\(\left(x-\frac{1}{2}\right)\cdot\frac{5}{2}+\frac{1}{2}=\frac{7}{4}\)
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( 1/7 . x - 2/7 ) . ( -1.5 . x + 3/5 ) . ( 1/ 3 . x + 4/3) + 0
<=> +) 1/7 . x - 2/7 = 0 +) (- 1 / 5) . x +3/5 = 0 +) 1/ 3 . x + 4/ 3 = 0
x = 2 x = 3 x = 4
Vậy x = 2 : x = 3 ; x=4
Vi \(\left(\frac{1}{7}x-\frac{2}{7}\right)\cdot\left(-\frac{1}{5}x+\frac{3}{5}\right)\cdot\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{7}x-\frac{2}{7}=0\\-\frac{1}{5}x+\frac{3}{5}=0\\\frac{1}{3}x+\frac{4}{3}=0\end{cases}\Rightarrow\hept{\begin{cases}\frac{1}{7}x=\frac{2}{7}\\-\frac{1}{5}x=-\frac{3}{5}\\\frac{1}{3}x=-\frac{4}{3}\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\x=3\\x=-4\end{cases}}}\)
Vậy \(x\in\left\{-4;3;2\right\}\)
\(\Rightarrow\frac{1}{7}x-\frac{2}{7}=0\text{ hoặc }-\frac{1}{5}x+\frac{3}{5}=0\text{ hoặc }\frac{1}{3}x+\frac{4}{3}=0\)
\(\Rightarrow x=2\text{ hoặc }x=3\text{ hoặc }x=-4\)
Vậy tập nghiệm của pt là \(S=\left\{2;3;-4\right\}\)
A= \(\left(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}\right)+\left(\frac{-6}{13}+\frac{1}{2}+\frac{4}{3}\right)\)
A= \(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}-\frac{6}{13}+\frac{1}{2}+\frac{4}{3}\)
A= \(\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{7}{13}+\frac{6}{13}\right)-\left(\frac{1}{3}-\frac{4}{3}\right)\)
A= \(1-1-\left(-1\right)\)
A= \(1\)
B= \(0,75+\frac{2}{5}+\left(\frac{1}{9}-\frac{7}{5}+\frac{5}{4}\right)\)
B= \(\frac{3}{4}+\frac{2}{5}+\frac{1}{9}-\frac{7}{5}+\frac{5}{4}\)
B= \(\left(\frac{3}{4}+\frac{5}{4}\right)+\left(\frac{2}{5}-\frac{7}{5}\right)+\frac{1}{9}\)
B= \(2-1+\frac{1}{9}\)
B= \(\frac{9}{9}+\frac{1}{9}\)
B= \(\frac{10}{9}\)
C= \(\left(\frac{-3}{2}.\frac{4}{3}\right).\left(\frac{-9}{2}\right)-\frac{1}{4}\)
C = \(-2.\left(\frac{-9}{2}\right)-\frac{1}{4}\)
C = \(9-\frac{1}{4}\)
C = \(\frac{36}{4}-\frac{1}{4}\)
C = \(\frac{35}{4}\)
D = \(\frac{5}{4}.\left(\frac{-7}{10}.\frac{5}{4}-\frac{7}{8}.\frac{7}{10}\right)\)
D = \(\frac{5}{4}.\left(\frac{-7}{8}-\frac{49}{80}\right)\)
D = \(\frac{-35}{32}-\frac{49}{64}\)
D = \(\frac{-70}{64}-\frac{49}{64}\)
D = \(\frac{-119}{64}\)
k mk nha ^_^
kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
Ta có \(\left(x-\frac{1}{2}\right).\frac{5}{2}+\frac{1}{2}=\frac{7}{4}\)
\(\Rightarrow\left(x-\frac{1}{2}\right).\frac{5}{2}=\frac{7}{4}-\frac{1}{2}\)
\(\Rightarrow\left(x-\frac{1}{2}\right).\frac{5}{2}=\frac{5}{4}\)
\(\Rightarrow x-\frac{1}{2}=\frac{5}{4}\div\frac{5}{2}\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow x=1\)
(x-\(\frac{1}{2}\)).\(\frac{5}{2}\)+\(\frac{1}{2}\)=\(\frac{7}{4}\)
(x-\(\frac{1}{2}\)).\(\frac{5}{2}\)=\(\frac{7}{4}\)- \(\frac{1}{2}\)
(x-\(\frac{1}{2}\)).\(\frac{5}{2}\)= \(\frac{7}{4}\)- \(\frac{2}{4}\)
(x-\(\frac{1}{2}\)).\(\frac{5}{2}\)\(\frac{5}{2}\)= \(\frac{5}{4}\)
(x-\(\frac{1}{2}\)) = \(\frac{5}{4}\): \(\frac{5}{2}\)= \(\frac{5}{4}\). \(\frac{2}{5}\)= \(\frac{1}{2}\)
x = \(\frac{1}{2}\)+ \(\frac{1}{2}\)=\(\frac{1}{4}\)
vậy x\(\frac{7}{4}\)\(\frac{7}{4}\)