C = 48.(52 + 1) (54 + 1) (58 + 1)(516 + 1)(532 + 1)(564 + 1)
E= (x - 2)3- (x + 1).(x2 - x + 1)+6.(x - 1)2
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Ta có: \(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Rightarrow P=\dfrac{24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{5^{32}-1}{2}\)
Ta có:
( 5 2 - 1).P = ( 5 2 – 1).12.( 5 2 + 1)( 5 4 + 1)( 5 8 + 1)( 5 16 + 1)
= 12.( 5 2 – 1).( 5 2 + 1)( 5 4 + 1)( 5 8 + 1)( 5 16 + 1)
= 12.( 5 4 - 1)( 5 4 + 1)( 5 8 + 1)( 5 16 + 1)
= 12.( 5 8 - 1)( 5 8 + 1)( 5 16 + 1)
= 12.( 5 16 - 1)( 5 16 + 1)
= 12.( 5 32 - 1)
Bài 2b cho mình hỏi tận cùng dãy số 3x3x3x.....x là 3 hay 2
Nếu là 3 thì ko tính được
Còn nếu là 2 thì B = 3^2016 x 2 = 81^ 504 x 2 . 81 tận cùng là 1 thì x 504 lần tận cùng cũng là 1 . x 2 thì tận cùng là 2
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)
\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)
\(=\dfrac{1}{2}x^3+x^2-15x-18\)
2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)
\(=4x^3+6x^2-6x^2-9x+10x+15\)
\(=4x^3+x+15\)
3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)
\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)
\(=3x^5-x^4+5x^3+10x^2+26x-5\)
4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=x^3-2x^2-x+2\)
Bài 5:
a. 1 - 2y + y2
= (1 - y)2
b. (x + 1)2 - 25
= (x + 1)2 - 52
= (x + 1 - 5)(x + 1 + 5)
= (x - 4)(x + 6)
c. 1 - 4x2
= 12 - (2x)2
= (1 - 2x)(1 + 2x)
d. 8 - 27x3
= 23 - (3x)3
= (2 - 3x)(4 + 6x + 9x2)
e. (đề hơi khó hiểu ''x3'' !?)
g. x3 + 8y3
= (x + 2y)(x2 - 2xy + y2)
Giải phương trình:
a) x+1 /9 + x+2 /8 = x+3 /7 + x+4 /6
b) x+43 /57 + x+46 /54 = x+49 /51 + x+52 /48
a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+2\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
Mà \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\)
Vậy x = -10
b) \(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)
\(\Rightarrow\left(\frac{x+43}{57}+1\right)+\left(\frac{x+46}{54}+1\right)=\left(\frac{x+49}{51}+1\right)+\left(\frac{x+52}{48}+1\right)\)
\(\Rightarrow\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
\(\Rightarrow\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)
Mà \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)
\(\Rightarrow x+100=0\)
\(\Rightarrow x=-100\)
Vậy x = -100
a.\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
=>\(\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
<=> \(\frac{x+1+9}{9}+\frac{x+2+8}{8}=\frac{x+3+7}{7}+\frac{x+4+6}{6}\)
<=>\(\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)
<=> \(\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
<=> \(\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
<=> x+10=0
<=> x=-10
Vậy tập nghiệm của phương trình trên là S=\(\left\{-10\right\}\)
b. \(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)
=> \(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)<=>\(\frac{x+43+57}{57}+\frac{x+46+54}{54}=\frac{x+49+51}{51}+\frac{x+52+48}{48}\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
<=>(x+100)\(\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)\)=0
<=>x+100=0
<=>x= -100
Vậy tập nghiệm của phương trình trên là S=\(\left\{-100\right\}\)
có rút gọn phân số 35/6 về phân số đc ko ? Nếu có thì mn rút gọn cho mik nhé !
Ai xong tr tui k cho
\(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\left(5^{128}-1\right)=2.5^{128}-2\)
c: Ta có: \(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^2-1\right)\left(5^2+1\right)\cdot\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{16}-1\right)\cdot\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{64}-1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{128}-1\right)\)
\(=2\cdot5^{128}-2\)