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\(\Leftrightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)^2+\dfrac{1}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{2}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
1.
\(\left(\dfrac{-2}{3}\right).0,75+1\dfrac{2}{3}:\left(\dfrac{-4}{9}\right)+\left(\dfrac{-1}{2}\right)^2\)
\(=\left(\dfrac{-2}{3}\right).\dfrac{3}{4}+\dfrac{5}{3}.\left(\dfrac{9}{-4}\right)+\dfrac{1}{4}\)
\(=-\dfrac{1}{2}+\dfrac{45}{-12}+\dfrac{1}{4}\)
\(=-\dfrac{6}{12}+\dfrac{-45}{12}+\dfrac{3}{4}\)
\(=\dfrac{-48}{12}\)
\(=-4\)
2.
a) \(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{10}{20}\)
\(\Leftrightarrow x=\dfrac{-11}{20}\)
b) \(\left|x-\dfrac{2}{5}\right|+\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{11}{4}-\dfrac{3}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=-2\Rightarrow x=-2+\dfrac{2}{5}=\dfrac{-8}{5}\\x-\dfrac{2}{5}=2\Rightarrow x=2+\dfrac{2}{5}=\dfrac{12}{5}\end{matrix}\right.\)
3.
a) \(\dfrac{16}{2^n}=2\)
\(\Leftrightarrow2^n=16:2\)
\(\Leftrightarrow2^n=8\)
\(\Leftrightarrow2^n=2^3\)
\(\Leftrightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{81}=-27\)
\(\Leftrightarrow\left(-3\right)^n=\left(-27\right).81\)
\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^3.\left(-3\right)^4\)
\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^7\)
\(\Leftrightarrow n=7\)
4. Ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\) (1)
\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\) (2)
Từ (1) và (2) suy ra \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Vì \(x-y+x=-49\) ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)
Vậy \(\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
1: =>1/3:x=3/5-2/3=9/15-10/15=-1/15
=>x=-1/3:1/15=5
2: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)
=>x*2/3=-1
=>x=-3/2
3: \(\Leftrightarrow\dfrac{8}{3}:x=\dfrac{25}{12}:\dfrac{-3}{50}=\dfrac{25}{12}\cdot\dfrac{-50}{3}\)
hay x=-48/625
9: =>x=-2*3/1,5=-4
8: =>2/3:x=5/2:-3/10=5/2*(-10)/3=-50/6=-25/3
=>x=-2/3:25/3=-2/3*3/25=-2/25
Giải:
\(\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2-\dfrac{1}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=-\dfrac{3}{40}+\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}\)
\(\Leftrightarrow\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}:\dfrac{1}{2}\)
\(\Leftrightarrow\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{5}=\dfrac{1}{2}\\\dfrac{1}{3}x-\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{7}{10}\\\dfrac{1}{3}x=-\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{21}{10}\\x=-\dfrac{9}{10}\end{matrix}\right.\)
Vậy ...
Chúc bạn học tốt!
\(\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2-\dfrac{1}{5}=-\dfrac{3}{40}\\ \dfrac{1}{2}\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=-\dfrac{3}{40}+\dfrac{1}{5}\\ \dfrac{1}{2}\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}\\ \left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}:\dfrac{1}{2}\\\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{4}\\ \left(\dfrac{1}{3}x-\dfrac{1}{5}\right)=\left(\pm\dfrac{1}{2}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{5}=\dfrac{1}{2}\\\dfrac{1}{3}x-\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x=\dfrac{7}{10}\\\dfrac{1}{3}x=\dfrac{3}{10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{21}{10}\\x=\dfrac{9}{10}\end{matrix}\right. \)
Vậy \(x=\dfrac{21}{10}\) hoặc \(x=\dfrac{9}{10}\)
a) \(\left[\left(\dfrac{3}{5}\right)^2-\left(\dfrac{2}{5}\right)^2\right]\cdot X=\left(\dfrac{1}{5}\right)^3\)
\(\left(\dfrac{3}{5}-\dfrac{2}{5}\right)\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\cdot X=\dfrac{1}{125}\)
\(\dfrac{1}{5}\cdot1\cdot X=\dfrac{1}{125}\)
\(X=\dfrac{1}{125}:\dfrac{1}{5}=\dfrac{1}{25}\)
b) \(1\dfrac{2}{5}\cdot x+\dfrac{3}{7}=\dfrac{-4}{5}\)
\(1\dfrac{2}{5}\cdot x=\dfrac{-4}{5}-\dfrac{3}{7}\)
\(1\dfrac{2}{5}\cdot x=-\dfrac{43}{35}\)
\(x=-\dfrac{43}{35}:1\dfrac{2}{5}=-\dfrac{43}{49}\)
c) \(\left(3x-2\right)^2=9\)
*Nếu \(9=3^2\) thì:
\(3x-2=3\)
\(3x=5\Rightarrow x=\dfrac{5}{3}\)
*Nếu \(9=\left(-3\right)^2\) thì
\(3x-2=-3\)
\(3x=-1\Rightarrow x=-\dfrac{1}{3}\)
d) \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Chúc bạn học giỏi.
a)\(\dfrac{3^2-2^2}{5^2}.x=\dfrac{1}{5^3}\)
\(\Leftrightarrow\dfrac{5}{5^2}.x=\dfrac{1}{5^3}\)
\(\Leftrightarrow\dfrac{1}{5}.x=\dfrac{1}{5^3}\)
\(\Leftrightarrow x=\dfrac{1}{25}\)
b)\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{7}{5}x=-\dfrac{43}{35}\)
\(\Leftrightarrow x=\dfrac{-43}{49}\)
c)\(9x^2-12x+4=9\)
\(\Leftrightarrow9x^2-12x-5=0\)
\(\Leftrightarrow9x^2-15x+3x-5=0\)
\(\Leftrightarrow3x\left(3x-5\right)+3x-5=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
d)\(\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
a) |x - 4| = 6
\(\Rightarrow\) \(\pm\)x = \(\pm\)10
\(\Rightarrow\) x = {10, -10}
b) \(\dfrac{x}{5}=\dfrac{9}{7}\)
\(\Rightarrow7x=45\)
\(\Rightarrow x=\dfrac{45}{7}\)
Ta có:
9 - (2/3 - x) = -5/4
2/3 - x = 9 - (-5/4)
2/3 - x = 9 + 5/4
2/3 + x = 41/4
x = 41/4 - 2/3
x = 123/12 - 8/12
x = 115/12