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\(B=\left[\frac{4}{11}\cdot\left(\frac{1}{25}\right)^0+\frac{7}{22}\cdot2\right]^{2010}-\left(\frac{1}{2^2}:\frac{8^2}{4^4}\right)^{2009}\)
\(B=\left[\frac{4}{11}\cdot1+\frac{7}{22}\cdot2\right]^{2010}-\left(\frac{1}{2^2}:\frac{\left(2^3\right)^2}{\left(2^2\right)^4}\right)^{2009}\)
\(B=\left[\frac{4}{11}+\frac{7}{11}\right]^{2010}-\left(\frac{1}{2^2}:\frac{2^6}{2^8}\right)^{2009}\)
\(B=1^{2010}-\left(\frac{1}{2^2}\cdot\frac{2^8}{2^6}\right)^{2009}\)
\(B=1-1^{2009}=1-1=0\)
\(A=\left[\dfrac{4}{11}.\left(\dfrac{1}{25}\right)^0+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}:\dfrac{8^2}{4^2}\right)^{2009}\)
....=.\(\left(\dfrac{4}{11}.1+\dfrac{7}{11}\right)^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^2}{8^2}\right)^{2009}\)
.....\(=\left(\dfrac{4}{11}+\dfrac{7}{11}\right)^{2010}-0\)
....\(=1-0=1\)
\(a,A=\left[\frac{4}{11}.\left(\frac{1}{25}\right)^0+\frac{7}{22}.2\right]^{2010}-\left(\frac{1}{2^2}:\frac{8^2}{4^4}\right)^{2009}\)
\(A=\left(\frac{4}{11}.1+\frac{7}{11}\right)^{2010}-\left(\frac{1}{2^2}.2^2\right)^{2009}\)
\(A=1-1=0\)
\(b,B=\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)
\(B=\frac{0,8:1}{\frac{3}{5}}+\frac{\left(1\right):\frac{4}{7}}{\left(\frac{59}{9}-\frac{13}{4}\right).36}\)
\(B=0,8.\frac{5}{3}+\frac{\frac{7}{4}}{\frac{119}{36}.36}\)
\(B=\frac{4}{3}+\frac{7}{4}.\frac{1}{119}\)
\(B=\frac{4}{3}+\frac{1}{68}=\frac{275}{204}\)
ai lam guip toi cau nay voi mai toi nop bai roi
so sanh 2 phan so sau bang cach nahnh nhat: 2007/2008 voi 2008/2009
câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)
\(B=1^{2010}-1^{2009}=1-1=0\)
câu 2
a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)
\(\Leftrightarrow2x=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{24}\)
b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
\(\text{Bài 4:}\)
\(a.\left|x-\frac{3}{5}\right|< \frac{1}{3}\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}< \frac{1}{3}\\x-\frac{3}{5}>-\frac{1}{3}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x< \frac{14}{15}\\x>\frac{4}{15}\end{cases}\Rightarrow\frac{4}{15}< x< \frac{14}{15}}\)
\(b.\left|-5,5\right|=5,5\)
\(\Rightarrow\left|x+\frac{11}{2}\right|>5,5\Rightarrow\orbr{\begin{cases}x+\frac{11}{2}>5,5\\x+\frac{11}{2}< -5,5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>0\\x< -11\end{cases}}\)