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a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)
\(\Rightarrow x+3=-3\)
\(\Rightarrow x=-6\)
b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)
\(\Rightarrow2x+2=-2\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)
\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)
\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
\(a,\left[\left(-\frac{1}{2}\right)^3-\left(\frac{3}{4}\right)^3.\left(-2\right)^2\right]:\left[2.\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}\right]\)
\(=\left[\left(-\frac{1}{8}\right)-\frac{27}{64}.4\right]:\left[2.\left(-1\right)+\frac{9}{16}-\frac{3}{8}\right]\)
\(=\left[\left(-\frac{1}{8}-\frac{27}{16}\right)\right]:\left[-2+\frac{9}{16}-\frac{3}{8}\right]\)
\(=\frac{-2-27}{16}:\frac{-32+9-6}{16}\)
\(=-\frac{29}{16}:\frac{-29}{16}=1\)
\(b,\left[\left(\frac{4}{3}\right)^{-2}\left(\frac{3}{2}\right)^4\right]:\left(\frac{3}{2}\right)^6\)
\(=\left(\frac{9}{16}.\frac{81}{16}\right):\frac{729}{64}\)
\(=\frac{729}{64}:\frac{729}{64}=1\)
\(a,\frac{2^3.2^4}{2^5}=\frac{2^7}{2^5}=2^2=4\)
\(b,\frac{\left(0,2\right)^5.\left(0,6\right)^4}{\left(0,2\right)^7.\left(0,3\right)^4}=\frac{\left(0,2\right)^5.\left(0,3\right)^4.2^4}{\left(0,2\right)^7.\left(0,3\right)^4}=\frac{2^4}{\left(0,2\right)^2}\)
\(c,\frac{3^3.12^4}{6^5.9^4}=\frac{3^3.6^4.2^4}{6^5.3^8}=\frac{2^4}{6.3^5}=\frac{2^4}{2.3.3^5}=\frac{2^3}{3^6}\)
\(d,\frac{2^3+2^4+2^5}{7^2}=\frac{8+16+32}{49}=\frac{56}{49}=\frac{8}{7}\)
\(e,\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)
f) \(\frac{3^3.\left(0,5\right)^5}{\left(1,5\right)^4}=\frac{3^3.\left(0,5\right)^5}{\left[3.\left(0,5\right)\right]^4}=\frac{3^3.\left(0,5\right)^5}{3^4.\left(0,5\right)^4}=\frac{0,5}{3}=\frac{1}{6}\)
b) \(\frac{2^3+3.2^6-4^3}{2^3+3^2}=\frac{2^3.\left(1+3.2^3-2^3\right)}{2^3+3^2}=\frac{2^3.17}{17}=2^3=8\)
Các phần còn lại tương tự, bạn tự làm nhé !
(*) Lưu ý ở những bài rút gọn có chứa lũy thừa thì bạn đưa số đó về số nguyên tố rồi thực hiện như bình thường .
VD : \(4^3=\left(2^2\right)^3=2^6\) ( đưa về số nguyên tố là 2 )
\(6^3=\left(2.3\right)^3=2^3.3^3\) ( đưa về tích hai số nguyên tố )
\(A=\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
\(B=\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=2\)
\(C=\frac{4^5\cdot9^{4-2\cdot6^9}}{2^{10}\cdot3^8+6^8\cdot20}=0\)
A=\(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
Giải:
1) \(7^8.\left(-\dfrac{1}{7}\right)^8\)
\(=7^8.\left(\dfrac{1}{7}\right)^8\)
\(=7^8.\dfrac{1^8}{7^8}\)
\(=1\)
2) \(\left(\dfrac{4}{3}\right)^{10}.\left(-\dfrac{3}{4}\right)^{10}\)
\(=\left(\dfrac{4}{3}\right)^{10}.\left(\dfrac{3}{4}\right)^{10}\)
\(=\dfrac{4^{10}}{3^{10}}.\dfrac{3^{10}}{4^{10}}\)
\(=1\)
3) \(\left(-\dfrac{7}{2}\right)^{2006}.\left(-\dfrac{2}{7}\right)^{2006}\)
\(=\left(\dfrac{7}{2}\right)^{2006}.\left(\dfrac{2}{7}\right)^{2006}\)
\(=1\)
4) \(\left(-\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\left(\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\dfrac{5^{2007}.13^{2006}}{13^{2007}.5^{2006}}\)
\(=\dfrac{5}{13}\)
Vậy ...
a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}+\dfrac{5}{48}=\dfrac{6}{48}=\dfrac{1}{8}\)
\(\Leftrightarrow-\dfrac{1}{12}< x< \dfrac{1}{8}\)
=>x=0
c: \(\Leftrightarrow x=\dfrac{-1}{2}\cdot\dfrac{1}{4}=\dfrac{-1}{8}\)
d: \(\Leftrightarrow x^8=x^7\)
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
e: \(\Leftrightarrow3^x=\dfrac{3^{10}}{3^9}=3\)
hay x=1
f: =>x-1=20
hay x=21
\(\left(\frac{2}{3}\right)^6\)NHA CÁC BN