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a) Vì x< 0 nên x= \(-\sqrt{7}\)
b) x-2 =\(\sqrt{2}\)hoặc x-2 = -\(\sqrt{2}\)
suy ra x= \(\sqrt{2}\)+2 hoặc x= \(-\sqrt{2}\)+2
c)
x+\(\sqrt{3}\) =\(\sqrt{5}\)hoặc x+\(\sqrt{3}\) = -\(\sqrt{5}\)
suy ra x= \(\sqrt{5}-\sqrt{3}\)hoặc x= \(-\sqrt{5}-\sqrt{3}\)
Các bạn tự kết luận nhé
a) 273 : 32 = (33)3 : 32
= 39 : 32
= 37
b) (3/5)15 : (9/25)5 = (3/5)15 : [(3/5)2]5
= (3/5)15 : (3/5)10
= (3/5)2
a) \(\left(\frac{1}{3}-\frac{1}{5}\right)^2:\left(\frac{1}{5}\right)^2=\left[\left(\frac{1}{3}-\frac{1}{5}\right):\frac{1}{5}\right]^2=\left(\frac{2}{15}:\frac{1}{5}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c)\(7\frac{1}{23}+\frac{10}{27}-5\frac{1}{23}+\frac{17}{27}+2^3=\left(7\frac{1}{23}-5\frac{1}{23}\right)+\left(\frac{10}{27}+\frac{17}{27}\right)+2^3=2+1+8=11\)
d)\(5.\left(-\frac{5}{2}\right)^2+\frac{1}{5}.\left(-3\right)^2=5.\frac{25}{4}+\frac{1}{5}.9=\frac{125}{4}+\frac{9}{5}=\frac{661}{20}\)
A) \(=\frac{\left(-1\right).2^{17}.5^6.3^{12}}{2^{16}.5^53^{13}}=\frac{10}{3}\)
B) Tương tự câu A bạn tự làm nha
a)\(\left(0,25^{10}\right).4^{10}.\sqrt{5^2-3^2}=\left(0,25.4\right)^{10}.\sqrt{25-9}=1^{10}.\sqrt{16}=1.4=4\)
b)\(\frac{\left(-3\right)^6.15^5+9^3.\left(-15\right)^6}{\left(-3\right)^{10}.5^5.2^3}=\frac{3^6.15^5+3^6.15^6}{3^{10}.5^5.2^3}=\frac{3^6.15^5.\left(1+15\right)}{3^{10}.5^5.2^3}\)\(=\frac{3^{11}.5^5.16}{3^{10}.5^5.2^3}=3.2=6\)
2)a)\(4-\left|x+\frac{2}{3}\right|=-1\Rightarrow\left|x+\frac{2}{3}\right|=5\Rightarrow\orbr{\begin{cases}x+\frac{2}{3}=5\\x+\frac{2}{3}=-5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=\frac{-17}{3}\end{cases}}\)
b)\(\frac{x-2}{-9}=\frac{16}{2-x}\Rightarrow\left(x-2\right)^2=144\Rightarrow\orbr{\begin{cases}x-2=12\\x-2=-12\end{cases}\Rightarrow\orbr{\begin{cases}x=14\\x=-10\end{cases}}}\)
c)\(\frac{2}{3}x+\frac{1}{7}=\frac{5}{3}\Rightarrow\frac{2}{3}x=\frac{32}{21}\Rightarrow x=\frac{16}{7}\)
=>(-a^5(-a^5))^2 + (-a^2.a^2)^5
=>(a^10)^2+(-a^4)^5
=>a^20-(a^4)^5
=>a^20-a^20
=>0(tmđk)
Em tham khảo nhé!
\(\left[-a^5\left(-a\right)^5\right]^2+\left[-a^2\left(-a\right)^2\right]^5\)
\(=\left[\left(-a\right)^{5+5}\right]^2+\left[\left(-a\right)^{2+2}\right]^5\)
\(=\left[\left(-a\right)^{10}\right]^2+\left[\left(-a\right)^4\right]^5\)
\(=\left(-a\right)^{20}+\left(-a\right)^{20}\)
\(=2\left(-a\right)^{20}\)