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\(0,\left(34\right)=0\left(01\right).34=\dfrac{1}{99}\)
\(0,\left(5\right)=0,\left(1\right).5=\dfrac{1}{9}.5=\dfrac{5}{9}\)
\(0,\left(123\right)=0,\left(001\right).123=\dfrac{1}{999}.123=\dfrac{123}{999}=\dfrac{41}{333}\)
\(25^{10}\cdot\dfrac{1}{5^{20}}+\left(-\dfrac{3}{4}\right)^8\cdot\left(-\dfrac{4}{3}\right)^8-2011^0\\ =\left(5^2\right)^{10}\cdot\dfrac{1}{5^{20}}+\dfrac{\left(-3\right)^8}{4^8}\cdot\dfrac{\left(-4\right)^8}{3^8}-1\\ =5^{20}\cdot\dfrac{1}{5^{20}}+\dfrac{\left(-3\right)^8}{4^8}\cdot\dfrac{\left(-4\right)^8}{3^8}-1\\ =\dfrac{5^{20}}{5^{20}}+\dfrac{\left(-3\right)^8\cdot\left(-4^8\right)}{4^8\cdot3^8}-1\\ =1+1-1\\ =1\)
cứ phan tích cho hết đi là đc 9^6. 9^10 = (3^2)^6...................
tự làm đi
1. Tính:
a. \(\dfrac{9^6.9^{10}}{3^{32}}=\dfrac{\left(3^2\right)^6.\left(3^2\right)^{10}}{3^{32}}=\dfrac{3^{12}.3^{20}}{3^{32}}=\dfrac{3^{32}}{3^{32}}=1\)
b. \(\dfrac{25^8.25^{10}}{5^{34}}=\dfrac{\left(5^2\right)^8.\left(5^2\right)^{10}}{5^{34}}=\dfrac{5^{16}.5^{20}}{5^{34}}=\dfrac{5^{36}}{5^{34}}=5^{36}:5^{34}=5^2=25\)
c. \(\dfrac{7^{56}}{49^9.49^{20}}=\dfrac{7^{56}}{\left(7^2\right)^9.\left(7^2\right)^{20}}=\dfrac{7^{56}}{7^{18}.7^{40}}=\dfrac{7^{56}}{7^{58}}=7^{56}:7^{58}=\dfrac{7^{56}}{7^{56}.7^2}=\dfrac{1}{7^2}=\dfrac{1}{49}\)
d. \(\dfrac{4^2.4^3}{2^{10}}=\dfrac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\dfrac{2^4.3^6}{2^{10}}=\dfrac{2^{10}}{2^{10}}=1\)
e. \(\dfrac{2^{17}.25^5}{10^8.8^3}=\dfrac{2^{17}.\left(5^2\right)^5}{\left(2.5\right)^8.\left(2^3\right)^3}=\dfrac{2^{17}.5^{10}}{2^8.5^8.2^9}=\dfrac{2^{17}.5^{10}}{2^{17}.5^8}=\dfrac{5^{10}}{5^8}=5^{10}:5^8=5^2=25\)
f. \(\dfrac{3^{15}.25^4}{15^6.27^3}=\dfrac{3^{15}.\left(5^2\right)^4}{\left(3.5\right)^6.\left(3^3\right)^3}=\dfrac{3^{15}.5^8}{5^6.3^6.3^9}=\dfrac{3^{15}.5^8}{5^6.3^6.3^9}=\dfrac{5^8}{5^6}=5^8:5^6=5^2=25\)
2. Tính lũy thừa âm:
a. 3-2 = \(\dfrac{1}{3^2}\) = \(\dfrac{1}{9}\)
b. 2-3 = \(\dfrac{1}{2^3}\) = \(\dfrac{1}{8}\)
3. Tính :
a. \(\dfrac{\left(0,8\right)^4}{\left(0,4\right)^3}=\dfrac{\left(0,8\right)^3.0,8}{\left(0,4\right)^3}=\left(\dfrac{0,8}{0,4}\right)^3.0,8=2^3.0,8=8.0,8=6,4\)
b. \(\dfrac{\left(0,8\right)^3}{\left(0,4\right)^4}=\dfrac{\left(0,8\right)^3}{\left(0,4\right)^3:0,4}=\left(\dfrac{0,8}{0,4}\right)^3.\dfrac{1}{0,4}=2^3.2,5=8.2,5=20\)
c. \(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,6\right)^5}{\left(0,2\right)^5.\left(0,2\right)}=\left(\dfrac{\left(0,6\right)}{\left(0,2\right)}\right)^5.\dfrac{1}{0,2}=3^5.\dfrac{1}{0,2}=\dfrac{3^5}{0,2}=1215\)
P/s : Chế Mai Ngọc Trâm thử tham khảo thử đi!!!
a: \(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\cdot\dfrac{18}{5}-\dfrac{6}{5}:\dfrac{-9}{5}+4\)
\(=\dfrac{18}{5}-\dfrac{6}{5}\cdot\dfrac{-5}{9}+4\)
\(=\dfrac{18}{5}+\dfrac{2}{3}+4\)
\(=\dfrac{124}{15}\)
b: \(=\dfrac{9}{25}\cdot\left(\dfrac{3}{5}-\dfrac{1}{5}+\dfrac{1}{2}\right)-\dfrac{3}{8}:\dfrac{9}{8}\)
\(=\dfrac{9}{25}\cdot\dfrac{4}{10}-\dfrac{1}{3}\)
\(=-\dfrac{71}{375}\)
c: \(=\dfrac{7}{10}:\dfrac{4}{5}+\dfrac{2}{9}:\dfrac{5}{9}+\dfrac{1}{8}\)
\(=\dfrac{7}{10}\cdot\dfrac{5}{4}+\dfrac{2}{5}+\dfrac{1}{8}\)
=1+2/5
=7/5
d: \(=\dfrac{3}{7}\left(19+\dfrac{1}{3}-33-\dfrac{1}{3}\right)-\dfrac{2}{7}=\dfrac{3}{7}\cdot\left(-14\right)-\dfrac{2}{7}=-6-\dfrac{2}{7}=\dfrac{-44}{7}\)
e: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{11}\cdot3^{11}-2^{12}\cdot3^{12}}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{-2^{11}\cdot3^{11}\left(1+2\cdot3\right)}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\)
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 \(0;\dfrac{1}{10}\)
Bài giải
Ta có : \(9^{99}=\left(9^{11}\right)^9\)
Vì \(\left(9^{11}\right)^9>99^9\text{ }\left[\left(81\cdot9^9\right)^9>99^9\right]\text{ }\Rightarrow\text{ }9^{99}>99^9\)
2) a) \(\left(x+\dfrac{4}{5}\right)^2=\dfrac{9}{25}\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{3}{5}\\x+\dfrac{4}{5}=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{5}\\x=\dfrac{-7}{5}\end{matrix}\right.\) vậy \(x=\dfrac{-1}{5};x=\dfrac{-7}{5}\)
b) \(\left|x-\dfrac{3}{7}\right|=-2\) vì giá trị đối không âm được nên phương trình này vô nghiệm
c) điều kiện : \(x\ge-7\) \(\sqrt{x+7}-2=4\Leftrightarrow\sqrt{x+7}=4+2=6\)
\(\Leftrightarrow x+7=6^2=36\Leftrightarrow x=36-7=29\) vậy \(x=29\)
d) \(x^2-\dfrac{7}{9}x=0\Leftrightarrow x\left(x-\dfrac{7}{9}\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-\dfrac{7}{9}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{7}{9}\end{matrix}\right.\) vậy \(x=0;x=\dfrac{7}{9}\)
1) tìm GTNN
a) \(B=\left|x-2017\right|+\left|x-20\right|\)
B \(\ge\left|x-2017-x+20\right|=\left|-1997\right|=1997\)
Dấu " = " xảy ra khi và chỉ khi 20 \(\le x\le2017\)
Vậy MinB = 1997 khi 20 \(\le x\le2017\)
b) \(C=\left|x-3\right|+\left|x-5\right|\)
\(C\ge\left|x-3-x+5\right|=\left|2\right|=2\)
Dấu " = " xảy ra khi 3 \(\le x\le5\)
Vậ MinC = 2 khi và chỉ khi 3 \(\le x\le5\)
c) \(C=\left|x^2+4\right|+3\)
Ta thấy \(x^2+4\ge0\) với mọi x
nên \(\left|x^2+4\right|+3=x^2+4+3=x^2+7\)\(\ge\) 7
Dấu " =" xảy ra khi x = 0
MinC = 7 khi và chỉ khi x = 0
\(\dfrac{\left(13\dfrac{1}{4}-1\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(=\dfrac{1\dfrac{25}{108}.230\dfrac{1}{25}+46\dfrac{3}{4}}{4\dfrac{16}{21}:\left(-1\dfrac{20}{21}\right)}=\dfrac{330\dfrac{1}{25}}{-2\dfrac{18}{41}}=-135,3164\)
\(\left\{{}\begin{matrix}99^{29}=\left(99^2\right)^{10}=9801^{10}\\9999^{10}=9999^{10}\end{matrix}\right.\)
\(9801^{10}< 9999^{10}\Leftrightarrow99^{20}< 9999^{10}\)
\(\left\{{}\begin{matrix}\left|-\dfrac{25}{46}\right|=\dfrac{25}{46}>0\\\left(-\dfrac{25}{46}\right)^{2005}< 0\end{matrix}\right.\)
\(\Rightarrow\left(-\dfrac{25}{46}\right)^{2005}< \left|-\dfrac{25}{46}\right|\)
a/ Ta có :
\(99^{20}=\left(99^2\right)^{10}=9081^{10}\)
Vì \(9081^{10}< 9999^{10}\Leftrightarrow99^{20}< 9999^{10}\)
b/ Ta có :
\(\left|\dfrac{-25}{46}\right|=\dfrac{25}{46}>0\)
\(\left(\dfrac{-25}{46}\right)^{2005}< 0\)
\(\Leftrightarrow\left|\dfrac{-25}{46}\right|>\left(\dfrac{-25}{46}\right)^{2005}\)