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L
2
NA
7 tháng 11 2018
\(\frac{-3}{4}.{5}\frac{3}{13}-0,75.\frac{36}{13}\)
\(=\frac{-3}{4}.\frac{68}{13}-0,75.\frac{36}{13}\)
\(=\frac{-3}{4}.\frac{68}{13}-\frac{3}{4}.\frac{36}{13}\)
\(=\frac{-3}{4}.(\frac{68}{13}-\frac{36}{13})\)
\(=\frac{-3}{4}.\frac{32}{13}\)
\(=\frac{-24}{13}\)
7 tháng 11 2018
−34.5313−0,75.3613−34.5313−0,75.3613
=−34.6813−0,75.3613=−34.6813−0,75.3613
=−34.6813−34.3613=−34.6813−34.3613
=−34.(6813−3613)=−34.(6813−3613)
=−34.3213=−34.3213
=−2413
NT
1
2 tháng 3 2020
\(\left[\left(2+2\frac{1}{3}\right)-0,75\right]\left[3\frac{1}{2}-0,5:\left(\frac{3}{5}+\frac{1}{3}-\frac{1}{2}\right)\right]\)
\(=\left[\left(2+\frac{7}{3}\right)-\frac{75}{100}\right]\left[\frac{7}{2}-\frac{5}{10}:\left(\frac{3}{5}+\frac{1}{3}-\frac{1}{2}\right)\right]\)
\(=\left[\frac{2\cdot3+7}{3}-\frac{3}{4}\right]\left[\frac{7}{2}-\frac{1}{2}:\frac{13}{30}\right]\)
\(=\left[\frac{13}{3}-\frac{3}{4}\right]\left[\frac{7}{2}-\frac{1}{2}\cdot\frac{30}{13}\right]\)
\(=\left[\frac{13}{3}-\frac{3}{4}\right]\left[\frac{7}{2}-\frac{1}{1}\cdot\frac{15}{13}\right]\)
\(=\left[\frac{13}{3}-\frac{3}{4}\right]\left[\frac{7}{2}-\frac{15}{13}\right]\)
\(=\frac{43}{12}\cdot\frac{61}{26}=\frac{2623}{312}\)
ST
13 tháng 12 2017
\(\dfrac{-4}{13}\cdot\dfrac{5}{17}+\dfrac{-12}{13}\cdot\dfrac{4}{17}+\dfrac{4}{13}\)
\(=\dfrac{-4}{17}\cdot\dfrac{5}{17}+\dfrac{-4}{13}\cdot3\cdot\dfrac{4}{17}+\dfrac{4}{13}\)
\(=\dfrac{4}{13}\left(\dfrac{-5}{17}+\dfrac{-12}{17}+1\right)\)
\(=\dfrac{4}{13}\left(-1+1\right)=\dfrac{4}{13}\cdot0=0\)
TT
1
16 tháng 10 2018
\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)
\(\Rightarrow2B-B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(2,\)
\(a,\dfrac{45^{10}.2^{10}}{75^{15}}\)
\(=\dfrac{5^{10}.9^{10}.2^{10}}{25^{15}.3^{15}}\)
\(=\dfrac{5^{10}.3^{20}.2^{10}}{5^{30}.3^{15}}\)
\(=\dfrac{5^{10}.3^{15}.\left(3^5.2^{10}\right)}{5^{10}.3^{15}.\left(5^{20}\right)}\)
\(=\dfrac{3^5.2^{10}}{5^{20}}\)
\(b,\dfrac{2^{15}.9^4}{6^3.8^3}\)
\(=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
\(c,\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{4^{10}.2^{10}+4^{10}}{4^4.2^4+4^4.4^7}=\dfrac{4^4.\left(4^6.2^{10}+4^6\right)}{4^4.\left(2^4+4^7\right)}\)
\(=\dfrac{4^{11}+4^6}{4^8.4^7}=\dfrac{4^6.\left(4^5+1\right)}{4^6.\left(4^2-4\right)}=\dfrac{1024+1}{16-4}=\dfrac{1025}{12}\)
\(d,\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
\(3,\)
\(a,\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x+4=\dfrac{1}{2}\\2x+4=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-4=\dfrac{-7}{2}\\2x=\dfrac{-1}{2}-4=\dfrac{-9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-7}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-7}{4};\dfrac{-9}{4}\right\}\)
\(b,\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2=\left(-6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=6+3=9\\2x=-6+3=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{9}{2};\dfrac{-3}{2}\right\}\)
\(c,5^{x+2}=628\)
\(5^{x+2}=5^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=4-2=2\)
Vậy \(x=2\)
\(d,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
16 tháng 10 2018
Bài 1:
B= \(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{99}\)
2B= \(2.[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}]\)
2B= \(1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{98}\)
⇒2B-B= \(1-\left(\dfrac{1}{2}\right)^{99}\)
B= 1
Vậy B=1
Bài 2:
a, \(\dfrac{45^{10}.2^{10}}{75^{15}}\)= \(\dfrac{\left(3^2.5\right)^{10}.2^{10}}{\left(3.5^2\right)^{15}}=\dfrac{3^{20}.5^{10}.2^{10}}{3^{15}.5^{30}}=\dfrac{3^5.2^{10}}{5^{20}}\)
b, \(\dfrac{2^{15}.9^4}{6^3.8^3}=\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
c,\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2.4\right)^{10}+4^{10}}{\left(2.4\right)^4+4^{11}}=\dfrac{2^{10}.4^{10}+4^{10}}{2^4.4^4+4^{11}}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.4^5}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(4^5+1\right)}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(2^{10}+1\right)}=4^4=256\)
d, \(\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
Bài 3:
a, \(\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2\)
\(2x+4=\dfrac{1}{2}\)
\(2x=\dfrac{1}{2}-4\)
\(2x=-\dfrac{7}{2}\)
\(x=-\dfrac{7}{2}:2\)
\(x=-\dfrac{7}{2}.\dfrac{1}{2}\)
\(x=-\dfrac{7}{4}\)
b, \(\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2\)
\(2x-3=6\)
\(2x=9\)
\(x=\dfrac{9}{2}\)
c, \(5^{x+2}=625\)
\(5^{x+2}=5^4\)
\(x+2=4\)
\(x=2\)
VH
16 tháng 6 2019
câu a) mình chịu (dùng kiến thức lớp 12 chắc làm đc haha)
b) gt ⇒ \(\frac{1}{6}.6^{x+2}-6^x=6^{14}-6^{13}\)
⇒ \(6^{x+1}-6^x=6^{14}-6^{13}\)
⇒ \(6^x\left(6-1\right)=6^{13}\left(6-1\right)\)
⇒ \(x=13\)
c) gt ⇒ \(\frac{1}{2}.2^{x+4}-2^x=2^{13}-2^{10}\)
⇒ \(2^{x+3}-2^x=2^{13}-2^{10}\)
⇒ \(2^x\left(2^3-1\right)=2^{10}\left(2^3-1\right)\)
⇒ \(x=10\)
d) gt ⇒ \(\frac{1}{3}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^{x+3}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)
⇒ \(x=13\)
V
21 tháng 7 2018
*Trả lời :
a) \(-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)
= \(-\dfrac{3}{4}.\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
=\(\dfrac{3}{4}.\dfrac{-68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
=\(\dfrac{3}{4}.\cdot\left(\dfrac{-68}{13}-\dfrac{36}{13}\right)\)
=\(\dfrac{3}{4}.\left(-8\right)\)
= \(-6\)
b)\(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
=\(\dfrac{41}{9}-\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
=\(\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\left(-\dfrac{5}{7}\right)\)
=\(\dfrac{90}{9}:\left(-\dfrac{5}{7}\right)\)
=\(10:\left(-\dfrac{5}{7}\right)\)
=\(-14\)
c)\(\left(-\dfrac{3}{5}+\dfrac{4}{9}\right):\dfrac{7}{11}+\left(-\dfrac{2}{5}+\dfrac{5}{9}\right):\dfrac{7}{11}\)
=\(\left(-\dfrac{3}{5}\right)+\dfrac{4}{9}:\dfrac{7}{11}+\left(-\dfrac{2}{5}\right)+\dfrac{5}{9}:\dfrac{7}{11}\)(áp dụng tính chất phá ngoặc )
=\(\left\{\left[-\dfrac{3}{5}+\left(-\dfrac{2}{5}\right)\right]+\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\right\}:\dfrac{7}{11}\)
=\(\left(-\dfrac{5}{5}+\dfrac{9}{9}\right):\dfrac{7}{11}\)
=\(\left(-1+1\right):\dfrac{7}{11}\)
\(=0:\dfrac{7}{11}\)
=0.
d)\(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}:\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)
=\(\dfrac{6}{7}:\left[\dfrac{3}{26}+\left(-\dfrac{6}{26}\right)\right]+\dfrac{6}{7}:\left[\dfrac{1}{10}+\left(-\dfrac{16}{10}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{3}{26}\right)+\dfrac{6}{7}:\left(-\dfrac{3}{2}\right)\)
=\(\dfrac{6}{7}:\left[\left(-\dfrac{3}{26}\right)+\left(-\dfrac{39}{26}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{21}{13}\right)\)
=\(-\dfrac{26}{49}\)
TT
1 tháng 11 2017
a. Áp dụng t/c dãy tỉ sô bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{3y}{12}=\dfrac{x-3y}{3-12}=\dfrac{36}{-9}=-4\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-4\Rightarrow x=-12\\\dfrac{y}{4}=-4\Rightarrow y=-16\end{matrix}\right.\)
Vậy.............
b. Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x+3y}{4+9}=\dfrac{39}{13}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=3\Rightarrow x=6\\\dfrac{y}{3}=3\Rightarrow y=9\end{matrix}\right.\)
Vậy.........
c. Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{4x}{12}=\dfrac{3y}{15}=\dfrac{4x-3y}{12-15}=\dfrac{12}{-3}=-4\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-4\Rightarrow x=-12\\\dfrac{y}{5}=-4\Rightarrow y=-20\end{matrix}\right.\)
Vậy............
DV
1 tháng 11 2017
a, \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{3}=\dfrac{3y}{12}\)
Áp dụng t/c dãy tỉ số = nhau ,ta có :
\(\dfrac{x}{3}=\dfrac{3y}{12}=\dfrac{x-3y}{3-12}=\dfrac{36}{-9}=-4\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=-4\Rightarrow\left\{{}\begin{matrix}x=-12\\y=-16\end{matrix}\right.\)
Vậy ...
b,c tương tự
\(-\frac{3}{4}\cdot\frac{68}{13}-0,75\cdot\frac{36}{15}=-\frac{51}{13}-\frac{9}{5}=5\frac{47}{65}\)
\(-\frac{3}{4}.\frac{68}{13}-0,75.\frac{36}{15}\)
\(=-\frac{3.68}{4.13}-\frac{3}{4}.\frac{36}{15}\)
\(=-\frac{204}{52}-\frac{3.36}{4.15}\)
\(=-\frac{51}{13}-\frac{108}{60}\)
\(=-\frac{51}{13}-\frac{9}{5}\)
\(=-\frac{255}{65}-\frac{117}{65}\)
\(=-\frac{372}{65}\)
~ Rất vui vì giúp đc bn ~