Ta có: \(5^{x+1}-5^{x}=500\)

=>\(5^{x}\cdot5-5^{x}=500\)

=>\(5^{x}\cdot4=500\)

=>\(5^{x}=\frac{500}{4}=125=5^3\)

=>x=3

\(5^{x+1}-5^{x}=500\)

\(5^{x}\cdot\left(5-1\right)=500\)

\(5^{x}\cdot4=500\)

\(5^{x}=100:4\)

\(5^{x}=125\)

\(5^{x}=5^3\)

\(\Rightarrow x=3\)

Vậy x = 3

ko đăng linh tinh nha bn!

a: Ta có: \(x^2-6x=27\)

=>\(x^2-6x-27=0\)

=>\(x^2-9x+3x-27=0\)

=>x(x-9)+3(x-9)=0

=>(x-9)(x+3)=0

=>\(\left[\begin{array}{l}x-9=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=9\\ x=-3\end{array}\right.\)

b: \(x^2+6x+9=0\)

=>\(x^2+2\cdot x\cdot3+3^2=0\)

=>\(\left(x+3\right)^2=0\)

=>x+3=0

=>x=-3

c: \(x^2-10x+25=0\)

=>\(x^2-2\cdot x\cdot5+5^2=0\)

=>\(\left(x-5\right)^2=0\)

=>x-5=0

=>x=5

Ta có: \(\left(2x-1\right)^2=100\)

=>\(\left[\begin{array}{l}2x-1=10\\ 2x-1=-10\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=11\\ 2x=-9\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{11}{2}\\ x=-\frac92\end{array}\right.\)

a: Hiệu vận tốc của hai xe là 60-50=10(km/h)

Hai xe gặp nhau sau khi đi được:

100:10=10(giờ)

b: Chỗ gặp nhau cách A:

\(60\cdot10=600\left(\operatorname{km}\right)\)

Số số hạng trong dãy số 1;2;3;...;120 là:

\(\left(120-1\right):1+1=120-1+1=120\) (số)

Ta có: \(1-2+3-4+5-6+\cdots+119-120+121\)

=(1-2)+(3-4)+(5-6)+...+(119-120)+121

=(-1)+(-1)+...+(-1)+121

=-60+121

=61

Bài nào em làm được, bài nào em chưa làm được em?

Bài 6:

1: \(\left(1-3x\right)^3=-8\)

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

=>3x-1=2

=>3x=3

=>x=1

2: \(x\cdot\left(0,9\right)^3=\left(0,81\right)^2\)

=>\(x\cdot\left(0,9\right)^3=\left(0,9\right)^4\)

=>\(x=\left(0,9\right)^4:\left(0,9\right)^3=0,9\)

3: \(4x^8=1024\)

=>\(x^8=\frac{1024}{4}=256\)

=>\(\left[\begin{array}{l}x=2\\ x=-2\end{array}\right.\)

4: \(x^5:\left(-9\right)=27\)

=>\(x^5=27\cdot\left(-9\right)=-243=\left(-3\right)^5\)

=>x=-3

Bài 5:

a: \(4^4=\left(2^2\right)^4=2^{2\cdot4}=2^8\)

b: \(2^{30}=\left(2^3\right)^{10}=8^{10}\)

\(3^{20}=\left(3^2\right)^{10}=9^{10}\)

mà 8<9

nên \(2^{30}<3^{20}\)

c: \(9^5=\left(3^2\right)^5=3^{2\cdot5}=3^{10}\)

\(27^3=\left(3^3\right)^3=3^{3\cdot3}=3^9\)

mà \(3^{10}>3^9\)

nên \(9^5>27^3\)

d: \(243^3=\left(3^5\right)^3=3^{15};125^5=\left(5^3\right)^5=5^{3\cdot5}=5^{15}\)

mà 3<5

nên \(243^3<125^5\)

bài 3:

1) \(-0,25+\frac14+4,2:\frac{-15}{12}\)

\(=-\frac14+\frac14+\frac{21}{5}\cdot\frac{-12}{15}\)

\(=\frac{21\cdot\left(-12\right)}{5\cdot15}=\frac{-252}{75}=-\frac{84}{25}\)

2) \(-2\frac14-\left(-0,75\right):\frac12\)

\(=-\frac94+\frac32=-\frac94+\frac64=-\frac34\)

3) \(0,8+\left(-1,5\right)\cdot\frac12+\frac{-1}{2}\)

\(=\frac45-\frac34-\frac12=\frac{16}{20}-\frac{15}{20}-\frac{10}{20}=-\frac{9}{20}\)

4) \(1,25\cdot\frac38+\left(-\frac58\right)+\frac{-5}{2}\)

\(=\frac54\cdot\frac38-\frac58-\frac52=\frac{15}{32}-\frac58-\frac52\)

\(=\frac{15}{32}-\frac{20}{32}-\frac{80}{32}=-\frac{85}{32}\)

bài 4:

1) \(\frac32\cdot\left(-\frac{37}{10}\right)+\frac{17}{2}\cdot\left(-\frac{37}{10}\right)\)

\(=\left(-\frac{37}{10}\right)\cdot\left(\frac32+\frac{17}{2}\right)=\left(-\frac{37}{10}\right)\cdot10=-37\)

2) \(\frac{7}{13}\cdot\left(-\frac{3}{11}\right)-\frac{7}{13}\cdot\frac{8}{11}\)

\(=\frac{7}{13}\cdot\left(-\frac{3}{11}\right)+\frac{7}{13}\cdot\left(-\frac{8}{11}\right)\)

\(=\frac{7}{13}\cdot\left\lbrack\left(-\frac{3}{11}\right)+\left(-\frac{8}{11}\right)\right\rbrack\)

\(=\frac{7}{13}\cdot\left(-1\right)=-\frac{7}{13}\)

3) \(\frac35\cdot\frac67+\frac37:\frac53-\frac27:1\frac23\)

\(=\frac35\cdot\frac67+\frac37\cdot\frac35-\frac27\cdot\frac35\)

\(=\frac35\cdot\left(\frac67+\frac37-\frac27\right)=\frac35\cdot1=\frac35\)

4) \(\frac19:\frac45+\frac19:\frac47\)

\(=\frac19\cdot\frac54+\frac19\cdot\frac74=\frac19\cdot\left(\frac54+\frac74\right)\)

\(=\frac19\cdot3=\frac13\)

GP là gold point , để tăng lên các cấp bậc khác nhau