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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 ~
mk chưa học nhưng mk nghĩ là bỏ phân số đi thay bằng phép tính rồi nhóm zô
=>\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)\left(\frac{x+5}{324}+1\right)=-4+4\)
=>\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}=0\)
=>\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\right)=0\)
Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\)
=>x+329=0
=>x=-329
thêm mỗi số hạng 1 đơn vị
\(1=\frac{327}{327}=\frac{326}{326}=\frac{325}{325}=\frac{324}{324}\)
Cứ thế sẽ ra (x + 329) (1/327 + 1/326 + 1/325 + 1/324) = -4
Cứ làm theo sẽ ra x = -656
Mong bạn là người đầu tiên k mk
k người hôm trước , Hôm sau k người
Câu 1
4 p/s cộng thêm 1,p/s cuối trừ 4 rồi nhóm vs nhau
d/s la x= - 329
Câu 2
NHân vs 7 thành 7S rồi rút gọn là đc
Câu 1 :
a) \(\Leftrightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Dễ thấy \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\) \(\Rightarrow x+329=0\Rightarrow x=-329\)
a) ta có:
\(|2x-6|+5x=9\Leftrightarrow|2x-6|=9-5x\)
\(2x-6=9-5x\Leftrightarrow7x=15\Leftrightarrow x=\frac{15}{7}\)
\(2x-6=5x-9\Leftrightarrow3x=3\Leftrightarrow x=1\)
b) Ta có:
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329}{5}+4=0\)
\(\Leftrightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
do \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)nên \(x+329=0\Leftrightarrow x=-329\)
Vậy ............................................. chúc bn hok tốt ^-^
\(a,\frac{x+5}{2010}+\frac{x+6}{2009}+\frac{x+7}{2008}=-3\)
\(\Rightarrow\left(\frac{x+5}{2010}+1\right)+\left(\frac{x+6}{2009}+1\right)+\left(\frac{x+7}{2008}+1\right)=0\)
\(\Rightarrow\frac{x+2016}{2010}+\frac{x+2016}{2009}+\frac{x+2006}{2008}=0\)
chỉ bt lm v thoi "(
a) \(\frac{x+5}{2010}+\frac{x+6}{2009}+\frac{x+7}{2008}=-3\)
<=> \(\frac{x+5}{2010}+1+\frac{x+6}{2009}+1+\frac{x+7}{2008}+1=0\)
<=> \(\frac{x+2015}{2010}+\frac{x+2015}{2009}+\frac{x+2015}{2008}=0\)
<=> \(\left(x+2015\right)\left(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)
<=> \(x+2015=0\) (do 1/2010 + 1/2009 + 1/2008 # 0 )
<=> \(x=-2015\)
Vậy...
b) mạo phép chỉnh đề
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+344}{5}=0\)
<=> \(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+344}{5}-3=0\)
<=> \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{5}=0\)
làm tương tự a
\(\Rightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)=0\)
\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}=0\)
\(\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\right)=0\)
MÀ\(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\)
NÊN \(x+329=0\)
\(\Rightarrow x=-329\)
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
a) -325