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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/ \(4\left(\dfrac{1}{2}-x\right)-5\left(x-\dfrac{3}{10}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow2-4x-5x-\dfrac{3}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow2+\left(-4x\right)+\left(-5x\right)+\left(\dfrac{-3}{2}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow-9x+\dfrac{1}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow-9x=\dfrac{5}{4}\)
\(\Leftrightarrow x=-\dfrac{5}{36}\)
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
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: \(C=\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\)
Dấu '=' xảy ra khi x=-1 và y=1/3
b: \(\left(2x-1\right)^2+3>=3\)
Do đó: D<=5/3
Dấu '=' xảy ra khi x=1/2
2: (3x-4)^2+2>=2
=>5/(3x-4)^2+2<=5/2
=>B>=-5/2
Dấu = xảy ra khi x=4/3
4: D=(3x^2+7-4)/(3x^2+7)=1-4/3x^2+7
3x^2+7>=7
=>4/3x^2+7<=4/7
=>-4/3x^2+7>=-4/7
=>D>=3/7
Dấu = xảy ra khi x=0
2) B = \(\dfrac{-5}{\left(3x-4\right)^2+2}\)
Ta có: ( 3x-4)2 \(\ge\) 0 , \(\forall\) x
=> ( 3x-4)2 +2 \(\ge\) 2, \(\forall\) x
=> \(\dfrac{1}{\left(3x-4\right)^2+2}\) \(\le\) \(\dfrac{1}{2}\) , \(\forall\) x
=> \(\dfrac{-5}{\left(3x-4\right)^2+2}\) \(\ge\) \(\dfrac{-5}{2}\) , \(\forall\) x
=> B \(\ge\) \(\dfrac{-5}{2}\)
Vậy B đạt GTNN khi bằng \(\dfrac{-5}{2}\)
Dấu "= " xảy ra khi 3x - 4 = 0
4) D=\(\dfrac{3x^2+3}{3x^2+7}\)
= 1 - \(\dfrac{4}{3x^2+7}\)
Ta có: 3x2 \(\ge\) 0, \(\forall\) x
=> 3x2 +7 \(\ge\) 7, \(\forall\) x
=> \(\dfrac{1}{3x^2+7}\) \(\le\) \(\dfrac{1}{7}\)
=> \(\dfrac{4}{3x^2+7}\) \(\le\) \(\dfrac{4}{7}\)
=> 1 - \(\dfrac{4}{3x^2+7}\) \(\ge\) \(\dfrac{3}{7}\)
Vậy D đạt GTNN khi bằng \(\dfrac{3}{7}\)
Dấu "=" xảy ra khi x = 0