viết biểu thức sau thành tích 2 thừa số: 1, a+a x 4 +a x 2+a x 3 2, b+b x 2+b x 4
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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
a)5x5+5x3+5x2-10x5
5x(5+3+2-10)
5x1=5
b)(24+6x5+6)-(12+6x3)
6x(4+5+1)-[6x(2+3)]
60-30=30
c)23+39+37+21+34+26
(23+37)+(39+21)+(34+26)
60+60+60=60x3=180
15 x 5 + 3 x 5 + 5 x 2 – 10 x 5
= 5 x (15 + 3 + 2 – 10)
= 5 x 10
= 50
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) \(a^3+4a^2-29a+24=\left(a^3-a^2\right)+\left(5a^2-5a\right)+\left(-24a+24\right)\)
\(=\left(a-1\right)\left(a^2+5a-24\right)=\left(a-1\right)\left(a^2+8a-3a-24\right)=\left(a-1\right)\left(a+8\right)\left(a-3\right)\)
b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
Ta có \(\left(a+b+c\right)^3=a^3+b^3+c^3+3a^2b+3ab^2+3ac^2+3bc^2+3a^2c+3b^2c+6abc\)
\(\Rightarrow\left(a+b+c\right)^3-a^3-b^3-c^3=3a^2b+3ab^2+3ac^2+3bc^2+3a^2c+3b^2c+6abc\)
\(=3\left(a^2b+ab^2\right)+3\left(bc^2+ac^2\right)+3\left(a^2c+abc\right)+3\left(bc^2+abc\right)\)
\(=3\left(a+b\right)\left(ab+bc+ac+bc\right)=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
c) Theo trên ta có
\(a^3+b^3+c^3-3abc=\left(a+b+c\right)^3-3\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+3abc\right)\)
\(=\left(a+b+c\right)^3-3\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ca-3ab-3bc-3ca\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
d) \(x^5+x-1=\left(x^5-x^4+x^3\right)+\left(x^4-x^3+x^2\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)
(24 + 6 x 5 + 6 ) – (12 + 6 x 3)
= (6 x 4 + 6 x 5 + 6 x 1) – (6 x 2 + 6 x 3)
= 6 x (4 + 5 + 1) – 6 x (2 + 3)
= 6 x 10 – 6 x 5
= 6 x (10 – 5)
= 6 x 5
= 30
b: \(3^4\cdot3^5:\dfrac{1}{27}==3^9\cdot3^3=3^{12}\)
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, \(x^1\).\(x^2\).\(x^3\)....\(x^{100}\)
= \(x^{1+2+3+...+100}\)
= \(x^{\left(1+100\right).100:2}\)
= \(x^{5050}\)