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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.
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
a/ \(\dfrac{x}{9}=\dfrac{16}{x}\)
\(\Leftrightarrow x^2=9.16\)
\(\Leftrightarrow x^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right.\)
Vậy ...
b/ \(x^3+27=0\)
\(\Leftrightarrow x^3=-27\)
\(\Leftrightarrow x^3=\left(-3\right)^3\)
\(\Leftrightarrow x=-3\)
Vậy ...
c/ \(\left|x\left(x^2-\dfrac{5}{4}\right)=x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-\dfrac{5}{4}\right)=x\\x\left(x^2-\dfrac{5}{4}\right)=-x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^3-\dfrac{5}{4}x=x\\x^3-\dfrac{5}{4}x=-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^3-\left(\dfrac{5}{4}x+x\right)=0\\x^3-\left(\dfrac{5}{4}x-x\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^3-\dfrac{9}{4}x=0\\x^3-\dfrac{1}{4}x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-\dfrac{9}{4}\right)=0\\x\left(x^2-\dfrac{1}{4}\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x^2-\dfrac{9}{4}=0\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{4}=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
c/ Với mọi x ta có :
\(\left|x-5\right|=\left|5-x\right|\)
\(\Leftrightarrow\left|x+3\right|+\left|x-5\right|=\left|x+3\right|+\left|5-x\right|\)
\(\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge\left|\left(x+3\right)+\left(5-x\right)\right|\)
\(\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge\left|8\right|\)
\(\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge8\)
Dấu "=" xảy ra khi :
\(\left(x+3\right)\left(5-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0\\5-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0\\5-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3\\5\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3\\5\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-3\le x\le5\\x\in\varnothing\end{matrix}\right.\)
Vậy ...
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
Bạn tính hai vế à.!? Hay tính vế thứ nhất rồi với vế thứ 2.!???
a, \(125^3:5^7=\left(5^3\right)^3:5^7=5^9:5^7=5^2\)
b, \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{4}{49}\right)^5:\left(\dfrac{8}{343}\right)^2\)
= \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2^2}{7^2}\right)^5:\left(\dfrac{2^3}{7^3}\right)^2\)
= \(\left(\dfrac{2}{7}\right)^{18}:\left[\left(\dfrac{2}{7}\right)^2\right]^5:\left[\left(\dfrac{2}{7}\right)^3\right]^2\)
=\(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2}{7}\right)^{10}:\left(\dfrac{2}{7}\right)^6\)
= \(\left(\dfrac{2}{7}\right)^{18-10-6}=\left(\dfrac{2}{7}\right)^2\)
c, \(3-\left(\dfrac{-7}{9}\right)^0+\left(\dfrac{1}{3}\right)^5.3^5\)
= 3 - 1 +\(\left[\left(\dfrac{1}{3}\right)^5.3^5\right]\)
= 2 + 1=3
d, \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(9.5\right)^{10}.5^{20}}{\left(25.3\right)^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}\)
= \(\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5\)
c. \(^{ }\left(2x+3\right)^2=\dfrac{9}{121}\)
=> \(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
=> 2x +3 = \(\dfrac{3}{11}\) hoặc 2x+3 = \(\dfrac{-3}{11}\)
=> x= \(\dfrac{-15}{11}\) hoặc x = \(\dfrac{-18}{11}\)
d. \(\left(2x-1\right)^3=\dfrac{-8}{27}\)
=> \(\left(2x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)
=> 2x-1 = \(\dfrac{-2}{3}\)
=> x= \(\dfrac{1}{6}\)
\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)
\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)
\(\Rightarrow x=\dfrac{100}{401}\)
a) (-5/9)^10 : x = (-5/9)^8
=> x = (-5/9)^10 : (-5/9)^8
=> x = (-5/9)^10-8 = (-5/9)^2
=> x = 25/81
b ) x : (-5/9)^8 = (-9/5)^8
=> x = (-9/5)^8 . (-5/9)^8
=> x = ( (-9)^8.(-5)^8 )/(5^8 . 9^8 )
=> x = 1
C) x^3 = -8 =(-2)^3
=> x = -2
a) (-5/9)¹⁰ : x = (-5/9)⁸
x = (-5/9)¹⁰ : (-5/9)⁸
x = (-5/9)²
x = 25/81
b) x : (-5/9)⁸ = (-9/5)⁸
x = (-9/5)⁸ . (-5/9)⁸
x = [-9/5 . (-5/9)]⁸
x = 1⁸
x = 1
c) x³ = -8
x³ = (-2)³
x = -2