Sửa đề: \(\dfrac{1}{5}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)

Đặt \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(\dfrac{1}{5}-\dfrac{1}{6}< \dfrac{1}{5\cdot6}< \dfrac{1}{5^2}< \dfrac{1}{4\cdot5}=\dfrac{1}{4}-\dfrac{1}{5}\)

\(\dfrac{1}{6}-\dfrac{1}{7}< \dfrac{1}{6\cdot7}< \dfrac{1}{6^2}< \dfrac{1}{5\cdot6}=\dfrac{1}{5}-\dfrac{1}{6}\)

...

\(\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{100\cdot101}< \dfrac{1}{100^2}< \dfrac{1}{100\cdot99}=\dfrac{1}{99}-\dfrac{1}{100}\)

Do đó: \(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=>\(\dfrac{1}{5}-\dfrac{1}{101}< A< \dfrac{1}{4}-\dfrac{1}{100}\)

=>\(\dfrac{1}{5}< A< \dfrac{1}{4}\)

A = \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + \(\dfrac{1}{7^2}\) + ... + \(\dfrac{1}{100^2}\)

\(\dfrac{1}{5.6}\) < \(\dfrac{1}{5^2}\) < \(\dfrac{1}{4.5}\)

\(\dfrac{1}{6.7}\) < \(\dfrac{1}{6^2}\) < \(\dfrac{1}{5.6}\)

\(\dfrac{1}{7.8}\) < \(\dfrac{1}{7^2}\) < \(\dfrac{1}{6.7}\)

......................

\(\dfrac{1}{100.101}\) < \(\dfrac{1}{100^2}\) < \(\dfrac{1}{99.100}\)

Cộng vế với vế ta có:

\(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + ... + \(\dfrac{1}{100.101}\)< \(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+...+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{99.100}\)

\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\) < \(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+...+\(\dfrac{1}{100^2}\)< \(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)

\(\dfrac{1}{5}\) - \(\dfrac{1}{101}\) < \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\)+...+\(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\) - \(\dfrac{1}{100}\)

\(\dfrac{6}{30}\) - \(\dfrac{1}{101}\) < \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\)+ .... + \(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\) - \(\dfrac{1}{100}\) < \(\dfrac{1}{4}\)

\(\dfrac{5}{30}\) +( \(\dfrac{1}{30}\) - \(\dfrac{1}{101}\)) < \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\)

\(\dfrac{1}{6}\) + (\(\dfrac{1}{30}\) - \(\dfrac{1}{101}\)) < \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\)

Vì \(\dfrac{1}{30}\) > \(\dfrac{1}{101}\) ⇒  \(\dfrac{1}{30}\) - \(\dfrac{1}{101}\) > 0 ⇒ \(\dfrac{1}{6}\) + (\(\dfrac{1}{30}\) - \(\dfrac{1}{101}\)) > \(\dfrac{1}{6}\)

Vậy  \(\dfrac{1}{6}\) < \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{100^2}\) < \(\dfrac{1}{4}\) (đpcm)

a: \(-\dfrac{15}{19}=-1+\dfrac{4}{19}\)

\(-\dfrac{37}{41}=-1+\dfrac{4}{41}\)

\(-\dfrac{5}{9}=-1+\dfrac{4}{9}\)

\(\dfrac{23}{-27}=-\dfrac{23}{27}=-1+\dfrac{4}{27}\)

\(-\dfrac{7}{11}=-1+\dfrac{4}{11}\)

mà \(\dfrac{4}{41}< \dfrac{4}{27}< \dfrac{4}{19}< \dfrac{4}{11}< \dfrac{4}{9}\)

nên \(-\dfrac{37}{41}< -\dfrac{23}{27}< -\dfrac{15}{19}< -\dfrac{7}{11}< -\dfrac{5}{9}\)

mà \(-\dfrac{37}{41}< -\dfrac{76}{89}< -\dfrac{23}{27}\)

nên  \(-\dfrac{37}{41}< -\dfrac{76}{89}< -\dfrac{23}{27}< -\dfrac{15}{19}< -\dfrac{7}{11}< -\dfrac{5}{9}\)

Em có nhầm đề ko nhỉ lớp 7 ko giải được bài này

\(Sai\) \(Thầy\) \(Ơi\)

\(P=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2017}\)

\(\left(-\dfrac{1}{7}\right).P=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2017}+\left(-\dfrac{1}{7}\right)^{2018}\)

\(P-\left(-\dfrac{1}{7}\right)P=\left(-\dfrac{1}{7}\right)^0-\left(-\dfrac{1}{7}\right)^{2018}\)

\(\dfrac{8}{7}P=1-\dfrac{1}{7^{2018}}\)

\(\dfrac{8}{7}P=\dfrac{7^{2018}-1}{7^{2018}}\)

\(P=\dfrac{7^{2018}-1}{8.7^{2017}}\)

\(\dfrac{x-1}{69}+\dfrac{x-2}{68}=\dfrac{x-5}{65}+\dfrac{x-6}{64}\)

\(\Leftrightarrow\left(\dfrac{x-1}{69}-1\right)+\left(\dfrac{x-2}{68}-1\right)-\left(\dfrac{x-5}{65}-1\right)-\left(\dfrac{x-6}{64}-1\right)=0\)

\(\Leftrightarrow\dfrac{x-70}{69}+\dfrac{x-70}{68}-\dfrac{x-70}{65}-\dfrac{x-70}{64}=0\)

\(\Leftrightarrow\left(x-70\right)\left(\dfrac{1}{69}+\dfrac{1}{68}-\dfrac{1}{65}-\dfrac{1}{64}\right)=0\)

\(\Leftrightarrow x-70=0\) (do \(\dfrac{1}{69}+\dfrac{1}{68}-\dfrac{1}{65}-\dfrac{1}{64}\ne0\))

\(\Rightarrow x=70\)

\(\dfrac{49^5+49^7+49^9}{7^{11}+7^{13}+7^{15}+7^{17}+7^{19}+7^{21}}\)

\(=\dfrac{7^{10}+7^{14}+7^{18}}{7^{11}\left(1+7^2\right)+7^{15}\left(1+7^2\right)+7^{19}\left(1+7^2\right)}\)

\(=\dfrac{7^{10}\left(1+7^4+7^8\right)}{7^{11}\left(1+7^2\right)\left(1+7^4+7^8\right)}=\dfrac{1}{7\left(1+7^2\right)}=\dfrac{1}{7\cdot50}=\dfrac{1}{350}\)

3.4

a) Ta có: 

`|5-2/3x|>=0` với mọi x

`|2/3y-4|>=0` với mọi y 

`=>|5-2/3x|+|2/3y-4|>=0` với mọi x

Mà: `|5-2/3x|+|2/3y-4|=0`

Dấu "=" xảy ra: `5-2/3x=0` và `2/3y-4=0`

`<=>2/3x=5` và `2/3y=4`

`<=>x=5:2/3=15/2` và `y=4:2/3=6`

b) 

`|2/3-1/2+3/4x|+|1,5-3/4-3/2y|=0`

`=>|1/6+3/4x|+|3/4-3/2y|=0`

Ta có:

`|1/6+3/4x|>=0` với mọi x

`|3/4-3/2y|>=0` với mọi y

`=>|1/6+3/4x|+|3/4-3/2y|>=0` với mọi x 

Mà: `|1/6+3/4x|+|3/4-3/2y|=0`

Dấu "=" xảy ra: `1/6+3/4x=0` và `3/4-3/2y=0`

`<=>3/4x=-1/6` và `3/2y=3/4`

`<=>x=-1/6:3/4=-2/9` và `y=3/4:3/2=1/2` 

5A:

b: \(47-\dfrac{\left(45\cdot24-5^2\cdot12\right)}{14}\)

\(=47-\dfrac{1080-25\cdot12}{14}\)

\(=47-\dfrac{1080-300}{14}=47-\dfrac{780}{14}=-\dfrac{61}{7}\)

d: \(2345-1000:\left[19-2\left(2\cdot1-18^2\right)\right]\)

\(=2345-1000:\left(19+2\cdot322\right)\)

\(=2345-\dfrac{1000}{19+644}=2345-\dfrac{1000}{663}=\dfrac{1553735}{663}\)

5B:

b: \(50-\left[\left(20-2^3\right)\right]:2+34\)

\(=84-\dfrac{20-8}{2}=84-6=78\)

c: \(20-\left[30-\left(5-1\right)^2\right]:3\)

\(=20-\dfrac{\left[30-4^2\right]}{3}\)

\(=20-\dfrac{14}{3}=\dfrac{46}{3}\)

d: \(205-\left[1200-\left(4^2-23\right)^3\right]:40\)

\(=205-\left[1200-\left(16-23\right)^3\right]:40\)

\(=205-\dfrac{\left[1200-\left(-7\right)^3\right]}{40}\)

\(=205-\dfrac{1200+343}{40}\)

\(=205-38,575=166,425\)

Lời giải:

a. $A(x) = 6x^3-7x^2-x+m=3x^2(2x+1)-5x(2x+1)+2(2x+1)+m-2$

$=(2x+1)(3x^2-5x+2)+m-2$

$=B(x)(3x^2-5x+2)+m-2$

Vậy $A(x):B(x)$ được thương $3x^2-5x+2$ và dư $m-2$

b.

Để dư bằng 4 thì $m-2=4$

$\Leftrightarrow x=6$

a: \(\dfrac{A\left(x\right)}{B\left(x\right)}=\dfrac{6x^3-7x^2-x+m}{2x+1}\)

\(=\dfrac{6x^3+3x^2-10x^2-5x+4x+2+m-2}{2x+1}\)

\(=3x^2-5x+2+\dfrac{m-2}{2x+1}\)

b: Để phép chia \(\dfrac{A\left(x\right)}{B\left(x\right)}\) có dư là 4 thì m-2=4

=>m=6

\(=\left(2x^3-6x^2\right)+\left(5x^2-15x\right)+\left(2x-6\right)\)

\(=2x^2\left(x-3\right)+5x\left(x-3\right)+2\left(x-3\right)\)

\(=\left(x-3\right)\left(2x^2+5x+2\right)\)

\(=\left(x-3\right)\left(2x^2+4x+x+2\right)\)

\(=\left(x-3\right)\left[2x\left(x+2\right)+\left(x+2\right)\right]\)

\(=\left(x-3\right)\left(x+2\right)\left(2x+1\right)\)