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\(B=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(7B=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\)
\(7B-B=\left(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)\)
\(6B=1-\frac{1}{7^{100}}\)
\(B=\frac{1-\frac{1}{7^{100}}}{6}\)
\(B=\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7B=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\)
\(\Rightarrow7B-B=\left(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6B=1-\frac{1}{7^{99}}\)
\(\Rightarrow B=\left(1-\frac{1}{7^{99}}\right):6\)
a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)
b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)
\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)
c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)
\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)
e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)
f) \(\left|12-x\right|-7=5\)
th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)
th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)
i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)
k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)
a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)
b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)
Làm nốt nha
a)
- \(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{59}.3\)
\(=3\left(2+2^3+...+2^{59}\right)⋮3\)
- \(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=2.7+2^4.7+...+2^{58}.7\)
\(=7\left(2+2^4+2^{58}\right)⋮7\)
- \(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=2.15+2^5.15+...+2^{57}.15\)
\(=15\left(2+2^5+2^{57}\right)⋮15\)
b) \(B=1+5+5^2+5^3+...+5^{96}+5^{97}+5^{98}\)
\(=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{96}+5^{97}+5^{98}\right)\)
\(=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+..+5^{96}\left(1+5+5^2\right)\)
\(=31+5^3.31+...+5^{96}.31\)
\(=31\left(1+5^3+...+5^{96}\right)⋮31\)
Ta có B = 3 + 32 + 33 + ... + 32014 + 32015
=> 3B = 32 + 33 + 34 + .... + 32015 + 32016
Lấy 3B trừ B theo vế ta có
3B - B = (32 + 33 + 34 + .... + 32015 + 32016) - (3 + 32 + 33 + ... + 32014 + 32015)
2B = 32016 - 3
Khi đó 2B + 3 = 3x
<=> 32016 - 3 + 3 = 3x
=> 32016 = 3x
=> x = 2016
Vậy x = 2016
Bg
Ta có: B = 3 + 32 + 33 +...+ 32014 + 32015
=> 3B = 3.(3 + 32 + 33 +...+ 32014 + 32015)
=> 3B = 3.3 + 3.32 + 3.33 +...+ 3.32014 + 3.32015
=> 3B = 32 + 33 + 34 +...+ 32015 + 32016
=> 3B - B = (32 + 33 + 34 +...+ 32015 + 32016) - (3 + 32 + 33 +...+ 32014 + 32015)
=> 2B = 32016 - 3
2B + 3 = 3x
=> 32016 - 3 + 3 = 3x
=> 32016 = 3x
=> x = 2016