[Toán 7]Tuyển chọn một số bài toán nâng cao lớp 7

[camatngutapbay]

[Toán 7] Tìm x

Tìm x, biết:

[TEX]\frac{x-1}{2015}+\frac{x-2}{2014}=\frac{x-3}{2013}+\frac{x-4}{2014}[/TEX]

[TEX]\frac{x+1}{9}+\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+4}{6}=-4[/TEX]
Chú ý đặt tiêu đề và gõ Latex

 

[su10112000a]

giải luôn vậy):
a/ $\dfrac{x-1}{2015} + \dfrac{x-2}{2014} = \dfrac{x-3}{2013} + \dfrac{x-4}{2012}$
$\rightarrow \dfrac{x-1}{2015} - 1 + \dfrac{x-2}{2014} -1 = \dfrac{x-3}{2013} - 1 + \dfrac{x-4}{2012} - 1$
$\rightarrow (x-2016)(\dfrac{1}{2015} + \dfrac{1}{2014} - \dfrac{1}{2013} - \dfrac{1}{2012}) = 0$
$\rightarrow x=2016$
câu b làm tương tự

 

[baochauhn1999]

Câu b:
$\frac{x+1}{9}+\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+4}{6}=-4$
$<=>(\frac{x+1}{9}+1)+(\frac{x+2}{8}+1)+(\frac{x+3}{7}+1)+(\frac{x+4}{6}+1)=0$
$<=>\frac{x+10}{9}+\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{6}=0$
$<=>(x+10)(\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+\frac{1}{6})=0$
$<=>x+10=0$
$<=>x=-10$

 

[tathi]

1.Chứng minh rằng:$\dfrac{1}{2^2}$-$\dfrac{1}{2^4}$+$\dfrac{1}{2^6}$-...+$\dfrac{1}{2^{4n-2}}$-$\dfrac{1}{2^{4n}}$+...+$\dfrac{1}{2^{2002}}$-$\dfrac{1}{2^{2004}}$<0,2
$$Giải$$
Ta có :
A=$\dfrac{1}{2^2}$-$\dfrac{1}{2^4}$+$\dfrac{1}{2^6}$-...+$\dfrac{1}{2^{4n-2}}$-$\dfrac{1}{2^{4n}}$+...+$\dfrac{1}{2^{2002}}$-$\dfrac{1}{2^{2004}}$<$\dfrac{1}{5}$
\Rightarrow A=$\dfrac{1}{2.2}$-$\dfrac{1}{4.4}$+$\dfrac{1}{8.8}$-...+$\dfrac{1}{2^{4n-2}}$-$\dfrac{1}{2^{4n}}$+...+$\dfrac{1}{2^{2002}}$-$\dfrac{1}{2^{2004}}$
\Rightarrow A=$\dfrac{1}{2^2.2^4}$+$\dfrac{1}{2^6.2^8}$+...+ $\dfrac{1}{2^{4n-2}.2^{4n}}$+...+$\dfrac{1}{2^{2002}.2^{2004}}$
\Rightarrow A=$\dfrac{1}{2^6}$+$\dfrac{1}{2^{14}}$+...+ $\dfrac{1}{2.^{4n.(-1)}}$+...+$\dfrac{1}{2^{4006}}$
Đó