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\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{x+4-x-1}{\left(x+4\right)\left(x+1\right)}=\dfrac{3}{10}\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=10\)
\(\Leftrightarrow x^2+5x-6=0\)
=>(x+6)(x-1)=0
=>x=-6 hoặc x=1

7x-2x=6^17:6^15+44:11
x.(7-2)=6^2+4
x.5=36+4
x.5=40
x=40;5
x=8
vậy x=8
k mk nha

1) \(2^{x+1}\cdot2^{2014}=2^{2015}\)\(\Leftrightarrow2^{2014x+2014}=2^{2015}\)\(\Leftrightarrow2014x+2014=2015\)\(\Leftrightarrow x=\frac{1}{2014}\)
2) \(7x-2x=\frac{6^{17}}{6^{15}}+\frac{44}{11}\)\(\Leftrightarrow5x=6^2+4=36+4=40\)\(\Leftrightarrow x=\frac{40}{5}=8\)
3) \(3^x=9\)\(\Leftrightarrow3^x=3^2\)\(\Leftrightarrow x=2\)
4) \(7x-x=\frac{5^{21}}{5^{19}}+3\cdot2^2-7^0\)\(\Leftrightarrow6x=5^2+3\cdot4-1=25+12-1=36\)\(\Leftrightarrow x=6\)
5) \(4^x=64\)\(\Leftrightarrow4^x=4^3\)\(\Leftrightarrow x=3\)
6) \(9^{x-1}=9\)\(\Leftrightarrow x-1=1\)\(\Leftrightarrow x=0\)
7) \(\frac{2^x}{2^5}=1\)\(\Leftrightarrow2^{x-5}=2^0\)\(\Leftrightarrow x-5=0\)\(\Leftrightarrow x=5\)
8) \(\left(5x-9\right)^3=216\)\(\Leftrightarrow\left(5x-9\right)^3=6^3\)\(\Leftrightarrow5x-9=6\)\(\Leftrightarrow5x=15\)\(\Leftrightarrow x=3\)
9) \(5\cdot3^{7x-11}=135\)\(\Leftrightarrow5.3^{7x-11}=5.3^3\)\(\Leftrightarrow3^{7x-11}=3^3\)\(\Leftrightarrow7x-11=3\)\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
10) \(2.3^x=19\cdot3^8-81^2\)\(\Leftrightarrow2.3^x=19\cdot3^8-3^8=18.3^8=2.3^{11}\)\(\Leftrightarrow3^x=3^{11}\Leftrightarrow x=11\)
Đây là cách làm của mình. Bạn có thể chỉnh sửa tuỳ ý theo cách làm của bạn nhé ^^
Học tốt ^3^

a, Ta có: \(142-\left(12x+30\right)=10^{5-3}\)
\(=>142-12x-30=10^2\)
\(112-12x=100\)
\(=>12x=112-100=12=>x=1\)
Vậy số cần tìm là 1;
b, Ta có: \(=>\left(5x+3^4\right)=6^9:6^8.3^4\)
\(=>5x+3^4=6.3^4=>5x=6.3^4-3^4\)
\(=>5x=5.3^4=>x=3^4=81\)
Vậy x=81;
CHÚC BẠN HỌC TỐT.......
Ta có: \(71.2-6(2x+5)=10^5:10^3\)
\(\Rightarrow142-12x-30=10^2\)
\(\Rightarrow142-30-10^2=12x\)
\(\Rightarrow142-30-100=12x\)
\(\Rightarrow12=12x\)
\(\Rightarrow x=\dfrac{12}{12}\)
\(\Rightarrow x=1\)
Vậy \(x=1\)

Bài 1 :
a/ \(a^3.a^9=a^{3+9}=a^{12}\)
b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)
c/ \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)
d/ \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)
e/ \(5^6:5^3+3^3.3^2\)
\(=5^3+3^5=125+243=368\)
i/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
(7x+1).68=610
=> 7x+1=62=36
=> 7x=35=> x=5
Tìm x:( 7x + 1 ) .68 = 610
Giải:Ta có:\(\left(7x+1\right).6^8=6^{10}\)
\(\Rightarrow7x+1=\frac{6^{10}}{6^8}\Rightarrow7x+1=6^2\)
\(\Rightarrow7x+1=36\Rightarrow7x=35\Rightarrow x=5\)
Vậy x=5 thỏa mãn