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 ~
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 \(x\in\left\{\frac{5}{2};-\frac{2}{3}\right\}\)
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
| x | 1 | 40 | 8 | 5 | 20 | 2 | 10 | 4 | -1 | -40 | -2 | -20 | -5 | -8 | -4 | -10 |
| 2y - 1 | 40 | 1 | 5 | 8 | 2 | 20 | 4 | 10 | -40 | -1 | -20 | -2 | -8 | -5 | -10 | -4 |
| y | 41/2 | 1(tm) | 3(tm) | 4,5 | 1,5 | 10,5 | 2,5 | 5,5 | -39/2 | 0(tm) | -9,5 | -0,5 | -3,5 | -2 | -4,5 | -1,5 |
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
