jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

bài 1:

bn lấy giá trị của √(4^2-3,9^2) là dc

bài 2

AB+BC=2√(3^2+4^2)=??

A                   =      \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\)+ \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)+......+\(\dfrac{1}{1024}\)

A \(\times\)2           = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+...+\(\dfrac{1}{512}\)

A\(\times\)2 - A     =   1 - \(\dfrac{1}{1024}\)

A               = \(\dfrac{1023}{1024}\)

B = \(\dfrac{1}{1\times2}\)+\(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+...+\(\dfrac{1}{98\times99}\)+\(\dfrac{1}{99\times100}\)

B = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+...+\(\dfrac{1}{98}\)-\(\dfrac{1}{99}\)+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)

B = 1 - \(\dfrac{1}{100}\)

B = \(\dfrac{99}{100}\)

C= \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+\(\dfrac{1}{42}\)+...+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)

C = \(\dfrac{1}{2\times3}\) +\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+...+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)

C = \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{11}\)

C = \(\dfrac{1}{2}\) - \(\dfrac{1}{11}\)

C = \(\dfrac{9}{22}\)

Ta có : \(12a+7b=64\)

Do \(64⋮4,12a⋮4\) \(\Rightarrow7b⋮4\) mà \(\left(7,4\right)=1\)

\(\Rightarrow b⋮4\) (1)

Từ giả thiết \(\Rightarrow7b\le64\) \(\Leftrightarrow b\le9\) kết hợp với (1)

\(\Rightarrow b\in\left\{4,8\right\}\)

+) Với \(b=4\) thì : \(12a+7\cdot4=64\)

\(\Leftrightarrow12a=36\)

\(\Leftrightarrow a=3\) ( thỏa mãn )

+) Với \(b=8\) thì \(12a+7\cdot8=64\)

\(\Leftrightarrow12a=8\)

\(\Leftrightarrow a=\frac{8}{12}\) ( loại )

Vậy : \(\left(a,b\right)=\left(3,4\right)\)

\(P=2^{100}-2^{99}-2^{98}-...-2^3-2^2-2\)

\(\Rightarrow2P=2^{101}-2^{100}-...-2^2\)

\(\Rightarrow P=2P-P=2^{101}-2^{100}-...-2^2-2^{100}+2^{99}+2^{98}+...+2=2^{101}-2.2^{100}+2=2\)

a) \(x-3=0\)

  \(\Leftrightarrow\)  \(x=0+3\)

  \(\Leftrightarrow\)  \(x=3\)

b) \(2x+6=0\)

 \(\Leftrightarrow\) \(2x=0-6\)

 \(\Leftrightarrow2x=-6\)

 \(\Leftrightarrow x=-3\)   

c) \(2x+3=x+9\)

\(\Leftrightarrow2x+3-x=9\)

\(\Leftrightarrow x+3=9\)

\(\Leftrightarrow x=9-3\)

\(\Leftrightarrow x=6\)

d) \(\left(x-4\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\2x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\2x+4+\left(-4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

Bài nhiều quá nên tôi chỉ làm bài 1, 2 thôi nhé :v

1.

\(a,x-3=0\)

\(\Leftrightarrow x=3\)

Vậy : S = {3}

\(b,2x+6=0\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\)

Vậy : S = {-3}

\(c,2x+3=x+9\)

\(\Leftrightarrow x=6\)

Vậy : S = {6]

\(d,\left(x+4\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)

Vậy : S = {-4; -2}

2.

\(a,\dfrac{3x+1}{4}=\dfrac{x+2}{3}\)

\(\Leftrightarrow3\left(3x+1\right)=4\left(x+2\right)\)

\(\Leftrightarrow9x+3=4x+8\)

\(\Leftrightarrow5x=5\)

\(\Leftrightarrow x=1\)

Vậy : S = {1}

\(b,\dfrac{4}{x}-\dfrac{5}{x+1}=0\) (ĐKXĐ : x ≠ 0; x ≠ -1)

\(\Rightarrow4\left(x+1\right)-5x=0\)

\(\Leftrightarrow4x+4-5x=0\)

\(\Leftrightarrow-x=-4\)

\(\Leftrightarrow x=4\) (N)

Vậy : S = {4}

không có câu hỏi ạ

câu ??