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a.\(\frac{5}{4}x^2y.\left(\frac{-5}{6}xy\right)^0\left(\frac{-7}{3}xy\right)\)= \(\frac{5}{4}x^2y.1.\left(\frac{-7}{3}xy\right)\)= \(\frac{-35}{12}x^3.y^2\)
câu b, c,d làm tương tự như trên nha ^.^
a) \(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{7}x=\frac{1}{4}\\\frac{-3}{4}x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)
Vậy \(x=\frac{7}{20}\) hoặc x=\(\frac{2}{3}\)
b) \(\left(\frac{4}{5}+x\right)\left(x-\frac{8}{13}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{4}{5}+x=0\\x-\frac{8}{13}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{8}{13}\end{cases}}\)
Vậy x=-4/5 hoặc x=8/13
c) \(\left(2x-\frac{1}{2}\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=3\end{cases}}\)
Vậy x=1/4 hoặc x=3
\(x+\frac{7}{2}x+x=\frac{1}{2}\)
\(2x+\frac{7}{2}x=\frac{1}{2}\)
\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)
\(\frac{11}{2}x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{11}{2}\)
\(x=\frac{1}{11}\)
a) \(4.\left(\frac{-1}{3}\right)^3-6.\left(\frac{-1}{2}\right)^2+\left(\frac{1}{2}\right)^0\)
= \(4.\left(\frac{-1}{27}\right)-6.\left(\frac{1}{4}\right)+1\)
= \(\frac{-4}{27}-\frac{3}{2}+1\)
= \(\frac{-8}{54}-\frac{81}{54}+\frac{54}{54}\)
= \(\frac{-35}{54}\)
b) \(\left(3^2\right)^2-\left[\left(-5\right)^2\right]^2+\left[\left(-2\right)^3\right]^2\)
= \(3^{2.2}-\left(-5\right)^{2.2}+\left(-2\right)^{3.2}\)
= \(3^4-\left(-5\right)^4+\left(-2\right)^6\)
= \(81-625+64\)
= \(-480\)
a,\(\left(x-\frac{2}{3}\right),\left(x+\frac{1}{1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{2}{3}\\x+\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{-1}{4}\end{matrix}\right.\)
b,\(\left(x-\frac{2}{3}\right)\left(2x-\frac{3}{4}\right)=\left(3x+\frac{1}{2}\right)\left(x+\frac{2}{3}\right)\)
\(\Leftrightarrow2x^2-\frac{3}{4}x-\frac{4}{3}x+\frac{1}{2}=3x^2+2x+\frac{1}{2}x+\frac{1}{3}\)
\(\Leftrightarrow2x^2-\frac{25}{12}x+\frac{1}{2}=3x^2+\frac{5}{2}x+\frac{1}{3}\)
\(\Leftrightarrow24x^2-25x+6=36x^2+30x+4\)
\(\Leftrightarrow24x^2-25x+6-36x^2-30x-4=0\)
\(\Leftrightarrow-12x^2-55x+2=0\)
\(\Leftrightarrow12x^2+55x-2=0\)
\(\Leftrightarrow x=\frac{-55\pm\sqrt{55^2-4.12\left(-2\right)}}{2.12}\)
\(\Leftrightarrow\frac{-55\pm\sqrt{3025+96}}{24}\)
\(\Leftrightarrow\frac{-55\pm\sqrt{3121}}{24}\)
\(\Leftrightarrow\frac{-55+\sqrt{3121}}{24}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{-55+\sqrt{3121}}{24}\\\frac{-55-\sqrt{3121}}{24}\end{matrix}\right.\)
Gửi tạm trước 2 câu !
\(a,\text{ }3^2\cdot\frac{1}{243}\cdot81^2\cdot3^{-3}=3^2\cdot\frac{1}{3^5}\cdot\left(3^4\right)^2\cdot\frac{1}{3^3}=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=3^2=9\)\(b,\text{ }\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(-3\cdot3\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3^{15}\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3}{-5}\)
Trả lời :
\(a,\text{ }3^2\cdot\frac{1}{243}\cdot81^2\cdot3^{-3}=3^2\cdot\frac{1}{3^5}\cdot\left(3^4\right)^2\cdot\frac{1}{3^3}=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=3^2=9\)\(b,\text{ }\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(-3\cdot3\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3^{15}\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3}{-5}\)
a) \(\left(-\frac{1}{2}\right)^3+\left(\frac{5}{6}\right)^0-\left(-\frac{3}{2}\right)^0-\left(-1\right)^0\)
\(=-\frac{1}{8}+1-1-1\)
\(=-\frac{1}{8}-1\)
\(=-\frac{9}{8}\)
b) \(4\cdot\left(-\frac{1}{3}\right)^0-2\cdot\left(-\frac{1}{2}\right)^2+3\cdot\left(-\frac{1}{2}\right)+1\)
\(=4\cdot1-2\cdot\frac{1}{4}-\frac{3}{2}+1\)
\(=4-\frac{1}{2}-\frac{3}{2}+1\)
\(=\frac{9}{2}\)