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\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
~ 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 
\(P=\left(0,5-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right):\left(-2\right)\)
\(=\left(-\dfrac{1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{2}\right)\)
\(=\left(\dfrac{-5-6}{10}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=-\dfrac{11}{10}:\left(-3\right)+\dfrac{1}{4}\)
\(=-\dfrac{11}{10}.\left(-\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{37}{60}\)
Vậy \(P=\dfrac{37}{60}\)
\(Q=\left(\dfrac{2}{25}-1,008\right):\dfrac{4}{7}:\left[\left(3\dfrac{1}{4}-6\dfrac{5}{9}\right):2\dfrac{2}{17}\right]\)
\(=\left(\dfrac{2}{25}-\dfrac{126}{125}\right):\dfrac{4}{7}:\left[\left(\dfrac{13}{4}-\dfrac{59}{9}\right).\dfrac{36}{17}\right]\)
\(=-\dfrac{116}{125}.\dfrac{7}{4}:\left(-\dfrac{119}{36}.\dfrac{36}{17}\right)\)
\(=\dfrac{-29.7}{125}:\left(-7\right)=\dfrac{29}{125}\)
Vậy \(Q=\dfrac{29}{125}\)
a, \(\dfrac{20^5.5^{10}}{100^5}=\dfrac{20^5.5^{10}}{\left(20.5\right)^5}=\dfrac{20^5.5^{10}}{20^5.5^5}=5^5\)
b,\(\dfrac{\left(0,9\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3.3\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3\right)^5.3^5}{\left(0,3\right)^6}=\dfrac{3^5}{\left(0,3\right)}\)
1. \(A=2x^2-5x-5\)
* Tại \(x=-2\) giá trị của biểu thức là :
\(A=2.\left(-2\right)^2-5.\left(-2\right)-5\)
\(A=8-\left(-10\right)-5=13\)
*Tại \(x=\dfrac{1}{2}\)
\(A=2\left(\dfrac{1}{2}\right)^2-5.\dfrac{1}{2}-5\)
\(A=-7\)
Câu 3:
a) \(A=\left(x-3\right)^2+9\ge9,\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-3=0\)
..........................\(\Leftrightarrow x=3\)
Vậy MIN A = 9 \(\Leftrightarrow x=3\)
P/s: câu b coi lại đề
c) \(\left|x-1\right|+\left(2y-1\right)^4+1\ge1;\forall x,y\)
Dấu "='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\2y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{2}\end{matrix}\right.\)
Vậy .............................
Câu 5:
Ta có: \(A=\dfrac{x-5}{x-3}=\dfrac{x-3-2}{x-3}=1-\dfrac{2}{x-3}\)
Để A nguyên thì \(2⋮\left(x-3\right)\)
\(\Rightarrow\left(x-3\right)\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Do đó:
\(x-3=-2\Rightarrow x=1\)
\(x-3=-1\Rightarrow x=2\)
\(x-3=1\Rightarrow x=4\)
\(x-3=2\Rightarrow x=5\)
Vậy .....................
A) \(\dfrac{4^5.4^2}{16^4}=\dfrac{4^7}{\left(2^4\right)^4}=\dfrac{2^{14}}{2^{16}}=\dfrac{1}{4}\)
b)\(\dfrac{2^8.9^4}{6^6.8^3}=\dfrac{2^8.\left(3^2\right)^4}{2^6.3^6.\left(2^3\right)^3}=\dfrac{2^8.3^8}{2^{15}.3^6}=\dfrac{9}{128}\)
c) \(\dfrac{6^3+3.6^2+3^3}{-13}=\dfrac{2^3.3^3+3.2^2.3^2+3^3}{-13}=\dfrac{2^3.3^3+3^3.2^2+3^3}{-13}=\dfrac{3^3.\left(2^3+2^2+1\right)}{-13}=\dfrac{3^3.13}{-13}=-9\)
a. \(\frac{20^5.5^{10}}{100^5}\)
\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)
\(=\frac{20^5.25^5}{100^5}\)
\(=\frac{500^5}{100^5}\)
\(=\left(\frac{500}{100}\right)^5\)
\(=5^5=3125\)
b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)
\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)
\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)
\(=3^5.\frac{1}{0,3}\)
\(=810\)
c. \(\frac{6^3+3.6^2+3^3}{-13}\)
\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)
\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\frac{3^3.13}{-13}\)
\(=\left(-3\right)^3\)
\(=-27\)
a) = 4. 5/4 + 25. [ 3/2 : (5/4)2] : 27/8
= 5 + 25. 12/5: 27/8
=5 +160/9
=205/9
b) = 8+ 3- 1+2.8
=11-1+2.8
=10+2.8
=10+ 16
= 26
c)= 3+1+1/4:2
= 4+ 0,125
=4,125
tìm giá trị của biểu thức sau
b)\(\left(0,25\right)^6.\left(-4\right)^6-\dfrac{72^2}{36^2}\)
\(=\left[0.25.\left(-4\right)\right]^4.\left(72:36\right)^2\)
\(=-1.4\)
\(=-4\)
c)\(9.\left(\dfrac{1}{3}\right)^3:\left[\left(-\dfrac{2}{3}+0.5-1\dfrac{1}{2}\right)\right]\)
\(=9.\dfrac{1}{27}:\left[\dfrac{-8}{27}+\dfrac{1}{2}-\dfrac{3}{2}\right]\)
=\(9.\dfrac{1}{27}:\left[\dfrac{-8}{27}+\left(-1\right)\right]\)
\(=9.\dfrac{1}{27}.\dfrac{-27}{35}\)
\(=\dfrac{3.3.1.9.\left(-3\right)}{-3.\left(-9\right).35}=\dfrac{-9}{35}\)
a. \(\left(0,25\right)^6.\left(-4\right)^6-\dfrac{72^2}{36^2}\)
\(=\left[0,24.\left(-4\right)\right]^6-\left(\dfrac{72}{36}\right)^2\)
\(=\left(-1\right)^6-2^2\)
\(=1-4=-3\)
b. \(9.\left(\dfrac{1}{3}\right)^3:\left[\left(\dfrac{-2}{3}\right)^3+0,5-1\dfrac{1}{2}\right]\)
\(=9.\dfrac{1}{27}:\left[\left(\dfrac{-8}{27}\right)+\dfrac{1}{2}-\dfrac{3}{2}\right]\)
\(=9.\dfrac{1}{27}:\dfrac{-35}{27}\)
\(=\dfrac{-9}{35}\)
A\(=3\left|2x-1\right|-5\)
Do \(3\left|2x-1\right|\ge0\forall x\Rightarrow3\left|2x-1\right|-5\ge-5\forall x\)
Dấu"=" xảy ra \(\Leftrightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)
Vậy Min A\(=-5\Leftrightarrow x=\dfrac{1}{2}\)
a)Ta có:
\(|2x-1|\ge0,\forall x\)
=>\(3\left|2x-1\right|\ge0,\forall x\)
=>\(3\left|2x-1\right|-5\ge-5,\forall x\)
=>A\(\ge-5\)
Dấu"="xảy ra:
<=>\(\left|2x-1\right|=0\)
<=>2x=1
<=>x=\(\dfrac{1}{2}\)
Vậy Min(A)=-5<=>x=1/2
a) Có \(\dfrac{2^8.9^2}{6^4.8^2}\)=\(\dfrac{2^8.3^4}{3^4.2^42^6}\)=\(\dfrac{2^8.3^4}{3^4.2^{10}}\)=\(\dfrac{2^8}{2^{10}}\)=\(\dfrac{1}{4}\)
b) Có \(\left(2^{-1}+3^{-1}\right):\left(2^{-1}-3^{-1}\right)+\left(2^{-1}.2^0\right).2^3\)
= \(\dfrac{2^{-1}+3^{-1}}{2^{-1}-3^{-1}}+2^{-1}.2^3\)
=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{\dfrac{1}{2}-\dfrac{1}{3}}+\dfrac{1}{2}.2^3\)
= \(\dfrac{\dfrac{5}{6}}{\dfrac{1}{6}}+2^2\)=5+8=13