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a/ \(\frac{1}{3}.3^x+3^{x+2}=3^{16}+3^{13}\)
\(\Leftrightarrow3^{x-1}+3^{x+2}=3^{13}+3^{16}\)
\(\Leftrightarrow3^{x-1}\left(1+3^3\right)=3^{13}\left(1+3^3\right)\)
\(\Leftrightarrow3^{x-1}=3^{13}\Rightarrow x-1=13\Rightarrow x=14\)
b/ \(\frac{1}{6}6^x+6^{x+2}=6^{15}+6^{18}\)
\(\Leftrightarrow6^{x-1}+6^{x+2}=6^{15}+6^{18}\)
\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^{15}\left(1+6^3\right)\)
\(\Rightarrow x=16\)
c/ \(\frac{1}{2}2^{x+3}-2^x=2^{22}-2^{20}\)
\(\Leftrightarrow2^x\left(2^2-1\right)=2^{20}\left(2^2-1\right)\)
\(\Rightarrow x=20\)
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.
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. \(x:3\dfrac{1}{15}=1\dfrac{1}{2}\)
\(x:\dfrac{46}{15}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}.\dfrac{46}{15}=\dfrac{23}{5}\)
b. \(x.\dfrac{3}{2}=-\dfrac{7}{6}\)
\(x=-\dfrac{7}{6}:\dfrac{3}{2}=-\dfrac{7}{9}\)
c. \(\dfrac{5}{6}+\dfrac{1}{4}:x=-\dfrac{2}{3}\)
\(\dfrac{13}{12}:x=-\dfrac{2}{3}\)
\(x=\dfrac{13}{12}:\left(-\dfrac{2}{3}\right)=-\dfrac{13}{8}\)
Còn lại tương tự thôi
\(\)
1, \(x\left(x+\dfrac{2}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)
2, a, \(\left|x+\dfrac{4}{6}\right|\ge0\)
Để \(\left|x+\dfrac{4}{6}\right|\) đạt GTNN thì \(\left|x+\dfrac{4}{6}\right|=0\)
\(\Leftrightarrow x+\dfrac{4}{6}=0\Rightarrow x=\dfrac{-2}{3}\)
Vậy, ...
b, \(\left|x-\dfrac{1}{3}\right|\ge0\)
Để \(\left|x-\dfrac{1}{3}\right|\) đạt GTLN thì \(\left|x-\dfrac{1}{3}\right|=0\)
\(\Leftrightarrow x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)
Vậy, ...
1)
a)
\(x\cdot\left(x+\dfrac{2}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2)
a)
\(\left|x+\dfrac{4}{6}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x+\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{-4}{6}\Leftrightarrow x=\dfrac{-2}{3}\)
Vậy \(Min_{\left|x+\dfrac{4}{6}\right|}=0\text{ khi }x=\dfrac{-2}{3}\)
b)
\(\left|x-\dfrac{1}{3}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(Min_{\left|x-\dfrac{1}{3}\right|}=0\text{ khi }x=\dfrac{1}{3}\)
a: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
\(\Leftrightarrow2^x=2^{10}\cdot5:\dfrac{5}{2}=2^{10}\cdot5\cdot\dfrac{2}{5}=2^{11}\)
=>x=11
b: \(\Leftrightarrow3^x\cdot\dfrac{1}{3}+3^x\cdot9=3^{13}\cdot28\)
\(\Leftrightarrow3^x=3^{13}\cdot28:\dfrac{28}{3}=3^{14}\)
hay x=14
Bài 2:
1: =>5x+1=6/7 hoặc 5x+1=-6/7
=>5x=-1/7 hoặc 5x=-13/7
=>x=-1/35 hoặc x=-13/35
2: =>x-1=4
=>x=5
3: =>3x-1=3
=>3x=4
=>x=4/3
4: \(\Leftrightarrow\dfrac{5}{x+3}=\dfrac{-5}{6}+\dfrac{1}{2}=\dfrac{-5+3}{6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
=>x+3=-15
=>x=-18
7: \(\Leftrightarrow2^{2x+1}+2^{2x+6}=264\)
=>2^2x+1*(1+2^5)=264
=>2^2x+1=8
=>2x+1=3
=>x=1
9: =>x^4=8x
=>x^4-8x=0
=>x=2
a/ Đề?
b/ \(\frac{1}{6}6^x+6^{x+2}=6^{10}+6^7\)
\(\Leftrightarrow6^{x-1}+6^{x+2}=6^{10}+6^7\)
\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^7\left(1+6^3\right)\)
\(\Leftrightarrow6^{x-1}=6^7\Rightarrow x-1=7\Rightarrow x=8\)
c/ Hoàn toàn tương tự câu trên:
\(2^{x-1}+2^{x+1}=2^{12}+2^{10}\)
\(\Leftrightarrow2^{x-1}\left(1+2^2\right)=2^{10}\left(1+2^2\right)\)
\(\Leftrightarrow x=11\)
1)
a.\(\dfrac{1}{5}+x=\dfrac{13}{50}\)
\(\Leftrightarrow x=\dfrac{13}{50}-\dfrac{1}{5}=\dfrac{13-10}{50}=\dfrac{3}{50}\)
b.\(\dfrac{1}{6}-x=\dfrac{5}{12}\)
\(\Leftrightarrow x=\dfrac{1}{6}-\dfrac{5}{12}=\dfrac{2-5}{12}=-\dfrac{3}{12}=-\dfrac{1}{4}\)
c.\(x\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow x\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{4}.\left(-\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
d.\(x:\dfrac{7}{11}=\dfrac{9}{33}\)
\(\Leftrightarrow x=\dfrac{9}{33}.\dfrac{7}{11}=\dfrac{3}{11}.\dfrac{7}{11}=\dfrac{21}{121}\)
e.\(\dfrac{3}{5}.x=-\dfrac{21}{10}\)
\(\Leftrightarrow x=-\dfrac{21}{10}:\dfrac{3}{5}=-\dfrac{21}{10}.\dfrac{5}{3}=-\dfrac{7}{2}\)
câu a) mình chịu (dùng kiến thức lớp 12 chắc làm đc haha)
b) gt ⇒ \(\frac{1}{6}.6^{x+2}-6^x=6^{14}-6^{13}\)
⇒ \(6^{x+1}-6^x=6^{14}-6^{13}\)
⇒ \(6^x\left(6-1\right)=6^{13}\left(6-1\right)\)
⇒ \(x=13\)
c) gt ⇒ \(\frac{1}{2}.2^{x+4}-2^x=2^{13}-2^{10}\)
⇒ \(2^{x+3}-2^x=2^{13}-2^{10}\)
⇒ \(2^x\left(2^3-1\right)=2^{10}\left(2^3-1\right)\)
⇒ \(x=10\)
d) gt ⇒ \(\frac{1}{3}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^{x+3}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)
⇒ \(x=13\)
câu d chưa có đóng ngoặc kìa bn