\(pt\Leftrightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+...=0\)

\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)

\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)

Do \(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}>0\) nên 50 - x = 0 hay x = 50.

pt<=>29-x/21+1+27-x/23+1+...=0

<=>50-x/21+50-x/23+50-x/25+50-x/27+50-x/29=0

<=>(50-x).(1/21+1/23+1/25+1/27+1/29)=0

Do 1/21+1/23+1/25+1/27+1/29>0 nên 50-x=0 hay x=50

5 ( năm )

a, <=> (x-5/100) -1 +(x-4/101) -1 +(x-3/102) -1= (x-100/5) -1+(x-101/4) -1 +(x-102/3) -1
<=> (x-105)(1/100 +1/101 +1/102)= (x-105)(1/5+1/4+1/3)
<=> (x-105)(1/100+1/101+1/102-1/5-1/4-1/3)=0
vì 1/100+1/101+1/102-1/5-1/4-1/3 khác 0 <=> x-105=0
<=> x=105

b, 29-x/21 +1+27-x/23 +1+25-x/25 +1+23-x/27 +1+21-x/29 +1=0
<=> 50-x/21 +50-x/23 +50-x/25 +50-x/27 +50-x/29=0
<=> (50-x)(1/21 +1/23 +1/25 +1/27 +1/29)=0
vì 1/21+1/23+1/25+1/27+1/29 lớn hơn 0
nên 50-x=0
<=> x=50

a) \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)

\(=>\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)

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

\(=>4\left(2x+1\right)\left(x-2\right)=0\)

\(=>\orbr{\begin{cases}2x+1=0\\x-2=0\end{cases}}\)

\(=>\orbr{\begin{cases}x=-\frac{1}{2}\\x=2\end{cases}}\)

b)\(x^3-\frac{x}{49}=0=>x\left(x^2-\frac{1}{49}\right)=0=>x\left(x-\frac{1}{7}\right)\left(x+\frac{1}{7}\right)=0\)

\(=>x=0\)hoặc \(x=\frac{1}{7}\) hoặc \(x=-\frac{1}{7}\)

a)\(\(\left(3x-1\right)^2-\left(x+3\right)^2=0\)\)

\(\(\Leftrightarrow\left(3x-1-x-3\right)\left(3x-1+x+3\right)=0\)\)

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

\(\(\Leftrightarrow\orbr{\begin{cases}2x-4=0\\4x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}}\)\)

b)\(\(x^3-\frac{x}{49}=0\)\)

\(\(\Leftrightarrow\frac{49x^3-x}{49}=0\)\)

\(\(\Leftrightarrow x\left(49x^2-1\right)=0\)\)

\(\(\Leftrightarrow\orbr{\begin{cases}x=0\\49x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(7x-1\right)\left(7x+1\right)=0\end{cases}}}\)\)\

\(\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7};x=-\frac{1}{7}\end{cases}}\)\)

c)\(\(x^2-7x+12=0\)\)

\(\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)\)

\(\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}}\)\)

d) \(\(4x^2-3x-1=0\)\)

\(\(\Leftrightarrow4x^2-4x+x-1=0\)\)

\(\(\Leftrightarrow4x\left(x-1\right)+\left(x-1\right)=0\)\)

\(\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)\)

\(\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\4x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{4}\end{cases}}}\)\)

e) Tham khảo tại : [Toán 8]Giải phương trình | Cộng đồng học sinh Việt Nam - HOCMAI Forum

https://diendan.hocmai.vn/threads/toan-8-giai-phuong-trinh.290061/

_Y nguyệt_

a thiếu đề bạn nhé

b) \(x^3-\frac{x}{49}=0\) 

\(\Rightarrow x\left(x^2-\frac{1}{49}\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x^2-\frac{1}{49}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}}\) 

Vậy.........

c)  \(x^2-7x++12=0\)  

\(\Rightarrow\left(x-3,5\right)^2-0,5^2=0\) 

\(\Rightarrow\left(x-3,5+0,5\right)\left(x-3,5-0,5\right)=0\) 

\(\Rightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}}\) 

Vậy.....

d) \(4x^2-3x-1=0\) 

\(\Rightarrow4x^2-3x+0,5625-1,5625=0\)

\(\Rightarrow\left(2x-0,75\right)^2-1,25^2=0\) 

\(\Rightarrow\left(2x-0,75+1,25\right)\left(2x-0,75-1,25\right)=0\) 

\(\Rightarrow\orbr{\begin{cases}2x+0,5=0\\2x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-0,25\\x=1\end{cases}}}\) 

Vậy.....

1) \(A=\frac{7}{10\times11}+\frac{7}{11\times12}+\frac{7}{12\times13}+...+\frac{7}{69\times70}\)

    \(A=7\times\left(\frac{1}{10\times11}+\frac{1}{11\times12}+\frac{1}{12\times13}+...+\frac{1}{69\times70}\right)\)

    \(A=7\times\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

    \(A=7\times\left(\frac{1}{10}-\frac{1}{70}\right)\)

   \(A=7\times\frac{3}{35}\)

   \(A=\frac{3}{5}\)

2) \(B=\frac{1}{25\times27}+\frac{1}{27\times29}+\frac{1}{29\times31}+...+\frac{1}{73\times75}\)

    \(B=\frac{1}{2}\times\left(\frac{2}{25\times27}+\frac{2}{27\times29}+\frac{2}{29\times31}+...+\frac{2}{73\times75}\right)\).

    \(B=\frac{1}{2}\times\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

    \(B=\frac{1}{2}\times\left(\frac{1}{25}-\frac{1}{75}\right)\)

    \(B=\frac{1}{2}\times\frac{2}{75}\)

    \(B=\frac{1}{75}\)

3) \(C=\frac{4}{2\times4}+\frac{4}{4\times6}+\frac{4}{6\times8}+...+\frac{4}{2008\times2010}\)

    \(C=\frac{4}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+...+\frac{2}{2008\times2010}\right)\)

    \(C=2\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

    \(C=2\times\left(\frac{1}{2}-\frac{1}{2010}\right)\)

    \(C=2\times\frac{502}{1005}\)

    \(C=\frac{1004}{1005}\)

_Chúc bạn học tốt_

ai làm ơn trả lời hộ mình câu này với

a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x=105\)
b) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)

\(pt\Leftrightarrow\frac{29}{21}-\frac{x}{21}+\frac{27}{23}-\frac{x}{23}+\frac{25}{25}-\frac{x}{25}+\frac{23}{27}-\frac{x}{27}+\frac{21}{29}-\frac{x}{29}=-5\Leftrightarrow-x\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=-5-\frac{29}{21}-\frac{27}{23}-\frac{25}{25}-\frac{23}{27}-\frac{21}{29}\Leftrightarrow-x=\frac{-5-\frac{29}{21}-\frac{27}{23}-\frac{25}{25}-\frac{23}{27}-\frac{21}{29}}{\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}}=-50\Leftrightarrow x=50\\ \Rightarrow S=\left\{50\right\}\)

E = ( x - 29 ) / 1970 + ( x - 27 ) / 1972 + ( x - 25 ) / 1974 + ( x - 23 ) / 1976 + ( x - 21 ) / 1978 + ( x - 19 ) / 1980 = ( x - 1970 ) / 29 + ( x - 1972 ) / 27 + ( x - 1974 ) / 25 + ( x - 1976 ) / 23 + ( x - 1978 ) / 21 + ( x - 1980 ) / 19

( Trừ từng số hạng cho 1 ra như sau )

E = (x - 1999)/ 1970 + ( x - 1999 ) / 1972 + ( x - 1999) / 1974 + ( x - 1999)/ 1976 + ( x -1999) / 1978 + ( x - 1999)/ 1980 = ( x - 1999)/29 + ( x - 1999) / 27 + ( x - 1999 ) / 25 + ( x - 1999) / 23 + ( x - 1999)/21 + ( x - 1999) / 19

< = > ( x - 1999 ) / 1970 + ( x - 1999 ) / 1972 + ( x - 1999 ) / 1974 + ( x - 1999) / 1976 + ( x - 1999) / 1978 + ( x - 1999) / 1980 - ( x - 1999) / 29 - ( x - 1999)/ 27 - ( 1 - 1999) / 25 - ( x-1999) / 23 - ( x - 1999) / 21 - ( x - 1999) / 19 = 0 ( chuyển vế )

< = > ( x - 1999 ) ( 1/1970 + 1/ 1972 + 1/1974 + 1/1976 + 1/1978 + 1/1980 - 1/29 - 1/27 - 1/25 - 1/23 - 1/21 - 1/19) = 0

Vì ( 1/1970 + 1/1972 + 1/1974 + 1/1976 + 1/1978 + 1/1980 - 1/29 -1/27 - 1/25 - 123 - 1/21 - 1/19 ) khác 0 nên để đẳng thức bằng 0 thì bắt buộc x - 1999 = 0

< = > x = 0 + 1999 = 1999

Vậy tập nghiệm của phương trình là S = { 1999 }

`[158-x]/31+[185-x]/29+[208-x]/27+[227-x]/25=10`

`<=>[158-x]/31-1+[185-x]/29-2+[208-x]/27-3+[227-x]/25-4=0`

`<=>[127-x]/21+[127-x]/29+[127-x]/27+[127-x]/25=0`

`<=>(127-x)(1/21+1/29+1/27+1/25)=0`

`<=>127-x=0`

`<=>x=127`