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(x + 2) / 327 + (x + 3) / 326 + (x + 4) / 325 + (x + 5) / 324 + (x + 349) / 5 = 0
<=> (x + 2) / 327 +1+ (x + 3) / 326 +1+ (x + 4) / 325 +1+ (x + 5) / 324 +1+ (x + 349) / 5 -4 = 0
<=> (x+ 329)/327 + (x+ 329)/326 + (x+ 329)/325 + (x+ 329)//324 + (x+ 329)/5 =0
<=> (x+ 329).(1/327 + 1/ 326 + 1/325 + 1/324 +1/5) =0
Do (1/327 + 1/ 326 + 1/325 + 1/324 +1/5) >0 nên x+ 329 =0 => x= -329
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\frac{x+2+327}{327}+\frac{x+3+326}{326}+\frac{x+4+325}{325}+\frac{x+5+324}{324}+\frac{x+349-20}{5}=0\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\)
\(\Rightarrow x+329=0\)\(\Leftrightarrow x=-329\)
Vậy \(x=-329\)
\(\frac{x+2}{327}\) +\(\frac{x+3}{326}\) +\(\frac{x+4}{325\ }+\frac{x+5}{324}+\frac{x+349}{5}=0\)
=> \(\left(\frac{x+2\ }{327}+1\right)+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}\)+1+(\(\frac{x+349}{5}\) - 4) = 0
\(\frac{x+329}{327}\) + \(\frac{x+329}{326}+\frac{x+329}{325}\) + \(\frac{x+329}{324}\) +\(\frac{x+329\ }{5\ }\) = 0
(x+329).(1/327+1/326+1/325+1/324+1/5) = 0
=> x + 329 = 0
x = -329
có mấy chỗ mk quên đóng ngoặc bn sửa giúp mk nak
Ta co
\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5}-4\)=0
\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\)
\(\Rightarrow x+329=0\Rightarrow x=-329\)
Vay \(x=-329\)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5}-4=0\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\left[x+329\right]\cdot\left[\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right]=0\)
\(\Leftrightarrow x+329=0\) vì \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\)
\(\Leftrightarrow x=-329\)
Vậy x = -329
Trả lời:
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5}-4=0\)
\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow x+329=0\Leftrightarrow x=-329\)
Vậy \(x=-329\)
~ Học tốt ~
\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\frac{x+329}{5}=0\)
\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
(x + 329) . (1/327 + 1/326 + 1/325 + 1/324 + 1/5) = 0
1/327 + 1/326 + 1/325 + 1/324 + 1/5 khác 0
< = > x + 329 = 0
x= -329
ta co
x+2/327 +1+x+3/326+1+x+4/325+x+5/324+x+349/5 -4=0
x+329/327+x+329/326+x+329/325+x+329/324+x+329/5=0
(x+329)(1/327+1/326+1/325+1/324+1/5)=0
x+329=0 (vì 1/327+1/326+1/325+1/324+1/5 khác 0)
x=-329
mk chưa học nhưng mk nghĩ là bỏ phân số đi thay bằng phép tính rồi nhóm zô
ta có:\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\left(1+\frac{x+2}{327}\right)+\left(1+\frac{x+3}{326}\right)+\left(1+\frac{x+4}{327}\right)+\left(1+\frac{x+5}{324}\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\left(x+329\right)=0\left(vì\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)>0\right)\)
\(x+329=0\Rightarrow x=-329\)
b) \(\left|5x-3\right|-x=7\)
\(\Rightarrow\left|5x-3\right|=7+x\)
\(\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-\left(7+x\right)\end{cases}\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}\Rightarrow}\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{2}{3}\end{cases}}}\)
Vậy ....................
Bạn ơi !!! ý A tham khảo tại link này nè :
https://h.vn/hoi-dap/question/394208.html
~ Học tốt ~
Bạn ơi ko vào được
P/s : Thấy câu a nó hơi bị sai ạ :D Sửa đề cho đúng hơn nhé :)) Chứ theo đề cũ thì tớ chịu
a) \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329}{5}+4=0\)
\(\Leftrightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\frac{x+329}{5}=0\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)
\(\Leftrightarrow x+329=0\)
\(\Leftrightarrow x=-329\)
b) \(\left|5x-3\right|-x=7\)
\(\Leftrightarrow\left|5x-3\right|=7+x\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}\Leftrightarrow}\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{2}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{5}{2};-\frac{2}{3}\right\}\)
Thank tất cả nha
1) a) Ta có : \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}+4=0\)
=> \(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}\right)+\left(\frac{x+5}{324}\right)+\frac{x+349}{5}=0\)
=> \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
=> \(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
=> \(x+329=0\left(\text{vì }\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
=> x = - 329
3) Ta có : \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\Rightarrow\frac{20+xy}{4x}=\frac{1}{8}\)
=> 8(20 + xy) = 4x
=> 2(20 + xy) = x
=> 40 + 2xy = x
=> 40 + 2xy - x = 0
=> 2xy - x = - 40
=> x(2y - 1) = - 40
Vì \(x;y\inℤ;x\ne0\Rightarrow2y-1\inℤ\)
Khi đó 40 = 1.40 = (-1).(-40) = 2.20 = (-2).(-20) = (-5).(-8) = 5.8 = 4.10 = (-4).(-10)
Lập bảng xét 16 trường hợp
Vậy các cặp (x;y) thỏa mãn là (40 ; 1) ; (8 ; 3) ; (-40 ; 0)
Lê Minh Sơn ghi sai đề câu a
Phải là :(x + 329)/5 không phải (x + 349)/5