1/5-|1/5-x|=1/5.Giúp mik với
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\(\frac{2}{5}\times\frac{1}{2}-\frac{2}{5}\times\frac{1}{3}-\frac{2}{5}\times\)\(\frac{1}{6}\)
\(=\frac{2}{5}\times\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\frac{2}{5}\times\frac{0}{6}\)
\(=\frac{2}{5}\times0\)
\(=0\)
\(\frac{2}{5}\times\frac{1}{2}-\frac{2}{5}\times\frac{1}{3}-\frac{2}{5}\times\frac{1}{6}\)
\(=\frac{2}{5}\times\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\frac{2}{5}\times0\)
\(=0\)
\(3\cdot\left(\dfrac{1}{2}-x\right)-5\left(x-\dfrac{1}{10}\right)=-\dfrac{7}{4}\\ \Leftrightarrow-8x=-\dfrac{15}{4}\\ \Leftrightarrow x=\dfrac{15}{32}\)
\(4\left(\dfrac{3}{4}+x\right)-\dfrac{1}{5}=7x-\dfrac{1}{7}\\ \Leftrightarrow3x=\dfrac{103}{35}\\ \Leftrightarrow x=\dfrac{103}{105}\)
\(3\left(\dfrac{1}{2}-x\right)-5\left(x-\dfrac{1}{10}\right)=-\dfrac{7}{4}\)
\(\Rightarrow\dfrac{3}{2}-3x-5x+\dfrac{1}{2}=-\dfrac{7}{4}\)
\(\Rightarrow2-8x=-\dfrac{7}{4}\)
\(\Rightarrow8x=2+\dfrac{7}{4}\)
\(\Rightarrow8x=\dfrac{15}{4}\)
\(\Rightarrow x=\dfrac{15}{32}\)
___________
\(4\left(\dfrac{3}{4}+x\right)-\dfrac{1}{5}=7x-\dfrac{1}{7}\)
\(\Rightarrow3+4x-\dfrac{1}{5}=7x-\dfrac{1}{7}\)
\(\Rightarrow4x+\dfrac{14}{5}=7x-\dfrac{1}{7}\)
\(\Rightarrow7x-4x=\dfrac{14}{5}+\dfrac{1}{7}\)
\(\Rightarrow3x=\dfrac{103}{35}\)
\(\Rightarrow x=\dfrac{103}{105}\)
\(x\) x\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)= 55
\(x\) x \(\dfrac{11}{20}\) = 55
\(x\) = 55 : \(\dfrac{11}{20}\)
\(x\) =100
c: Ta có: \(\dfrac{2}{5}\cdot\left[\left(\dfrac{3}{5}\right)^2:\left(-\dfrac{1}{5}\right)^2-7\right]\cdot\left(1000\right)^0\cdot\left|-\dfrac{11}{15}\right|\)
\(=\dfrac{2}{5}\cdot\left(\dfrac{9}{25}:\dfrac{1}{25}-7\right)\cdot1\cdot\dfrac{11}{15}\)
\(=\dfrac{2}{5}\cdot\dfrac{11}{15}\cdot2\)
\(=\dfrac{44}{75}\)
3-/x/=5=>/x/=-2(vô lý) /2-x/=5=>2-x=5 hoặc 2-x=-5
=>ko có giá trị nào của x thỏa mãn điều kiện =>x=-3 =>x=7
/x+3/=0=>x+3=o=>x=-3 /x+5/+3=1=>/x+5/=-2(vô lý)
/x-3/=1=>x-3=1 hoặc x-3=-1 =>ko có giá trị nào của x thỏa mãn điều kiện
x-3=-1=>x=2
a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
⇔\(7\left(x-3\right)=5\left(x+5\right)\)
⇔\(7x-21=5x+25\)
⇔\(7x-21-5x-25=0\)
⇔\(2x-46=0\)
⇔\(2x=46\)
⇔\(x=23\)
a) 3/8 . x = 9/8 - 1
3/8 . x = 1/8
x = 1/8 : 3/8
x = 1/3
b) 4/5 . x = 7/5 - 1/5
4/5 . x = 6/5
x = 6/5 : 4/5
x = 3/2
c) 12/7 : x + 2/3 = 7/5
12/7 : x = 7/5 - 2/3
12/7 : x = 11/15
x = 12/7 : 11/15
x = 180/77
d) 3.(x + 7) - 15 = 27
3.(x + 7) = 27 + 15
3.(x + 7) = 42
x + 7 = 42 : 3
x + 7 = 14
x = 14 - 7
x = 7
a) \(\dfrac{3}{8}x=\dfrac{9}{8}-1\)
\(\Rightarrow\dfrac{3}{8}x=\dfrac{1}{8}\)
\(\Rightarrow x=\dfrac{1}{8}:\dfrac{3}{8}=\dfrac{1}{3}\)
b) \(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{1}{5}\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{6}{5}\)
\(\Rightarrow x=\dfrac{6}{5}:\dfrac{4}{5}=\dfrac{3}{2}\)
c) \(\dfrac{12}{7}:x+\dfrac{2}{3}=\dfrac{7}{5}\)
\(\Rightarrow\dfrac{12}{7}:x=\dfrac{7}{5}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{12}{7}:x=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{12}{7}:\dfrac{11}{15}=\dfrac{180}{77}\)
d) \(3\left(x+7\right)-15=27\)
\(\Leftrightarrow3\left(x+7\right)=42\)
\(\Leftrightarrow x+7=14\Leftrightarrow x=7\)
Ta có: \(\dfrac{1}{5}-\left|\dfrac{1}{5}-x\right|=\dfrac{1}{5}\)
\(\Leftrightarrow\left|\dfrac{1}{5}-x\right|=0\)
\(\Leftrightarrow x-\dfrac{1}{5}=0\)
hay \(x=\dfrac{1}{5}\)