a: \(\dfrac{1}{2}B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)

\(\Leftrightarrow-\dfrac{1}{2}B=\dfrac{1}{2^{100}}-\dfrac{1}{2}\)

hay \(B=\dfrac{-1}{2^{99}}+1\)

b: |3x+5|=10

=>3x+5=10 hoặc 3x+5=-10

=>3x=5 hoặc 3x=-15

=>x=-5 hoặc x=5/3

Bài 1:

a) [ (1/6 + 1/10 + 1/15) : (1/6 + 1/10 - 1/15) phần 1/2 - 1/3 + 1/4 - 1/5 ] : (1/4 - 1/6)

= [ (1/6 : 1/6) + (1/10 : 1/10) - (1/15 : 1/15) phần 30/60 - 20/60 + 15/60 - 12/60 ] : (3/12 - 2/12)

= [ 1 + 1 - 1 phần 13/60 ] : 1/12

= [ 1 : 13/60 ] x 12

= 60/13 x 12

=720/ 13

b) (3/20 + 1/2 - 1/15) x 12/49 phần 3 và 1/3 + 2/9

= (9/60 + 30/60 - 4/60) x 12/49 phần 10/3 + 2/9

= 7/12 x 12/49 phần 30/9 + 2/9

= 1/7 : 32/9

= 1/7 x 9/32

= 9/224

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

2/ Đặt biểu thức trên là B.Ta có: \(B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n+1}-\frac{1}{n}=1-\frac{1}{n}< 1\forall n\ge2\) (đpcm)

5/ (không chắc,quên cách làm mọe rồi) Gọi số viên bi của Hằng là x < 88 (viên). Suy ra: Số bi xanh là:\(\frac{x}{12}\)

Số viên bi vàng: \(\frac{x}{6}\) suy ra tổng số viên bi xanh và vàng là: \(\frac{x}{6}+\frac{x}{12}=x\left(\frac{1}{6}+\frac{1}{12}\right)=x\left(\frac{2}{12}+\frac{1}{12}\right)=x.\frac{3}{12}=\frac{x}{4}\)

đặt B=1/3 + 1/3^2 + 1/3^3 +.....+ 1/3^99 

=>1/3B=1/3^2 + 1/3^3 +.....+ 1/3^100

=>1/3B-B=1/3^2 + 1/3^3 +.....+ 1/3^100-1/3-1/3^2-1/3^3-...-1/3^100

=>-2/3=1/3^100-1/3

=>B=(1/3^100-1/3):(-2/3)<1/2 (vì kết quả ra số âm )

\(a.ĐKXĐ:\hept{\begin{cases}1-3x\ne0\\3x+1\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\...\\x\ge0\end{cases}}}\)

\(b,M=\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10}{1-6x+9x^2}\)

\(=\left(\frac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\frac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)

\(=\left(\frac{3x+9x^2+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)

\(=\frac{5x+3x^2}{1+3x}.\frac{1-3x}{2\left(3x^2+5\right)}\)

==>Sai đề không mem

\(P=\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{2x-5}{x^2+5x}+\frac{x}{5-x}\)

\(=\left(\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right):\frac{2x-5}{x\left(x+5\right)}+\frac{x}{5-x}\)

\(=\left(\frac{x^2}{x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{x\left(x+5\right)\left(x-5\right)}\right).\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)

\(=\frac{\left(x+x-5\right)\left(x-x+5\right)}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)

\(=\frac{5\left(2x-5\right)}{\left(x-5\right)}.\frac{1}{2x-5}+\frac{x}{5-x}\)

\(=\frac{5}{x-5}-\frac{x}{x-5}\)

\(=\frac{5-x}{x-5}\)

\(=\frac{-\left(x-5\right)}{x-5}\)

\(=-1\)

=> biểu thức  P k phụ thuộc vào x

1/a,

-Ta có: 

$B<1\Leftrightarrow B<\frac{10^{2005}+1+9}{10^{2006}+1+9}=\frac{10^{2005}+10}{10^{2006}+10}=\frac{10(10^{2004}+1)}{10(10^{2005}+1)}=\frac{10^{2004}+1}{10^{2005}+1}=A$

-Vậy: B<A

b,$A=1+(\frac{1}{2})^2+...+(\frac{1}{100})^2$

$\Leftrightarrow A=1+\frac{1}{2^2}+...+\frac{1}{100^2}$

$\Leftrightarrow A<1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}$

$\Leftrightarrow A<1+\frac{1}{1}-\frac{1}{2}+...+\frac{1}{99}-\frac{1}{100}$

$\Leftrightarrow A<1+1-\frac{1}{100}\Leftrightarrow A<2-\frac{1}{100}\Leftrightarrow A<2(đpcm)$
2,
a.
-Ta có:$\Rightarrow \frac{3x+7}{x-1}=\frac{3(x-1)+16}{x-1}=\frac{3(x-1)}{x-1}+\frac{16}{x-1}=3+\frac{16}{x-1}
-Để: 3x+7/x-1 nguyên
-Thì: $\frac{16}{x-1}$ nguyên
$\Rightarrow 16\vdots x-1\Leftrightarrow x-1\in Ư(16)\Leftrightarrow ....$
b, -Ta có:
$\frac{n-2}{n+5}=\frac{n+5-7}{n+5}=1-\frac{7}{n+5}$
-Để: n-2/n+5 nguyên
-Thì: \frac{7}{n+5} nguyên
$\Leftrightarrow 7\vdots n+5\Leftrightarrow n+5\in Ư(7)\Leftrightarrow ...$