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47 x 98 + 94
= 47 x 98 + 47 x 2
= 47 x ( 98 + 2 )
= 47 x 100
= 4700
học tốt ^^
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\(16\cdot4^{x+1}=64\)
\(\Leftrightarrow4^{x+1}=4\)
\(\Leftrightarrow x+1=1\)
\(\Leftrightarrow x=0\)
Ta có: 16 x \(4^{x+1}\)=64
Nên \(4^{x+1}\) =64 : 16 = 4=4\(4^1\)
Suy ra x+1 =1 =>x = 0
hc tot nha
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1:
1: \(A=\dfrac{3}{14}+\dfrac{2}{7}-\dfrac{3}{7}=\dfrac{3}{14}-\dfrac{1}{7}=\dfrac{3}{14}-\dfrac{2}{14}=\dfrac{1}{14}\)
2: \(\Leftrightarrow-\dfrac{2}{3}\cdot x=-\dfrac{2}{3}\)
=>x=1
3:
a: Cùng phía với điểm I: C,K,B
b: Tia đối của tia CK là CI hoặc CA
c: AC=2*1=2cm
=>CB=8-2=6cm
KB=6/2=3cm
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Bạn tách ra bớt được không ạ? (Tại vì để nhiều như vậy thì... mọi người hơi ngại làm á bạn :333)
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= \(\left(6-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{8}{5}:\dfrac{1}{4}\)
= \(\dfrac{16}{5}.\dfrac{25}{8}-\dfrac{32}{5}\)
=10-\(\dfrac{32}{5}\)
=\(\dfrac{18}{5}\)
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\(150-5\left(x-2\right)^2=25\)
\(5\left(x-2\right)^2=150-25=125\)
\(\left(x-2\right)^2=125:5=25\)
\(\Rightarrow\orbr{\begin{cases}x-2=5\\x-2=-5\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}}\)
Bài 1:
\(a)\left(x+\dfrac{2}{3}\right)^3=\dfrac{125}{64}.\\ \Leftrightarrow\left(x+\dfrac{2}{3}\right)^3=\left(\dfrac{5}{4}\right)^3.\\ \Rightarrow x+\dfrac{2}{3}=\dfrac{5}{4}.\\ \Leftrightarrow x=\dfrac{7}{12}.\)
\(b)\left(x-\dfrac{1}{2}\right)^3=\dfrac{8}{343}.\\\Leftrightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{7}\right) ^3.\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{2}{7}.\\ \Leftrightarrow x=\dfrac{11}{14}.\)
Bài 2:
\(a)\left(x-\dfrac{1}{3}\right)^2=\dfrac{25}{9}.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{3}\right)^2=\left(\dfrac{5}{3}\right)^2.\\\left(x-\dfrac{1}{3}\right)^2=\left(\dfrac{-5}{3}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{5}{3}.\\x-\dfrac{1}{3}=\dfrac{-5}{3}.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=\dfrac{-4}{3}.\end{matrix}\right.\)
\(b)\left(x-\dfrac{3}{4}\right)^2=\dfrac{49}{16}.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{3}{4}\right)^2=\left(\dfrac{7}{4}\right)^2.\\\left(x-\dfrac{3}{4}\right)^2=\left(\dfrac{-7}{4}\right)^2.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{7}{4}.\\x-\dfrac{3}{4}=\dfrac{-7}{4}.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}.\\x=-1.\end{matrix}\right.\)