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Bạn tham khảo nhé
\(a)\) \(\frac{x-1}{2003}+\frac{x-2}{2002}+\frac{x-3}{2001}-3=0\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2003}-1\right)+\left(\frac{x-2}{2002}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(-3+3\right)=0\)
\(\Leftrightarrow\)\(\frac{x-2004}{2003}+\frac{x-2004}{2002}+\frac{x-2004}{2001}=0\)
\(\Leftrightarrow\)\(\left(x-2004\right)\left(\frac{1}{2003}+\frac{1}{2002}+\frac{1}{2001}\right)=0\)
Vì \(\frac{1}{2003}+\frac{1}{2002}+\frac{1}{2001}\ne0\)
\(\Rightarrow\)\(x-2004=0\)
\(\Rightarrow\)\(x=2004\)
Vậy \(x=2004\)
Chúc bạn học tốt ~
\(b)\) \(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}=-4\)
\(\Leftrightarrow\)\(\left(\frac{315-x}{101}+1\right)+\left(\frac{313-x}{103}+1\right)+\left(\frac{311-x}{105}+1\right)+\left(\frac{309-x}{107}+1\right)=-4+4\)
\(\Leftrightarrow\)\(\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\)\(\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)
\(\Rightarrow\)\(416-x=0\)
\(\Rightarrow\)\(x=416\)
Vậy \(x=416\)
Chúc bạn học tốt ~
a)Mk ko hiểu làm gì có y đâu
b)Ta có:\(\frac{x-4}{y-3}=\frac{4}{3}\)
\(\Rightarrow3x-12=4y-12\)
\(\Rightarrow3x-4y=0\)
Mà \(x-y=5\Rightarrow x=5+y\)
Do đó:\(3\left(5+y\right)-4y=0\)
\(\Rightarrow15+3y-4y=0\)
\(\Rightarrow15-y=0\)
\(\Rightarrow y=15\)
Do đó:x=20
a) \(\frac{x}{7}=\frac{9}{7}\Rightarrow x=9\)
b) \(\frac{x-4}{y-3}=\frac{4}{3}\Rightarrow3\left(x-4\right)=4\left(y-3\right)\)
\(\Rightarrow3x-12=4y-12\)
\(\Rightarrow3x=4y\)
\(\Rightarrow3x=3y+y\)
\(\Rightarrow3x-3y=y\)
\(\Rightarrow3\left(x-y\right)=y\)
\(\Rightarrow3.5=y\)
\(\Rightarrow y=15\)
\(\Rightarrow x-15=5\)
\(\Rightarrow x=5+15\)
\(\Rightarrow x=20\)
Vậy \(y=15,x=20\)
a) \(\frac{3}{4}.x+40\%=\frac{-1}{4}\)
\(\frac{3}{4}.x+\frac{2}{5}=\frac{-1}{4}\)
\(\frac{3}{4}.x=\frac{-13}{20}\)
\(x=\frac{-13}{15}\)
Vậy \(x=\frac{-13}{15}\)
c) \(|x-1|=2^3+\left(-5\right)\)
\(|x-1|=3\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)
Vậy \(x\in\left\{-2;4\right\}\)
Giải:
Theo bài ra ta có:
\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{12}\)
\(\Rightarrow-3\le x\le\frac{23}{12}\)
\(\Rightarrow x\varepsilon\left\{-2;-1;0;1\right\}\)
\(\frac{-5}{6}+\frac{16}{6}+-\frac{29}{6}\le x\le\frac{-6}{12}+\frac{24}{12}+\frac{5}{12}\)
=>-3\(\le\) x\(\le\) 23/12
=> x thuộc{-2-1;0;1}
\(x+y+y+z+z+x=2\left(x+y+z\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}\)
\(\Rightarrow x+y+z=\frac{13}{24}\Rightarrow x=\frac{5}{24};y=\frac{7}{24};z=\frac{1}{24}\)
vậy \(\left(x;y;z\right)=\left(\frac{5}{24};\frac{7}{24};\frac{1}{24}\right)\)
x + y + z =( 1/2 + 1/3 + 1/4 ) : 2
= 13/24
từ đó tính x y z