Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh
a; [6.(- \(\dfrac{1}{3}\))3 - 3.(- \(\dfrac{1}{3}\) + 1)] - ( - \(\dfrac{1}{3}\) - 1)
= [6. \(\dfrac{-1}{3^3}\) - 3.\(\dfrac{2}{3}\)] - ( - \(\dfrac{1}{3}\) - \(\dfrac{3}{3}\))
= [\(\dfrac{-2}{9}\) - 2] + \(\dfrac{4}{3}\)
= [\(\dfrac{-2}{9}\) - \(\dfrac{18}{9}\)] + \(\dfrac{12}{9}\)
= - \(\dfrac{20}{9}\) + \(\dfrac{12}{9}\)
= \(\dfrac{-8}{9}\)
b; (63 + 3.62 + 33): 13
= (216 + 3.36 + 27) : 13
= (216 + 108 + 27): 13
= (324 + 27): 13
= 351 : 13
= 27
a/ \(\frac{1}{3}.3^x+3^{x+2}=3^{16}+3^{13}\)
\(\Leftrightarrow3^{x-1}+3^{x+2}=3^{13}+3^{16}\)
\(\Leftrightarrow3^{x-1}\left(1+3^3\right)=3^{13}\left(1+3^3\right)\)
\(\Leftrightarrow3^{x-1}=3^{13}\Rightarrow x-1=13\Rightarrow x=14\)
b/ \(\frac{1}{6}6^x+6^{x+2}=6^{15}+6^{18}\)
\(\Leftrightarrow6^{x-1}+6^{x+2}=6^{15}+6^{18}\)
\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^{15}\left(1+6^3\right)\)
\(\Rightarrow x=16\)
c/ \(\frac{1}{2}2^{x+3}-2^x=2^{22}-2^{20}\)
\(\Leftrightarrow2^x\left(2^2-1\right)=2^{20}\left(2^2-1\right)\)
\(\Rightarrow x=20\)
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 ⇒ \(\frac{1}{3}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^{x+3}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)
⇒ \(x=13\)
\(\left(1+1+1\right)!=6\)
\(2+2+2=6\)
\(3\cdot3-3=6\)
\(\sqrt{4}+\sqrt{4}+\sqrt{4}=6\)
\(5+\left(5:5\right)=6\)
\(6+6-6=6\)
\(7-\left(7:7\right)=6\)
\(\left(\sqrt{8+\left(8:8\right)}\right)!=6\)
\(\left(9-9\right)+\left(\sqrt{9}\right)!=6\)
\(\sqrt{10-\left(10:10\right)}!=6\)
\(x-\frac{3}{8}=\frac{1}{6}-\frac{1}{5}\)
=> \(x-\frac{3}{8}=\frac{5}{30}-\frac{6}{30}=-\frac{1}{30}\)
=> \(x=-\frac{1}{30}+\frac{3}{8}\)
=> \(x=\frac{41}{120}\)
\(-\frac{7}{10}\left(x+\frac{1}{3}\right)=\frac{4}{5}\)
=> \(-\frac{7}{10}x-\frac{7}{30}=\frac{4}{5}\)
=> \(-\frac{7}{10}x=\frac{4}{5}+\frac{7}{30}=\frac{31}{30}\)
=> \(x=\frac{31}{30}:\left(-\frac{7}{10}\right)=\frac{31}{30}\cdot\left(-\frac{10}{7}\right)=-\frac{31}{21}\)
\(x-\frac{4}{3}=\frac{5}{6}\Rightarrow x=\frac{5}{6}+\frac{4}{3}=\frac{5}{6}+\frac{8}{6}=\frac{13}{6}\)
Thiếu đề
\(\frac{6}{5}+\left(x-\frac{2}{3}\right)=\frac{4}{7}\)
=> \(\frac{6}{5}+x-\frac{2}{3}=\frac{4}{7}\)
=> \(\frac{6}{5}+x=\frac{4}{7}+\frac{2}{3}=\frac{26}{21}\)
=> \(x=\frac{26}{21}-\frac{6}{5}=\frac{4}{105}\)
a) 0,16 + 1,3
= \(\frac{4}{25}+\frac{13}{10}\)
= \(\frac{40}{250}+\frac{325}{250}\)
= \(\frac{73}{50}\)
b) 1,3 + 0,12 . \(2\frac{8}{11}\)
= \(\frac{13}{10}+\frac{3}{25}\times\frac{30}{11}\)
= \(\frac{13}{10}+\frac{18}{55}\)
= \(\frac{143}{110}+\frac{36}{110}\)
= \(\frac{179}{110}\)
c) 0,6 + 1,6
= \(\frac{3}{5}+\frac{8}{5}\)
= \(\frac{11}{5}\)
d) 3,6 + 1,36 \(\times\)\(2\frac{1}{5}\)
= \(\frac{18}{5}+\frac{34}{25}\times\frac{11}{5}\)
= \(\frac{18}{5}+\frac{374}{125}\)
= \(\frac{450}{125}+\frac{374}{125}\)
= \(\frac{824}{125}\)
\(\dfrac{\left(-\dfrac{1}{3}\right)^6}{\left(\dfrac{1}{6}\right)^2}=\dfrac{\left(\dfrac{1}{3}\right)^6}{\left(\dfrac{1}{6}\right)^2}\\ =\dfrac{\dfrac{1^6}{3^6}}{\dfrac{1^2}{6^2}}=\dfrac{\dfrac{1}{3^6}}{\dfrac{1}{6^2}}\\ =\dfrac{1}{3^6}:\dfrac{1}{6^2}=\dfrac{1}{3^6}.\dfrac{6^2}{1}\\ =\dfrac{6^2}{3^6}=\dfrac{2^2.3^2}{3^2.3^4}\\ =\dfrac{2^2}{3^4}=\dfrac{4}{81}\)
(-1/6)^3/(1/6)^2
= -1/18 / 1/36
= -1/18 ÷ 1/36
= -1/18 × 36/1
= -1/1 × 2/1
= -2