Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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)\(\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
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.
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
b, \(-x-2=\dfrac{5}{4}\Rightarrow-x=\dfrac{13}{4}\Rightarrow x=-\dfrac{13}{4}\)
c, \(\dfrac{4}{3}-\left(x-\dfrac{1}{5}\right)=\left|-\dfrac{3}{10}+\dfrac{1}{2}\right|-\dfrac{1}{6}\)
\(\Rightarrow\dfrac{4}{3}-x+\dfrac{1}{5}=\left|\dfrac{1}{5}\right|-\dfrac{1}{6}\)
\(\Rightarrow-x=\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{4}{3}-\dfrac{1}{5}\)
\(\Rightarrow-x=-\dfrac{3}{2}\Rightarrow x=\dfrac{3}{2}\)
d, \(\dfrac{1}{3}-\left(\dfrac{2}{3}-x+\dfrac{5}{4}\right)=\dfrac{7}{12}-\left(\dfrac{5}{2}-\dfrac{13}{6}\right)\)
\(\Rightarrow\dfrac{1}{3}-\dfrac{2}{3}+x-\dfrac{5}{4}=\dfrac{7}{12}-\dfrac{5}{2}+\dfrac{13}{6}\)
\(\Rightarrow x=\dfrac{7}{12}-\dfrac{5}{2}+\dfrac{13}{6}-\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{11}{6}\)
Chúc bạn học tốt!!!
a/ \(\dfrac{1}{3}-\dfrac{2}{5}+3x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1}{15}+3x=\dfrac{3}{4}\)
\(\Leftrightarrow3x=\dfrac{49}{60}\)
\(\Leftrightarrow x=\dfrac{49}{180}\)
Vậy....
b/ \(\dfrac{3}{2}-1+4x=\dfrac{2}{3}-7x\)
\(\Leftrightarrow\dfrac{1}{2}+4x=\dfrac{2}{3}-7x\)
\(\Leftrightarrow4x+7x=\dfrac{2}{3}-\dfrac{1}{2}\)
\(\Leftrightarrow11x=\dfrac{1}{6}\)
\(\Leftrightarrow x=\dfrac{1}{66}\)
Vậy....
c/ \(2\left(\dfrac{3}{4}-5x\right)=\dfrac{4}{5}-3x\)
\(\Leftrightarrow\dfrac{3}{2}-10x=\dfrac{4}{5}-3x\)
\(\Leftrightarrow-10x+3x=\dfrac{4}{5}-\dfrac{3}{2}\)
\(\Leftrightarrow-7x=-\dfrac{7}{10}\)
\(\Leftrightarrow x=-\dfrac{1}{10}\)
Vậy .....
d/ \(4\left(\dfrac{1}{2}-x\right)-5\left(x-\dfrac{3}{10}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow2-4x-5x-\dfrac{3}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow2+\left(-4x\right)+\left(-5x\right)+\left(\dfrac{-3}{2}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow-9x+\dfrac{1}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow-9x=\dfrac{5}{4}\)
\(\Leftrightarrow x=-\dfrac{5}{36}\)