Có gì sai sai đấy ạ, cho xin hỏi là có chép sai đề ko ạ?

\(\dfrac{x-2000}{22}\) +  \(\dfrac{x-2005}{17}\) + \(\dfrac{x}{674}\) = 5

\(\dfrac{x-2000}{22}\) + \(\dfrac{x-2005}{17}\) + \(\dfrac{x}{674}\) - 5 = 0

(\(\dfrac{x-2000}{22}\)  - 1) + (\(\dfrac{x-2005}{17}\) - 1) + (\(\dfrac{x}{674}\) - 3) = 0

\(\dfrac{x-2022}{22}\) + \(\dfrac{x-2022}{17}\) + \(\dfrac{x-2022}{674}\)  = 0

(\(x\) - 2022).(\(\dfrac{1}{22}\) + \(\dfrac{1}{17}\) + \(\dfrac{1}{647}\)) = 0 

Vì \(\dfrac{1}{22}\) + \(\dfrac{1}{17}\) + \(\dfrac{1}{647}\) > 0

Nên  \(x\) - 2022 = 0

         \(x\)            = 2022

Vậy \(x\)            = 2022

\(a.\dfrac{x-2}{2000}+\dfrac{x-3}{1999}=\dfrac{x-4}{1998}+\dfrac{x-5}{1997}\\ \Leftrightarrow\dfrac{x-2}{2000}-1+\dfrac{x-3}{1999}-1=\dfrac{x-4}{1998}-1+\dfrac{x-5}{1997}-1\\ \Leftrightarrow\dfrac{x-2}{2000}-\dfrac{2000}{2000}+\dfrac{x-3}{1999}-\dfrac{1999}{1999}=\dfrac{x-4}{1998}-\dfrac{1998}{1998}+\dfrac{x-5}{1997}-\dfrac{1997}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}=\dfrac{x-2002}{1998}+\dfrac{x-2002}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}-\dfrac{x-2002}{1998}-\dfrac{x-2002}{1997}=0\\ \Leftrightarrow\left(x-2002\right)\left(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\right)=0\\ \)

\(Do:\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\ne0\\ \Rightarrow x-2002=0\\ \Leftrightarrow x=2002\\ Vậy:S=\left\{2002\right\}\)

Mấy câu khác tương tự :v

b: \(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)

=>123-x=0

=>x=123

c: \(\Leftrightarrow\dfrac{x-2}{2017}+1=\dfrac{x-1}{2018}+\dfrac{x}{2019}\)

\(\Leftrightarrow\left(\dfrac{x-2}{2017}-1\right)=\left(\dfrac{x-1}{2018}-1\right)+\left(\dfrac{x}{2019}-1\right)\)

=>x-2019=0

=>x=2019

ĐKXĐ : a;b;c  \(\ne0\)

Ta có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2000}\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}-\dfrac{1}{a}\)

\(\Leftrightarrow\dfrac{b+c}{bc}=\dfrac{-\left(b+c\right)}{a\left(a+b+c\right)}\)

\(\Leftrightarrow\left(b+c\right)\left(\dfrac{1}{bc}+\dfrac{1}{a\left(a+b+c\right)}\right)=0\)

\(\Leftrightarrow\left(b+c\right).\dfrac{a\left(a+b+c\right)+bc}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left(b+c\right).\dfrac{a^2+ab+ac+bc}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\dfrac{\left(b+c\right)\left(a+b\right)\left(a+c\right)}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b+c=0\\a+b=0\\a+c=0\end{matrix}\right.\left(1\right)\)

Từ (1) kết hợp a + b + c = 2000 ta được điều phải chứng minh

1,

\(S=-\dfrac{7}{20}-\dfrac{7}{200}-\dfrac{7}{2000}-\dfrac{7}{20000}\\ =-\dfrac{7}{20}\left(1+\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}\right)\\ =-\dfrac{7}{20}\left(\dfrac{1000+100+10+1}{1000}\right)\\ =-\dfrac{7}{20}\cdot\dfrac{1111}{1000}\\ =\dfrac{7777}{20000}\)

2,

a, \(Tacó:\\ 9^{2000}=\left(3^2\right)^{2000}=3^{4000}\\ \Rightarrow9^{2000}=3^{4000}\)

b,

\(2^{225}=\left(2^{15}\right)^{15}=32768^{15}\\ 3^{150}=\left(3^{10}\right)^{15}=59049^{15}\\ Vì32768< 59049nên32768^{15}< 59049^{15}\\ \Rightarrow2^{225}< 3^{150}\)

3,

\(\left|x-7\right|=x-7\\ Vì\left|x-7\right|\ge0\forall x\\ \Rightarrow x-7\ge0\forall x\\ \Leftrightarrow x-7\ge0\\ \Leftrightarrow x\ge7\\ Vậyx\ge7\)

Bài 3:

\(\left|x-7\right|=x-7\)

Khi giá trị tuyệt đối của \(x-7\) bằng chính nó, thì \(x-7\) phải \(\ge0\)

Suy ra: \(x-7\ge0\Rightarrow x\ge7\)

Vậy \(x\ge 7\)

B

`=>` `B`

Vì các phân số cồn lại thuộc dạng `(x >= 1)`

B

Bài 1:

a) Ta có: \(13A=\dfrac{13^{16}+13}{13^{16}+1}=1+\dfrac{12}{13^{16}+1}\)

\(13B=\dfrac{13^{17}+13}{13^{17}+1}=1+\dfrac{12}{13^{17}+1}\)

Vì \(\dfrac{12}{13^{16}+1}>\dfrac{12}{13^{17}+1}\Rightarrow1+\dfrac{12}{13^{16}+1}>1+\dfrac{12}{13^{17}+1}\)

\(\Rightarrow13A>13B\)

\(\Rightarrow A>B\)

Vậy A > B

b) Ta có: \(1999C=\dfrac{1999^{2000}+1999}{1999^{2000}+1}=1+\dfrac{1998}{1999^{2000}+1}\)

\(1999D=\dfrac{1999^{1999}+1999}{1999^{1999}+1}=1+\dfrac{1998}{1999^{1999}+1}\)

\(\dfrac{1998}{1999^{2000}+1}< \dfrac{1998}{1999^{1999}+1}\Rightarrow1+\dfrac{1998}{1999^{2000}+1}< 1+\dfrac{1999}{1999^{1999}+1}\)

\(\Rightarrow1999C< 1999D\)

\(\Rightarrow C< D\)

Vậy C < D

   0,75 + \(\dfrac{9}{5}\) ( 1,5 - \(\dfrac{2}{3}\) )²

= 0,75 + \(\dfrac{9}{5}\) ( \(\dfrac{3}{2}\) - \(\dfrac{2}{3}\))²

= 0,75 + \(\dfrac{9}{5}\) (\(\dfrac{5}{6}\))²

= 0,75 + \(\dfrac{5}{4}\)

= 0,75 + 1,25

= 2 

\(\dfrac{-22}{25}\) + ( \(\dfrac{22}{7}\) - 0,12)

= \(\dfrac{-22}{25}\) + ( \(\dfrac{22}{7}\) - \(\dfrac{3}{25}\))

= \(\dfrac{-22}{25}\) + \(\dfrac{22}{7}\) - \(\dfrac{3}{25}\)

= - ( \(\dfrac{22}{25}\) + \(\dfrac{3}{25}\)) + \(\dfrac{22}{7}\)

= -1 + \(\dfrac{22}{7}\)

= \(\dfrac{-7}{7}\) + \(\dfrac{22}{7}\)

= \(\dfrac{15}{7}\)

a; 0,75 +9/5(1,5 - 2,3) = 

cvAjcfajshcfaSsjhcfzsjhcvZXVxhjcvzhjc

Bài 1: 

a: \(\left|2x-1\right|+3>=3\)

Dấu '=' xảy ra khi x=1/2

b: \(=-\left|8x+1\right|+1\le1\)

Dấu '=' xảy ra khi x=-1/8

đề là j vậy bạn 

a) \(\dfrac{9}{11}-\dfrac{3}{11}=\dfrac{9-3}{11}=\dfrac{6}{11}\)

b) \(\dfrac{10}{4}-\dfrac{5}{4}=\dfrac{10-5}{4}=\dfrac{5}{4}\)

c) \(\dfrac{22}{15}-\dfrac{8}{15}=\dfrac{22-8}{15}=\dfrac{14}{15}\)