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2.(x-3)+3x+0.5=\(\dfrac{3}{4}\)
4x+2+4x=272
(1,2-5x).(2\(\dfrac{1}{8}\) +1/2 x)=0
GIÚP MÌNH VỚI !!!!
\(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\\ \Leftrightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\\ \Leftrightarrow x\left(2+3\right)=\dfrac{3}{4}-\dfrac{1}{2}+6\\ \Leftrightarrow5x=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{25}{4}:5=\dfrac{5}{4}\\ ---\\ 4^{x+2}+4^x=272\\ \Leftrightarrow4^x\left(4^2+1\right)=272\\ \Leftrightarrow4^x.17=272\\ \Leftrightarrow4^x=\dfrac{272}{17}=16=4^2\\ Vậy:x=2\\ ----\\ \left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1,2-5x=0\\2,125+0,5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1,2\\0,5x=-2,125\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}=0,24\\x=\dfrac{-2,125}{0,5}=-4,25\end{matrix}\right.\)
a) \(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\)
\(\Rightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow5x-6=\dfrac{3}{4}-\dfrac{1}{2}\)
\(\Rightarrow5x-6=\dfrac{1}{4}\)
\(\Rightarrow5x=\dfrac{1}{4}+6\)
\(\Rightarrow5x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:5\)
\(\Rightarrow x=\dfrac{5}{4}\)
b) \(4^{x+2}+4^x=272\)
\(\Rightarrow4^x\cdot4^2+4^x\cdot1=272\)
\(\Rightarrow4^x\cdot\left(16+1\right)=272\)
\(\Rightarrow4^x\cdot17=272\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
c) \(\left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}1,2-5x=0\\\dfrac{15}{8}+\dfrac{1}{2}x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x=1,2\\\dfrac{1}{2}x=-\dfrac{15}{8}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}\\x=-\dfrac{15}{8}:\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{25}\\x=-\dfrac{15}{4}\end{matrix}\right.\)
Mọi người ơi giúp mik nhanh lên, mai mik KT rồi, mik lo quá ak, nghe nói là đê khó lắm!!!
\(âP\left(x\right)=13x^3+4x^2-11x-2\)
\(b.Q\left(x\right)=x^3+9x-5\)
\(c.A\left(x\right)=14x^3-x^2+10x+14\)
\(d.B\left(x\right)=2x^2+x+3\)
3 − 2 2 x − 3 = 2 5 + 2 9 − 6 x − 3 2 3 − 2 2 x − 3 = 2 5 − 2 6 x − 9 − 3 2 3 − 2 2 x − 3 = 2 5 + 2 3 2 x − 3 − 3 2 2 3 2 x − 3 − 2 2 x − 3 = 2 5 − 3 2 − 3 2 − 6 3 2 x − 3 = 4 − 15 − 30 10 − 4 3 2 x − 3 = − 41 10 4 3 2 x − 3 = 41 10 4.10 = 41.3. 2 x − 3 40 = 123. 2 x − 3 2 x − 3 = 40 123 2 x = 40 123 + 3 2 x = 40 + 369 123 2 x = 409 123 x = 409 246
\(-\dfrac{15}{23}:\dfrac{22x}{7}=-\dfrac{14x}{11}:\left(13+\dfrac{4}{5}\right)\)
=>\(-\dfrac{15}{23}\cdot\dfrac{7}{22x}=\dfrac{-14x}{11}:\dfrac{69}{5}\)
=>\(-\dfrac{105}{23\cdot22x}=\dfrac{-70x}{11\cdot69}\)
=>\(\dfrac{-3}{2x}=\dfrac{-2x}{3}\)
=>\(4x^2=9\)
=>\(x^2=\dfrac{9}{4}\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(\dfrac{-15}{23}:\dfrac{22x}{7}=\dfrac{-14x}{11}:\left(13+\dfrac{4}{5}\right)\)
\(\Leftrightarrow-\dfrac{15}{23}\cdot\dfrac{7}{22x}=\dfrac{-14x}{11}:\dfrac{69}{5}\)
=>\(\dfrac{-15\cdot7}{23\cdot22x}=\dfrac{-14x}{11}\cdot\dfrac{5}{69}\)
=>\(\dfrac{5\cdot3\cdot7}{23\cdot2\cdot11x}=\dfrac{2\cdot7x}{11}\cdot\dfrac{5}{3\cdot23}\)
=>\(\dfrac{3}{2x}=\dfrac{2x}{3}\)
=>\(4x^2=9\)
=>\(x^2=\dfrac{9}{4}\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
a) \(\left|-\frac{2}{11}+\frac{3}{22}x\right|-\frac{1}{2}=\frac{5}{7}\)
=> \(\left|-\frac{2}{11}+\frac{3}{22}x\right|=\frac{17}{14}\)
=> \(\orbr{\begin{cases}-\frac{2}{11}+\frac{3}{22}x=\frac{17}{14}\\-\frac{2}{11}+\frac{3}{22}x=-\frac{17}{14}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{215}{21}\\x=-\frac{53}{7}\end{cases}}\)
b) \(-\frac{7}{8}x-5\frac{3}{4}=3\)
=> \(-\frac{7}{8}x-\frac{23}{4}=3\)
=> \(-\frac{7}{8}x=3+\frac{23}{4}=\frac{35}{4}\)
=> \(x=\frac{35}{4}:\left(-\frac{7}{8}\right)=\frac{35}{4}\cdot\left(-\frac{8}{7}\right)=-10\)
c) \(2x+\left(-\frac{2}{7}\right)-7=-11\)
=> \(2x-\frac{2}{7}-7=-11\)
=> \(2x=-11+7+\frac{2}{7}=-\frac{26}{7}\)
=> \(x=\left(-\frac{26}{7}\right):2=-\frac{13}{7}\)
d) \(\frac{3}{7}+x:\frac{14}{15}=\frac{1}{2}\)
=> \(x:\frac{14}{15}=\frac{1}{2}-\frac{3}{7}=\frac{1}{14}\)
=> \(x=\frac{1}{14}\cdot\frac{14}{15}=\frac{1}{15}\)
\(2^{2x}+4^{x+2}=272\)
\(\Rightarrow4^x+4^x\cdot4^2=272\)
\(\Rightarrow4^x\left(1+4^2\right)=272\)
\(\Rightarrow4^x\cdot17=272\)
\(\Rightarrow4^x=16\)
\(4^x=4^2\)
\(\Rightarrow x=2\)
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