\(d,2,5.5^{n-3}.2.5+5^n-6.5^{n-1}=5.5.5^{n-3}+5^n-6.5^{n-1}=5^2.5^{n-3}+5^n-6.5^{n-1}\)

  \(=5^{n-3+2}+5^n-6.5^{n-1}=5^{n-1}\left(1+5-6\right)=5^{n-1}.0=0\)

a, \(10^{n+1}-6.10^n=10^n\left(10-6\right)=4.10^n\)

b. \(2^{n+3}+2^{n+2}-2^{n+1}+2^n=2^n\left(2^3+2^2-2+1\right)=2^n\left(8+4-2+1\right)=11.2^n\)

a) Ta có:

\(90.10^k-10^{k+2}+10^{k+1}\)

\(=90.10^k-10^k.10^2+10^k.10\)

\(=10^k\left(90-10^2+10\right)\)

\(=10^k.0=0\)

b) Ta có:

\(2,5.5^{n-3}.10+5^n-6.5^{n-1}\)

\(=2,5.10.5^{n-3}+5^n-6.5^{n-1}\)

\(=5.5.5^{n-3}+5^n-6.5^{n-1}\)

\(=5^2.5^{n-3}+5^n-6.5^{n-1}\)

\(=5^{n-3+2}+5^n-6.5^{n-1}\)

\(=5^{n-1}\left(1+5-6\right)\)

\(=5^{n-1}.0=0\)

a) Rút gọn biểu thức:

\(90\times10^k-10^{k+2}+10^{k+1}=90\times10^k-10^k\times10^2+10^k\times10\) \(=10^k\times\left(90-10^2+10\right)\) \(=10^k\times\left(90-100+10\right)\) \(=10^k\times0=0\)

b) Rút gọn biểu thức:

\(2,5\times5^{n-3}\times10+5^n-6\times5^{n-1}=2,5\times\dfrac{5^n}{5^3}\times10+5^n-6\times\dfrac{5^n}{5}\) \(=2,5\times\dfrac{5^n}{125}\times10+5^n-\dfrac{6}{5}\times5^n\) \(=0,2\times5^n+5^n-1,2\times5^n\) \(=5^n\times\left(0,2+1-1,2\right)=5^n\times0=0\)

a: \(10^{n+1}=10^n\cdot10\)

b: \(2^{n+3}+2^{n+1}-2^{n+1}+2^n\)

\(=2^n\cdot8+2^n=9\cdot2^n\)

c: \(90\cdot10^k-10^{k+2}+10^{k+1}\)

\(=90\cdot10^k+10^k\cdot10-10^k\cdot100=0\)

a) 10^{n + 1} - 6.10ⁿ

= 10ⁿ . 10 - 6 . 10ⁿ

= 10ⁿ . (10 - 6)

= 10ⁿ . 4

b) 2^{n + 3} + 2^{n + 2} - 2^{n + 1} + 2ⁿ

= 2ⁿ . 2³ + 2ⁿ . 2² - 2ⁿ . 2 + 2ⁿ . 1

= 2ⁿ . (8 + 4 - 2 + 1)

= 2ⁿ . 11

Bài làm

a) 10^(n + 1) - 6 * 10^n

= 10^n + 10 - 6 * 10^n

= 10^n * ( 10 - 6 )

= 10^n * 4

b) 90 * 10^n - 10(n + 2) + 10^(n + 1)

= 90 * 10^n - 10^n * 10^2 + 10^n * 10

= 10^n * ( 90 - 10^2 + 10 )

= 10^n * ( 90 - 100 + 10 )

= 10^n * 0

= 0

Bài làm :

\(a,10^{n+1}-6.10^n\)

\(=10^n.10-6.10^n\)

\(=10^n.\left(10-6\right)\)

\(=10^n.4\)

\(b,90.10^n-10^{n+2}+10^{n+1}\)

\(=90.10^n-10^n.10^2+10^n.10\)

\(=10^n.\left(90-100+10\right)\)

\(=10^n.0\)

\(=0\)

Học tốt nhé

a) Ta có:

90.10^k−10^k+2+10^k+1

=90.10^k−10^k.10²+10^k.10

=10^k(90−10²+10)

=10^k.0=0

b) \(x^2\left(x^2+4\right)-x^2-4=0\)

.\(\Leftrightarrow x\left(x^2+4\right)-\left(x^2+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+4\right)=0\)

\(\Rightarrow x-1=0\)(vì \(x^2+4>0\))

\(\Leftrightarrow x=1\)

Ta có : (2x - 1)² - (4x² - 1) = 0

<=> (2x - 1)² - [(2x)² - 1²] = 0

<=> (2x - 1)² - (2x - 1)(2x + 1) = 0

<=> (2x - 1)[2x - 1 - (2x + 1)] = 0

<=> (2x - 1)(-2) = 0

=> 2x - 1 = 0

=> 2x = 1

=> x = \(\frac{1}{2}\)

Vậy x = \(\frac{1}{2}\)

1a) \(10^{n+1}-6\cdot10^n\)

\(=10^n\cdot10-6\cdot10^n\)

= \(10^n\left(10-6\right)\)

\(=10^n\cdot4\)

b) \(2^{n+3}+2^{n+2}-2^{n+1}+2^n\)

\(=2^n\cdot2^3+2^n\cdot2^2-2^n\cdot2+2^n\)

\(=2^n\left(2^3+2^2-2+1\right)\)

\(=2^n\cdot11\)

c) \(90\cdot10^k-10^{k+2}+10^{k+1}\)

\(=90\cdot10^k-10^k\cdot10^2+10^k\cdot10\)

\(=10^k\left(90-10^2+10\right)=0\)

d) \(2,5\cdot5^{n-3}\cdot10+5^n-6\cdot5^{n-1}\)

\(=\dfrac{2,5\cdot10\cdot5^n}{5^3}+5^n-\dfrac{6\cdot5^n}{5}\)

\(=\dfrac{5^n}{5}+5^n-\dfrac{6\cdot5^n}{5}\)

\(=\dfrac{5^n+5^n\cdot5-6\cdot5^n}{5}=\dfrac{5^n\left(5-6\right)+5^n}{5}=0\)

2. \(M+\left(6x^2-4xy\right)=7x^2-8xy+y^2\)

\(M=\left(7x^2-8xy+y^2\right)-\left(6x^2-4xy\right)\)

\(M=7x^2-8xy+y^2-6x^2+4xy\)

\(M=7x^2-6x^2-8xy+4xy+y^2\)

\(M=x^2-4xy+y^2\)

Mk cảm ơn bn nhiều lắm ạ Lê Mỹ Linh

a)\(10^{n+1}-6.10^n\)

\(=10^n.\left(10-6\right)\)

\(=10^n.4\)

b)\(90.10^n-10^{n+2}+10^{n+1}\)

\(=10^n.\left(90-10^2+10\right)\)

\(=10^n.0=0\)

a, 10ⁿ⁺¹-6.10ⁿ=10.10ⁿ-6.10ⁿ=4.10ⁿ