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a, Ta có : \(-\dfrac{3}{2}-2x+\dfrac{3}{4}=-2\)
\(\Rightarrow-2x=-2+\dfrac{3}{2}-\dfrac{3}{4}=-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{5}{8}\)
Vậy ...
b, Ta có : \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\left(-\dfrac{3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)
\(\Rightarrow-\dfrac{29}{6}\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)=\dfrac{2}{5}\)
\(\Rightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=-\dfrac{12}{145}\)
\(\Rightarrow-\dfrac{2}{3}x=-\dfrac{12}{145}+\dfrac{3}{5}=\dfrac{15}{29}\)
\(\Rightarrow x=-\dfrac{45}{58}\)
Vậy...
a: Xét ΔABH và ΔACH có
AB=AC
AH chung
BH=CH
Do đó: ΔABH=ΔACH
a) Ta có: \(y=f\left(x\right)=4x^2-5\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(3\right)=4.3^2-5=31\\f\left(-\dfrac{1}{2}\right)=4.\left(-\dfrac{1}{2}\right)^2-5=-4\end{matrix}\right.\)
b) Ta có: \(f\left(x\right)=-1\)
\(\Rightarrow4x^2-5=-1\)
\(\Leftrightarrow4x^2=4\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{1;-1\right\}\) thì \(f\left(x\right)=-1\)
c) \(\forall x\in R,f\left(x\right)=f\left(-x\right)\Leftrightarrow f\left(-x\right)=4.\left(-x\right)^2-5=4x^2-5=f\left(x\right)\)
Vậy \(\forall x\in R\) thì \(f\left(x\right)=f\left(-x\right)\)
\(a.f\left(3\right)=4.3^2-5=31.\\ f\left(\dfrac{-1}{2}\right)=4.\left(\dfrac{-1}{2}\right)^2-5=-4.\)
\(b.f\left(x\right)=-1.\Rightarrow4x^2-5=-1.\\ \Leftrightarrow4x^2=4.\Leftrightarrow x^2=1.\\ \Leftrightarrow x=\pm1.\)
\(c.f\left(x\right)=f\left(-x\right).\\ \Rightarrow4x^2-5=4\left(-x\right)^2-5.\\ \Leftrightarrow4x^2-5=4x^2-5.\)
\(\Leftrightarrow0x=0\) (luôn đúng).
Vậy với mọi x ∈ R thì f (x)= f (-x).
\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\dfrac{x+200}{5}+\dfrac{20}{5}-4=0\)
\(\Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\)
\(\Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\)
\(\Leftrightarrow x=-200\)( do \(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}>0\))
\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\\ \Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\left(\dfrac{x+220}{5}-4\right)=0\\ \Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\\ \Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\\ \Leftrightarrow x=-200\)
a) -6 . (- \(\dfrac{2}{3}\) ) . 0,25 = 14 . 0,25 = 3,5
b) -\(\dfrac{15}{4}\) . ( - \(\dfrac{7}{15}\) ) . ( - \(\dfrac{22}{25}\) ) = \(\dfrac{7}{4}\) . ( - \(\dfrac{22}{25}\) ) = - \(\dfrac{77}{50}\) = - 1,54
c) -\(\dfrac{21}{5}\) . ( - \(\dfrac{9}{11}\) ) . ( - \(\dfrac{11}{14}\) ) . \(\dfrac{2}{5}\) = \(\dfrac{189}{55}\) . ( - \(\dfrac{11}{14}\) ) . \(\dfrac{2}{5}\) = - \(\dfrac{297}{110}\) . \(\dfrac{2}{5}\) = - \(\dfrac{297}{275}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3>=-5\\2x-3< =5\end{matrix}\right.\Leftrightarrow-1< =x< =4\)
\(\left(-\frac{2}{3}\right)^2.x=\left(-\frac{2}{3}\right)^5\)
\(\Rightarrow x=\left(-\frac{2}{3}\right)^5:\left(-\frac{2}{3}\right)^2\)
\(\Rightarrow x=\left(-\frac{2}{3}\right)^3\)
Vậy \(x=-\frac{8}{27}\).