\left(y-2z+3\right)=y^2-\left(2z-3\right)^2\)

d) \(\left(x+2y+3z\right)\left(2y+3z-x\right)=\left(2y+3z\right)^2-x^2\)

a: =3^4*3^9=3^13

b: =2^10:2^7*2^20*2^30=2^53

c: =2^12*2^10*2^30=2^52

a) \(9^2\cdot27^3\)

\(=\left(3^2\right)^2\cdot\left(3^3\right)^3\)

\(=3^4\cdot3^9\)

\(=3^{4+9}\)

\(=3^{13}\)

b) \(4^5:2^7\cdot4^{10}\cdot2^{30}\)

\(=\left(2^2\right)^5:2^7\cdot\left(2^2\right)^{10}\cdot2^{30}\)

\(=2^{10}:2^7\cdot2^{20}\cdot2^{30}\)

\(=2^3\cdot2^{20}\cdot2^{30}\)

\(=2^{3+20+30}\)

\(=2^{53}\)

c) \(64^2\cdot4^5\cdot2^{30}\)

\(=\left(2^6\right)^2\cdot\left(2^2\right)^5\cdot2^{30}\)

\(=2^{12}\cdot2^{10}\cdot2^{30}\)

\(=2^{12+10+30}\)

\(=2^{52}\)

1, a+ax2+ax3+ax4

=a x 1+ax2+ax3+ax4

= a x (1+2+3+4) = a x 10

2, b x 5 + b x 4 + b x 6

= b x (5+4+6) = b x 15 

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