19x+22x3:14=11-6^2
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\(=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{11.3^{29}-3^{29}.3}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.2^3}{2^2.3^{28}}=\frac{3.2}{1.1}=6\)
a) 11/12 + 5/6 x 3/2 - 2/3
= 11/12 + 15/12 - 2/3
= 26/12 - 2/3
= 26/12 - 8/12
= 18 /12 = 3/2
b) 4/9 : 4/6 x 3/4 +5/3
= 4/9 x 6/4 x 3/4 + 5/3
= 2/3 x 3/4 + 5/3
= 6/12 + 5/3
= 6/12 + 20/12
= 26/12 = 13/6
c) 5/7 : 11/14 : 15/22 x 3/5
= 5/7 x 14/11 x 22/15 x 3/5
= 10/11 x 22/15 x 3/5
= 4/3 x 3/5 = 20/15 = 4/3
a. \(\dfrac{11}{12}+\dfrac{5}{6}\)x\(\dfrac{3}{2}-\dfrac{2}{3}\) \(=\)\(\dfrac{11}{12}+\dfrac{5}{4}-\dfrac{2}{3}=\dfrac{11}{12}+\dfrac{15}{12}-\dfrac{8}{12}=\dfrac{11+15-8}{12}=\dfrac{18}{12}=\dfrac{3}{2}\)
b. \(\dfrac{4}{9}:\dfrac{4}{6}\)x\(\dfrac{3}{4}+\dfrac{5}{3}\) \(=\dfrac{4}{9}:\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{9}+\dfrac{5}{3}=\dfrac{8}{9}+\dfrac{15}{9}=\dfrac{8+15}{9}=\dfrac{23}{9}\)
c. \(\dfrac{5}{7}:\dfrac{11}{14}:\dfrac{15}{22}\)x\(\dfrac{3}{5}\) \(=\dfrac{10}{14}:\) \(\dfrac{11}{14}:\dfrac{15}{22}\)x\(\dfrac{3}{5}\) \(=\dfrac{10}{11}\)\(:\dfrac{15}{22}\)x\(\dfrac{3}{5}\) \(=\dfrac{20}{22}:\dfrac{15}{22}\)x\(\dfrac{3}{5}\) \(=\dfrac{4}{3}\)x\(\dfrac{3}{5}=\dfrac{4}{5}\)
\(B=\frac{2^{10}\cdot55-2^{10}}{2^8\cdot27}=\frac{2^{10}\left(55-1\right)}{2^8\cdot3^3}=\frac{2^{10}\cdot2\cdot3^3}{2^8\cdot3^3}=\frac{2^{11}\cdot3^3}{2^8\cdot3^3}=2^3=8.\)
\(C=\frac{\left(3\cdot4\cdot24\right)^2}{5\cdot2^5\cdot4^2-16^2}=\frac{288^2}{5\cdot16\cdot2^3.16-16^2}=\frac{2^5\cdot3^2}{16^2\left(5\cdot2^3-1\right)}=\frac{16\cdot2\cdot9}{16\cdot16\left(5\cdot8-1\right)}\)
\(=\frac{16\cdot2\cdot9}{16\cdot16\cdot39}=\frac{16\cdot2\cdot9}{16\cdot2^4\cdot13\cdot3}=\frac{1\cdot3}{2^3\cdot13\cdot1}=\frac{3}{8\cdot13\cdot1}=\frac{3}{104}\)
\(D=\frac{11\cdot3^{22}\cdot3^7-9^{14}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{30}-3^{28}}{2^2\cdot3^{28}}=\frac{3^{28}\left(11\cdot3^2-1\right)}{4\cdot3^{38}}=\frac{3^{28}\left(11\cdot9-1\right)}{4\cdot3^{28}}=\frac{99-1}{4}=\frac{98}{4}=24,5\)
Ta có:\(x=18\Rightarrow\hept{\begin{cases}x+1=19\\-x-1=-19\end{cases}}\)
Thay vào BT D ta được:
\(D=x^{12}+\left(-x-1\right)x^{11}+\left(x+1\right)x^{10}+\left(-x-1\right)x^9+...+\left(x+1\right)x^2+\left(-x-1\right)x+1\)
\(=x^{12}-x^{12}-x^{11}+x^{11}+x^{10}-x^{10}-x^9+...+x^3+x^2-x^2-x+1\)
\(=1-x\)
\(19x+22\times3:14=11-6^2\)
\(\Leftrightarrow19x+66:14=11-36\)
\(\Leftrightarrow19x+\dfrac{33}{7}=-25\)
\(\Leftrightarrow19x=-25-\dfrac{33}{7}\)
\(\Leftrightarrow19x=-\dfrac{208}{7}\)
\(\Leftrightarrow x=\dfrac{-\dfrac{208}{7}}{19}\)
\(\Leftrightarrow x=-\dfrac{208}{133}\)