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cảm ơn Nguyễn Thu Trang nha những ng đó hack nik tr khi olm đc đổi ms cơ lên....buồn
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Câu 2:
\(a,x+\dfrac{2}{5}=\dfrac{1}{6}\) \(b,\dfrac{1}{3}-\dfrac{5}{3}.x=\dfrac{2}{7}\)
\(x=\dfrac{1}{6}-\dfrac{2}{5}\) \(\dfrac{5}{3}.x=\dfrac{1}{3}-\dfrac{2}{7}\)
\(x=-\dfrac{7}{30}\) \(\dfrac{5}{3}.x=\dfrac{1}{21}\)
Vậy \(x=-\dfrac{7}{30}\) \(x=\dfrac{1}{21}:\dfrac{5}{3}\)
\(x=\dfrac{1}{35}\)
Vậy \(x=\dfrac{1}{35}\)
a, ⇒ \(x=\dfrac{1}{6}-\dfrac{2}{5}\)
\(\Rightarrow x=\dfrac{5}{30}-\dfrac{12}{30}\)
\(\Rightarrow x=-\dfrac{7}{30}\)
b, \(\Rightarrow\dfrac{5}{3}x=\dfrac{1}{3}-\dfrac{2}{7}\)
⇒ \(\dfrac{5}{3}x=\dfrac{7}{21}-\dfrac{6}{21}\)
\(\Rightarrow\dfrac{5}{3}x=\dfrac{1}{21}\)
\(\Rightarrow x=\dfrac{1}{21}:\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{1}{21}.\dfrac{3}{5}=\dfrac{1}{35}\)
c, \(\Rightarrow\left|x-\dfrac{2}{5}\right|=3+\dfrac{1}{4}\)
\(\Rightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{13}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=\dfrac{13}{4}\\x-\dfrac{2}{5}=-\dfrac{13}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{73}{20}\\x=-\dfrac{57}{20}\end{matrix}\right.\)
d, ⇒ x.x = 2.8
⇒ x2 = 16
⇒ x2 = (\(\pm4\))2
⇒ x = \(\pm4\)
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Bài 11:
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{20}}\)
=>\(3\cdot A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{19}}\)
=>\(3\cdot A-A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{19}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{19}}-\dfrac{1}{3^{20}}\)
=>\(2A=1-\dfrac{1}{3^{20}}=\dfrac{3^{20}-1}{3^{20}}\)
=>\(A=\dfrac{3^{20}-1}{2\cdot3^{20}}\)
Bài 6:
a: ĐKXĐ: x>=-2
\(\sqrt{x+2}>=0\forall x\) thỏa mãn ĐKXĐ
=>\(\sqrt{x+2}+2>=2\forall x\) thỏa mãn ĐKXĐ
=>\(A>=2\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x+2=0
=>x=-2
Vậy: \(A_{min}=2\) khi x=-2
b: ĐKXĐ: x>=-5
\(\sqrt{x+5}>=0\forall x\) thỏa mãn ĐKXĐ
=>\(5\sqrt{x+5}>=0\forall x\)thỏa mãn ĐKXĐ
=>\(5\sqrt{x+5}-\dfrac{3}{5}>=-\dfrac{3}{5}\forall x\) thỏa mãn ĐKXĐ
=>\(B>=-\dfrac{3}{5}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x+5=0
=>x=-5
vậy: \(B_{min}=-\dfrac{3}{5}\) khi x=-5
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left|x+\frac{1}{3}\right|=0\)
\(\Rightarrow x+\frac{1}{3}=0\)
\(\Rightarrow x=\frac{-1}{3}\)
b) \(\left|x\right|-\frac{3}{5}=\frac{5}{9}\)
\(\Rightarrow\left|x\right|=\frac{52}{45}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{52}{45}\\x=\frac{-52}{45}\end{cases}}\)
c) \(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=\frac{-5}{4}\end{cases}}\)
d) \(\left(\frac{-5}{9}\right)^{10}:x=\left(\frac{-5}{9}\right)^8\)
\(\Rightarrow\left(\frac{-5}{9}\right)^{10}:\left(\frac{-5}{9}\right)^8=x\)
\(\Rightarrow x=\left(\frac{-5}{9}\right)^2=\frac{25}{81}\)
e) \(x:\left(-\frac{5}{9}\right)^8=\left(-\frac{5}{9}\right)^8\)
\(\Rightarrow x=\left(-\frac{5}{9}\right)^8.\left(-\frac{5}{9}\right)^8\)
\(\Rightarrow x=\left(-\frac{5}{9}\right)^{16}=\left(\frac{5}{9}\right)^{16}\)
f) x3 = -8
=> x3 = ( -2 )3
=> x = -2
g) ( x + 5 )3 = -27
=> ( x + 5 )3 = ( -3 )3
=> x + 5 = -3
=> x = -8
h) ( 2x - 3 )3 = -64
=> ( 2x - 3 )3 = ( -4 )3
=> 2x - 3 = -4
=> 2x = -1
=> x = -0,5
i) ( 2x - 3 )2 = 25
=> ( 2x - 3 )2 = 52 = ( -5 )2
\(\Rightarrow\orbr{\begin{cases}2x-3=5\\2x-3=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=8\\2x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)