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\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(\left(3x-\frac{3}{2}\right)-\left(5x+3\right)=-x+\frac{1}{5}\)
\(3x-\frac{3}{2}-5x-3-\frac{1}{5}=-x\)
\(-2x-6,7=-x\)
\(-2x-x=6,7\)
\(-3x=6,7\)
\(x=-\frac{67}{30}\)
Vậy \(x=-\frac{67}{30}\)
vd câu 1:
ta có x-y=4 =>x=4+y
ta có pt:
4+y/y-2=3/2
=>8+2y=3y-6
=>-y=-14
=>y=14
=>x=4+y=4+14=18
các bài khác cũng tương tự thôi bạn
=>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6+2/3
=>-7x=-1/6
=>x=1/42
\(\Leftrightarrow-5x-1-\dfrac{1}{2}x+\dfrac{1}{3}-\dfrac{3}{2}x+\dfrac{5}{6}=0\)
\(\Leftrightarrow-7x+\dfrac{1}{6}=0\)
=>7x=1/6
hay x=1/42
Câu 1:
\(\frac{1}{3}+\frac{3}{35}<\frac{x}{210}<\frac{4}{7}+\frac{3}{5}+\frac{1}{3}\)
\(\Rightarrow\frac{44}{105}<\frac{x}{210}<\frac{158}{105}\)
\(\Rightarrow\frac{88}{210}<\frac{x}{210}<\frac{316}{210}\)
\(\Rightarrow x\in\left\{89;90;91;92;...;310;311;312;313;314;315\right\}\)
Câu 3:
\(\frac{5}{3}\)\(+\frac{-14}{3}\)\(<\)\(x\)\(<\)\(\frac{8}{5}+\frac{18}{10}\)
\(\Rightarrow\)\(-9\)\(<\)\(x\)\(<\)\(3,4\)
Mà \(x\in Z\)
\(\Rightarrow x\in\left\{-8;-7;-6;-5;...;1;2;3\right\}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{5}{6}\) -\(\frac{3}{4}\) + \(\frac{2}{3}\) -\(\frac{1}{2}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{10}{12}\)-\(\frac{9}{12}\)+\(\frac{8}{12}\)-\(\frac{6}{12}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\)= \(\frac{1}{4}\)=> x. (\(\frac{1}{2}\)- \(\frac{2}{3}\) + \(\frac{3}{4}\)- \(\frac{5}{6}\)) = \(\frac{1}{4}\)=> x.( \(\frac{6}{12}\)- \(\frac{8}{12}\)+\(\frac{9}{12}\)-\(\frac{10}{12}\))= \(\frac{1}{4}\)=> x. \(\frac{-1}{4}\)=\(\frac{1}{4}\)=> x = \(\frac{1}{4}\): \(\frac{-1}{4}\)=> x = -1=>x.(1/2-2/3+3/4)=1/4
=>x.7/12=1/4
=>x=1/4:7/12
=>x=1/4.12/7
=>x=3/7
\(\Leftrightarrow\)3x-\(\frac{3}{2}\)-5x-3=-x+\(\frac{1}{5}\)
\(\Leftrightarrow\)3x-5x+x=\(\frac{1}{5}+\frac{3}{2}\)+3
\(\Leftrightarrow\)-x=4,7
\(\Leftrightarrow\)x=-4,7
tick nah