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3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)
=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)
=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)
=> \(x\) = \(\dfrac{2}{5}\)
4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)
=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)
=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)
=> \(3x=\dfrac{1}{9}\)
=> \(x=\dfrac{1}{9}:3\)
=> \(x=\dfrac{1}{27}\)
a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)
=>2/5x=8/5
=>x=4
b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)
=>1/3x=-6
=>x=-18
c: =>2|x-1/3|=0,24-4/5=-0,56<0
a: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5
=>x=3/5:2/3=3/5x3/2=9/10
b: \(\Leftrightarrow x\cdot2.8-50=34\)
=>2,8x=84
=>x=30
c: \(\Leftrightarrow\dfrac{1}{6}x=\dfrac{5}{12}\)
hay x=5/2
d: \(\Leftrightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}+\dfrac{7}{4}=\dfrac{41}{4}\)
=>2x-3/4=41/4 hoặc 2x-3/4=-41/4
=>2x=44/4=11 hoặc 2x=-19/2
=>x=11/2 hoặc x=-19/4
a) \(\dfrac{-3}{8}-\dfrac{1}{8}:x=-5\dfrac{1}{4}-2\dfrac{1}{4}\)
\(\dfrac{-3}{8}-\dfrac{1}{8}:x=\dfrac{-21}{4}-\dfrac{9}{4}\)
\(\dfrac{-3}{8}-\dfrac{1}{8}:x=\dfrac{-15}{2}\)
\(\dfrac{1}{8}:x=\dfrac{-3}{8}-\dfrac{-15}{2}\)
\(\dfrac{1}{8}:x=\dfrac{57}{8}\)
\(x=\dfrac{1}{8}:\dfrac{57}{8}\)
\(x=\dfrac{1}{57}\)
b) \(\left(1\dfrac{1}{3}-25\%-\dfrac{5}{12}\right)-2x=1,6:\dfrac{3}{5}\)
\(\left(1\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{5}{12}\right)-2x=\dfrac{8}{5}:\dfrac{3}{5}\)
\(\left(\dfrac{4}{3}-\dfrac{1}{4}-\dfrac{5}{12}\right)-2x=\dfrac{8}{3}\)
\(\left(\dfrac{13}{12}-\dfrac{5}{12}\right)-2x=\dfrac{8}{3}\)
\(\dfrac{2}{3}-2x=\dfrac{8}{3}\)
\(2x=\dfrac{2}{3}-\dfrac{8}{3}\)
\(2x=-2\)
\(x=\left(-2\right):2\)
\(x=-1\)
Tìm \(x\), biết :
a) \(x=\dfrac{1}{4}+\dfrac{2}{13}\)
b) \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)
a) \(x=\dfrac{1}{4}+\dfrac{2}{13}\)
=> \(x=\dfrac{13}{52}+\dfrac{8}{52}\)
=> \(x=\dfrac{21}{52}\)
b) \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)
=> \(\dfrac{x}{3}=\dfrac{14}{21}+\dfrac{-3}{21}\)
=> \(\dfrac{x}{3}=\dfrac{11}{21}\)
=> \(x=\dfrac{11}{21}\)x \(3\)
=> \(x=\dfrac{11}{7}\)
Các bạn ơi giúp mk với các bạn ơi mk sắp phải đi học rồi giúp mk với
a) \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{36}\) \(\Leftrightarrow\left(\dfrac{-1}{6}\right)^{x-1}=\dfrac{1}{36}\)
\(\Leftrightarrow\left(\dfrac{-1}{6}\right)^{x-1}=\left(\dfrac{1}{6}\right)^2\)
\(\Leftrightarrow x-1=2\Rightarrow x=3\)
b) \(\dfrac{25}{5^x}=\dfrac{1}{125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{3125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{5^5}\Rightarrow x=5\)
a) \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{36}\Leftrightarrow\left(-\dfrac{1}{6}\right)^{x-1}=\left(-\dfrac{1}{6}\right)^2\)
\(\Leftrightarrow x-1=2\Rightarrow x=2+1=3\)
b) \(\dfrac{25}{5^x}=\dfrac{1}{125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{3125}\Leftrightarrow\dfrac{25}{5^x}=\dfrac{25}{5^5}\Rightarrow x=5\)
Giờ mới đúng thật nè
a) \(x-25\%x=0,5\)
\(\dfrac{3}{4}x=0,5\)
x = \(\dfrac{2}{3}\)
b) \(\left(50\%x+5\dfrac{1}{4}\right).\dfrac{-2}{3}=2\dfrac{5}{6}\)
\(\left(0,5x+\dfrac{21}{4}\right)=\dfrac{-17}{4}\)
\(0,5x=\dfrac{-19}{2}\)
x = -19
c) \(\left(1\dfrac{1}{3}-25\%-\dfrac{5}{12}\right)+2x=1,5:\dfrac{3}{5}\)
\(\dfrac{2}{3}+2x=\dfrac{5}{2}\)
\(2x=\dfrac{11}{6}\)
x= \(\dfrac{11}{12}\)
Bài 1: Tính ( hợp lý nếu có thể )
\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)
\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)
\(=-1+1+\dfrac{2}{-5}\)
\(=0+\dfrac{2}{-5}\)
\(=\dfrac{2}{-5}\)
\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)
\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)
\(=0+\dfrac{2}{3}\)
\(=\dfrac{2}{3}\)
\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)
\(=-1+1\)
\(=0\)
\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)
\(=\dfrac{1}{6}+\dfrac{-1}{6}\)
\(=0\)
Bài 2: Tìm x,biết:
a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}-\dfrac{2}{3}\)
\(x=\dfrac{2}{15}\)
Vậy \(x=\dfrac{2}{15}\)
b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)
\(\Rightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{2}{3}\)
\(x=\dfrac{3}{3}=1\)
Vậy \(x=1\)
c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!
\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)
\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)
\(x=\dfrac{1}{44}\)
Vậy \(x=\dfrac{1}{44}\)
d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\)
\(x-25\%.x=\dfrac{1}{2}\\ x.1-\dfrac{1}{4}.x=\dfrac{1}{2}\\ x\left(1-\dfrac{1}{4}\right)=\dfrac{1}{2}\\ x.\dfrac{3}{4}=\dfrac{1}{2}\\ x=\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{1}{2}.\dfrac{4}{3}\\ x=\dfrac{2}{3}\)