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a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
~ Học tốt ~
Bài 1:
1) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)
\(=3^2.\left(\dfrac{1}{3}\right)^5.\left(3^4\right)^2.\dfrac{1}{3^3}\)
\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)
\(=3^2=9\)
2) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)
\(=\left(2^2.2^5\right):[2^3.\left(\dfrac{1}{2}\right)^4]\)
\(=2^7:2^3:\dfrac{1}{2^4}\)
\(=2^4.2^4=256\)
3)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}.1:2^3\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2^4}\)
\(=\dfrac{43}{48}\)
4)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=-3-1+\dfrac{1}{4}.\dfrac{1}{2}\)
\(=-3-1+\dfrac{1}{8}\)
\(=-4+\dfrac{1}{8}\\ \)
\(=-\dfrac{31}{8}\)
5)\([\left(0,1\right)^2]^0+[\left(\dfrac{1}{7}\right)^{-1}]^2.\dfrac{1}{49}.[\left(2^2\right)^3:2^5]\\ =1+7^2.\dfrac{1}{7^2}.2^6:2^5\\ =1+1.2\\ =3\)
Chúc bạn học tốt 
a,
\(\left(4x-\dfrac{1}{3}\right)^6=1\\ \Rightarrow\left[{}\begin{matrix}4x-\dfrac{1}{3}=1\\4x-\dfrac{1}{3}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{-2}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{-1}{6}\end{matrix}\right.\)
b,
\(\left(5x-\dfrac{2}{3}\right)^2=0\\ \Rightarrow5x-\dfrac{2}{3}=0\\ 5x=\dfrac{2}{3}\\ x=\dfrac{2}{15}\)
c,
\(\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)^3=-8\\ \Rightarrow\dfrac{1}{3}x-\dfrac{1}{2}=-2\\ \dfrac{1}{3}x=\dfrac{-3}{2}\\ x=\dfrac{-9}{2}\)
d,
\(\dfrac{81}{3^n}=3\\ \Leftrightarrow3^4:3^n=3^1\\\Leftrightarrow3^{4-n}=3^1 \\ \Rightarrow n=3\)
e,
\(\dfrac{\left(-2\right)^x}{64}=-2\\ \Leftrightarrow\left(-2\right)^x:\left(-2\right)^6=\left(-2\right)^1\\ \Leftrightarrow\left(-2\right)^{x-6}=\left(-2\right)^1\\ \Rightarrow x=7\)
f,
\(\left(-20\right)^n:10^n=16\\ \left[\left(-20\right):10\right]^n=16\\ \left(-2\right)^n=\left(-2\right)^4\\ \Rightarrow n=4\)
Bài 1:
a) \(\left(4x-\dfrac{1}{3}\right)^6=1\)
\(\Rightarrow4x-\dfrac{1}{3}=1\)
\(4x=1+\dfrac{1}{3}\)
\(4x=\dfrac{4}{3}\)
\(x=\dfrac{4}{3}:4\)
\(x=\dfrac{1}{3}\)
b) \(\left(5x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow5x-\dfrac{2}{3}=0\)
\(5x=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}:5\)
\(x=\dfrac{2}{15}\)
c) \(\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)^3=-8\)
\(\Rightarrow\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)^3=\left(-2\right)^3\)
\(\dfrac{1}{3}x-\dfrac{1}{2}=-2\)
\(\dfrac{1}{3}x=-2+\dfrac{1}{2}\)
\(\dfrac{1}{3}x=\dfrac{-3}{2}\)
\(x=\dfrac{-3}{2}:\dfrac{1}{3}\)
\(x=\dfrac{-9}{2}\)
d) \(\dfrac{81}{3^n}=3\)
\(\Rightarrow\dfrac{3^4}{3^n}=3\)
\(\Rightarrow3^n.3=3^4\)
\(3^{n+1}=3^4\)
n + 1 = 4
n = 4 - 1
n = 3
e) \(\dfrac{\left(-2\right)^x}{64}=-2\)
\(\Rightarrow\dfrac{\left(-2\right)^x}{\left(-2\right)^6}=-2\)
\(\Rightarrow\left(-2\right)^x=\left(-2\right)^6.\left(-2\right)\)
\(\left(-2\right)^x=\left(-2\right)^7\)
x = 7
f) (-20)n : 10n = 16
(-20 : 10)n = 16
(-2)n = 16
(-2)n = (-2)4
n = 4.
a)
\(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=-\dfrac{1}{4}-y\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-\dfrac{1}{3}+x=-\dfrac{1}{4}-y\\\dfrac{1}{2}-\dfrac{1}{3}+x=\dfrac{1}{4}+y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=-\dfrac{5}{12}\\x-y=\dfrac{1}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
b)\(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
ta thấy : \(\left|x-y\right|\ge0\\ \left|y+\dfrac{9}{25}\right|\ge0\)\(\Rightarrow\left|x-y\right|+\left|y+\dfrac{9}{25}\right|\ge0\)
đẳng thửc xảy ra khi : \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow x=y=-\dfrac{9}{25}\)
vậy \(\left(x;y\right)=\left(-\dfrac{9}{25};-\dfrac{9}{25}\right)\)
c) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
ta thấy \(\left(\dfrac{1}{2}x-5\right)^{20}\:và\:\left(y^2-\dfrac{1}{4}\right)^{10}\) là các lũy thừa có số mũ chẵn
\(\Rightarrow\:\)\(\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\ \left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy cặp số x,y cần tìm là \(\left(10;\dfrac{1}{2}\right)\:hoặc\:\left(10;-\dfrac{1}{2}\right)\)
d)
\(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\\ \Leftrightarrow x\left(x^2-\dfrac{5}{4}\right)=x\left(vì\:x\ge0\right)\\ \Leftrightarrow x\left(x^2-\dfrac{9}{4}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{9}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy x cần tìm là \(-\dfrac{3}{2};0;\dfrac{3}{2}\)
e)\(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
ta thấy: \(x^2\ge0;\left(y-\dfrac{1}{10}\right)^4\ge0\)
\(\Rightarrow x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\)
đẳng thức xảy ra khi: \(\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
vậy cặp số cần tìm là \(0;\dfrac{1}{10}\)
a)
\(\left(\dfrac{1}{12}\right)^{27}\) và \(\left(\dfrac{1}{12}\right)^{18}\)
Ta có:
\(\left(\dfrac{1}{12}\right)^{27}\)=\(\left(\left(\dfrac{1}{12}\right)^3\right)^9=\left(\dfrac{1}{18}\right)^9\)(1)
\(\left(\dfrac{1}{12}\right)^{18}\)\(=\left(\left(\dfrac{1}{12}\right)^3\right)^6=\left(\dfrac{1}{8}\right)^6\)(2)
Từ (1) và (2) ta có:
\(\left(\dfrac{1}{18}\right)^9>\left(\dfrac{1}{8}\right)^6\)
Vậy \(\left(\dfrac{1}{12}\right)^{27}\)>\(\left(\dfrac{1}{12}\right)^{18}\)
b)
\(\left(10\right)^{20}\) và \(\left(-9\right)^{10}\)
Ta có:
\(10^{20}=\left(\left(10\right)^4\right)^5\) (1)
\(\left(-9\right)^{10}=\left(\left(-9\right)^2\right)^5\) (2)
Từ (1) và (2), ta có:
\(\left(10000\right)^5>\left(81\right)^5\)
Vậy \(\left(10\right)^{20}\)> \(\left(-9\right)^{10}\)
c)
\(64^3\) và \(16^2\)
Ta có:
\(64^3=\left(\left(4^{ }\right)^3\right)^3\)(1)
\(16^2=\left(\left(4\right)^2\right)^2\)(2)
Từ (1) và (2), ta có:
\(\left(\left(4\right)^3\right)^3>\left(\left(4\right)^2\right)^2\)
Vậy \(64^3\)> \(16^2\)
d)
\(\left(-16\right)^3\) và \(\left(-64\right)^2\)
Ta có:
\(\left(-16\right)^3=\left(\left(-4\right)^2\right)^3\)(1)
\(\left(-64\right)^2=\left(\left(-4\right)^3\right)^2=\left(\left(-4\right)^2\right)^3\) (2)
Từ (1) và (2), ta có:
\(16^3=16^3\)
Vậy \(\left(-16\right)^3\)= \(\left(-64\right)^2\)
Chúc bạn học tốt!
chỗ câu a) mình nhầm nha bạn,1/12 mình nhầm mất (1/12)^3=1/1728. Bạn sửa lại sẽ dễ hiểu hơn :))
Bạn sửa hộ mình nhé, cảm ơn!
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
Các câu đúng: b,e
Các câu sai: a, c, d; f.
a) \(\left(-5\right)^2.\left(-5\right)^3=\left(-5\right)^5\);
c) \(\left(0,2\right)^{10}:\left(0,2\right)^5=\left(0,2\right)^{10-5}=0,2^5\);
d) \(\left[\left(-\dfrac{1}{7}\right)^2\right]^4=\left(-\dfrac{1}{7}\right)^{2.4}=\left(-\dfrac{1}{7}\right)^8\)
f \(\dfrac{8^{10}}{4^8}=\dfrac{\left(2^3\right)^5}{\left(2^2\right)^8}=\dfrac{2^{15}}{2^{16}}=\dfrac{1}{2}\)
\(\left(\dfrac{1}{16}\right)^{10}=\left[\left(\dfrac{1}{2}\right)^4\right]^{10}=\left(\dfrac{1}{2}\right)^{40}< \left(\dfrac{1}{2}\right)^{50}\\ \left(\dfrac{1}{2}\right)^{300}=\left(\dfrac{1}{2}\right)^{3\cdot100}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\\ \left(\dfrac{1}{3}\right)^{200}=\left(\dfrac{1}{3}\right)^{2\cdot100}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\\ \dfrac{1}{8}>\dfrac{1}{9}\Rightarrow\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\Rightarrow\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\\ \left(0,3\right)^{20}=\left(0,3\right)^{2\cdot10}=\left[\left(0,3\right)^2\right]^{10}=\left(0,09\right)^{10}< \left(0,1\right)^{10}\)
a) \(\left[\left(\dfrac{1}{2}\right)^4\right]^{10}=\left(\dfrac{1}{2}\right)^{40}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{40}< \left(\dfrac{1}{2}\right)^{50}\)
Vì \(40< 50\)
b)\(\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
\(\Rightarrow\text{}\text{}\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
Vì \(\dfrac{1}{8}>\dfrac{1}{9}\)
c)\(\left[\left(0,3\right)^2\right]^{10}=\left(0,09\right)^{10}\)
\(\Rightarrow\left(0,1\right)^{10}>\left(0,3\right)^{20}\)
Vì \(0,1>0,09\)