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\(1,\\ a,\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\\ c,\Leftrightarrow2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\\ 2,\\ a,=1\\ b,=\left(\dfrac{13}{4}\right)^2=\dfrac{169}{16}\\ c,=\left(-\dfrac{7}{4}\right)^2=\dfrac{49}{16}\\ d,=\left(\dfrac{3}{7}\right)^{20}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^8=...\\ e,=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2=10^2=100\)
1)
a) \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)
=> \(x-2\) và \(x+\frac{2}{3}\) cùng dấu.
Ta có 2 trường hợp:
TH1:
\(\left\{{}\begin{matrix}x-2>0\\x+\frac{2}{3}>0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x>2\\x>-\frac{2}{3}\end{matrix}\right.\) => \(x>2\left(TM\right).\)
TH2:
\(\left\{{}\begin{matrix}x-2< 0\\x+\frac{2}{3}< 0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x< 2\\x< -\frac{2}{3}\end{matrix}\right.\) => \(x< -\frac{2}{3}\left(TM\right).\)
Vậy \(x>2\) và \(x< -\frac{2}{3}.\)
Mình chỉ làm được thế thôi nhé bạn.
Chúc bạn học tốt!
1.b)
Ta có \(VT=\left(x-2,5\right)^{20}+\left(y+3,2\right)^{10}\ge0\forall x,y\)
Nên để xảy ra đẳng thức tức là để tìm được x thỏa mãn đề bài thì:
\(\left\{{}\begin{matrix}\left(x-2,5\right)^{20}=0\\\left(y+3,2\right)^{10}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2,5\\y=-3,2\end{matrix}\right.\)
Vậy...
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
a) => \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}=\frac{24}{54}=\frac{4}{9}\)
=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\) => \(x=\frac{6}{5}.\left(\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\right)\)
b) \(\frac{1}{3}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\) => \(\left(\frac{1}{2}x-1\right)^4=\frac{3}{48}=\frac{1}{16}\)
=> \(\frac{1}{2}x-1=\frac{1}{2}\) hoặc \(\frac{1}{2}x-1=-\frac{1}{2}\)
=> \(\frac{1}{2}x=\frac{3}{2}\) hoặc \(\frac{1}{2}x=\frac{1}{2}\) => x = 3 hoặc x = 1
c) \(\left(1+5\right).\left(\frac{3}{5}\right)^{x-1}=\frac{54}{25}\) => \(\left(\frac{3}{5}\right)^{x-1}=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
=> x - 1= 2 => x = 3
d) \(\left(1+\left(\frac{2}{3}\right)^2\right).\left(\frac{2}{3}\right)^x=\frac{101}{243}\) => \(\frac{13}{9}.\left(\frac{2}{3}\right)^x=\frac{101}{243}\)
=> \(\left(\frac{2}{3}\right)^x=\frac{101}{243}:\frac{13}{9}=\frac{101}{351}\) (có lẽ đề sai)
2) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\); \(\frac{1}{81^8}=\frac{1}{\left(3^4\right)^8}=\frac{1}{3^{32}}\)
Vì 333 > 332 => \(\frac{1}{3^{33}}\) < \(\frac{1}{3^{32}}\) => \(\frac{1}{27^{11}}\) < \(\frac{1}{81^8}\)
b) \(\frac{1}{3^{99}}=\frac{1}{\left(3^3\right)^{33}}=\frac{1}{27^{33}}
\(5^{200}=\left(5^2\right)^{100}=25^{100}\)
\(3< 25=>3^{100}< 25^{100}=>3^{100}< 5^{200}\)
\(\frac{75^{20}}{45^{10}.25^{15}}=\frac{25^{20}.3^{20}}{3^{10}.3^{10}.5^{10}.25^{15}}=\frac{25^{20}}{25^5.25^{15}}=1\)
\(=>75^{20}=45^{10}.25^{15}\left(dpcm\right)\)
P/S:nếu a=b=>a:b=1 mk làm theo cách đó cho nhanh mà bn ghi sai đề r