Bài 5: Phép cộng và phép nhân. Luyện tập 1. Luyện tập 2

Sửa đề: A=1/3+1/9+1/27+...+1/6561

=1/3+1/3^2+1/3^3+...+1/3^8

=>3A=1+1/3+...+1/3^7

=>3A-A=1-1/3^8

=>\(2A=\dfrac{3^8-1}{3^8}\)

=>\(A=\dfrac{3^8-1}{2\cdot3^8}\)

Đặt \(S=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{6561}\)

\(3S=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{2187}\)

\(2S=\dfrac{2188}{2187}-\left(\dfrac{1}{27}+\dfrac{1}{6561}\right)\)

\(2S=\dfrac{2188}{2187}-\dfrac{244}{6561}\)

\(2S=\dfrac{4376}{6561}-\dfrac{244}{6561}\)

\(2S=\dfrac{4132}{6561}\)

\(S=\dfrac{2066}{6561}\)

\(\dfrac{2^{13}+2^5}{2^{10}+2^2}=\dfrac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=\dfrac{2^5}{2^2}=2^3=8\)

a) 120 - (x.33 - 132).20 = 100

(x.33 - 132).20 = 120-100

(x.33 - 132).20 = 20

x.33-132=1

x.33=133

x=133/33

b) Có : 30⋮x => x∈Ư(30)={1;2;3;5;6;10;15;30}

mà 5≤x<30 nên x ∈ {5;6;10;15}

Vậy...

\(\left(x+5\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=7\end{matrix}\right.\)

=>x-5=0 hoặc x-7=0

=>x=5 hoặc x=7

\(\text{(x-5)(x-7)=0}\)

\(+\)) \(\text{x - 5 = 0 }\)\(\Rightarrow\) \(\text{x = 5}\)

\(+\)) \(\text{x - 7 = 0}\) \(\Rightarrow\) \(\text{x = 7}\)

vậy \(x=5\) hoặc \(\text{x = 7}\) 

575-(6x+70)=445

<=> 6x+70=130

<=> 6x = 60

<=> x=10

\(575-\left(6x+70\right)=445\)

\(\Rightarrow6x+70=575-445\)

\(\Rightarrow6x+70=130\)

\(\Rightarrow6x=130-70\)

\(\Rightarrow6x=60\)

\(\Rightarrow x=\dfrac{60}{6}\)

\(\Rightarrow x=10\)

`A = 2 + 2^2+ ... + 2^2017`

`=> 2A = 2^2 + 2^3 + ... + 2^2018`

`=> 2A - A = (2^2 + 2^3 + ... + 2^2018) - (2 + 2^2 + ... +2^2017)`

`=> A         = 2^2018 - 2`

`B = 1 + 3^2 + ... + 3^2018`

`=> 3^2B = 3^2 + 3^4 + ... + 3^2020`

`=> 9B-B =(3^2 + 3^4 + ... + 3^2020) - (1 + 3^2 + ... + 3^2018`

`=> 8B     = 3^2020 - 1`

`=> B       = (3^2020 - 1)/8`

`C = 5 + 5^2 - 5^3 + ... + 5^2018`

`=> 5C = 5^2 + 5^3 - 5^4 + ... +5^2019`

`=> 5C + C = ( 5^2 + 5^3 - 5^4 + ... 5^2019) + (5 + 5^2 - 5^3 + ... + 5^2018)`

`=> 6C = 55 + 5^2019`

`=> C  = (5^2019 + 55)/6`

Cái đầu là 1 hay 2¹ em? Chứ 1 thì không chia hết

\(A=1+2^1+2^2+2^3+2^4+2^5+...+2^{96}+2^{97}+2^{98}\\ =1.\left(1+2^1+2^2\right)+2^3.\left(1+2^1+2^2\right)+...+2^{96}.\left(1+2^1+2^2\right)\\ =1.7+2^3.7+...+2^{96}.7\\ =\left(1+2^3+...+2^{96}\right).7⋮7\)

Đây mới đúng và đủ đề em hi

\(\left(9x-36\right)\left(2x-10\right)=0\\9.\left(x-4\right).2.\left(x-5\right)=0\\ 18.\left(x-4\right).\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow \left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(91-5\left(5+x\right)=61\)

\(\Rightarrow5\left(5+x\right)=91-61\)

\(\Rightarrow5\left(5+x\right)=30\)

\(\Rightarrow5+x=\dfrac{30}{5}\)

\(\Rightarrow5+x=6\)

\(\Rightarrow x=6-5\)

\(\Rightarrow x=1\)

`53.(9-x)=53`

`=>9-x=53:53`

`=>9-x=1`

`=>x=9-1`

`=>x=8`

\(53\left(9-x\right)=53\)

\(\Rightarrow9-x=\dfrac{53}{53}\)

\(\Rightarrow9-x=1\)

\(\Rightarrow x=9-1\)

\(\Rightarrow x=8\)

=>9-x=1

=>x=8