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Bài 1:
a)
\(\dfrac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\\ =\dfrac{\left(2^2\right)^2\cdot\left(5^2\right)^2+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^{2\cdot2}\cdot5^{2\cdot2}+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^4\cdot5^4}{2^3\cdot5^2}+\dfrac{2^5\cdot5^3}{2^3\cdot5^2}\\ =2\cdot5^2+2^2\cdot5\\ =2\cdot25+4\cdot5\\ =50+20\\ =70\)
c)
\(\dfrac{\left(1-\dfrac{4}{9}-2\right)\cdot16}{\left(2-3\right)^{-2}}+12\\ =\dfrac{\left(\dfrac{9}{9}-\dfrac{4}{9}-\dfrac{18}{9}\right)\cdot16}{\left(-1\right)^{-2}}+12\\ =\dfrac{\dfrac{-13}{9}\cdot16}{\dfrac{1}{\left(-1\right)^2}}+12\\ =\dfrac{\dfrac{-208}{9}}{1}+12\\ =\dfrac{-208}{9}+12\\ =\dfrac{-208}{9}+\dfrac{108}{9}\\ =\dfrac{100}{9}\)
Bài 2:
a)
\(\left(x+2\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
b)
\(\left(1,78^{2x-2}-1,78^x\right):1,78^x=0\\ \Leftrightarrow\dfrac{1,78^{2x-2}}{1,78^x}-\dfrac{1,78^x}{1,78^x}=0\\ \Leftrightarrow\dfrac{1,78^{2x-2}}{1,78^x}-1=0\\ \Leftrightarrow \dfrac{1,78^{2x-2}}{1,78^x}=1\\ \Leftrightarrow1,78^{2x-2}=1,78^x\\ \Leftrightarrow2x-2=x\\ \Leftrightarrow2x-x=2\\ \Leftrightarrow x=2\)
d) \(5^{\left(x-2\right)\left(x+3\right)}=1\)
\(\Rightarrow5^{\left(x-2\right)\left(x+3\right)}=5^0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(x_1=-3;x_2=2\)
a, x:(1/2)3=-1/2
x:1/8= -1/2
x= -1/2.1/8
x=-1/16
b,(3/4)5.x=(3/4)7
x=(3/4)7:(3/4)5
x= (3/4)2
c,(2/5)^8:x=(2/5)^6
x=.......
như cái trên nha lm giống thế
dễ
ai đi qua tick cho mình nha
ai tick thì may mắn trọn đời
2) x4 -16 =0 => x4 =16 => x4 = 44 hoặc (-4)4 => x = 4 hoặc -4
Bài 1:
a: \(=\dfrac{-1}{8}+1-\dfrac{9}{4}-1\)
\(=\dfrac{-1}{8}-\dfrac{18}{8}=\dfrac{-19}{8}\)
b: \(=4\cdot1-2\cdot\dfrac{1}{4}+3\cdot\dfrac{-1}{2}+1\)
\(=4-\dfrac{1}{2}-\dfrac{3}{2}+1\)
=5-2
=3
a)\(x^2\left(x+2\right)+4\left(x+2\right)=0\)
\(\Rightarrow\left(x^2+4\right)\left(x+2\right)=0\)
\(x^2+4>0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
b) \(3^{x+2}+4.3^{x+1}+3^{x-1}=6^6\)
\(\Rightarrow3^x.9+3^x.12+3^x.\dfrac{1}{3}=46656\)
\(\Rightarrow3^x\left(9+12+\dfrac{1}{3}\right)=46656\Leftrightarrow3^x.\dfrac{64}{3}=46656\Leftrightarrow3^x=2187\Leftrightarrow x=7\)
Giải:
a) \(x^2\left(x+2\right)+4\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+4\right)=0\)
Vì \(x^2+4>0;\forall x\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy ..
b) \(3^{x+2}+4.3^{x+1}+3^{x-1}=6^6\)
\(\Leftrightarrow3^{x-1+3}+4.3^{x-1+2}+3^{x-1}=6^6\)
\(\Leftrightarrow3^{x-1}\left(3^3+4.3^2+3\right)=6^6\)
\(\Leftrightarrow3^{x-1}.66=6^6\)
\(\Leftrightarrow3^{x-1}=\dfrac{6^6}{66}\)
\(\Leftrightarrow3^x-3=\dfrac{7779}{11}\)
\(\Leftrightarrow3^x=\dfrac{7809}{11}\)
Tìm x rồi kết luận
( \(\dfrac{1}{125}\) - \(x^3\) ) ( \(x^2\) + 22.66) = 0
\(\left[{}\begin{matrix}\dfrac{1}{125}-x^3=0\\x^2-2^2.6^6=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x^3=\dfrac{1}{125}\\x^2=2^2(6^3)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x^2=(2.6^3)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=432^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=432\\x=-432\end{matrix}\right.\)
\(x\) ϵ { -432; \(\dfrac{1}{5}\); 432; }