\(\frac{\left(-27\right)^{10}.16^{25}}{-3.\left(-32\right)^{15}}\)các bạn ơi giúp mình với, mình cần gấp
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A) \(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)\)
\(=\frac{1}{2}\cdot\frac{1}{2}\)
\(=\frac{1}{4}\)
B) \(\left(\frac{-5}{28}+1.75+\frac{8}{35}\right):\left(-3\frac{9}{20}\right)\)
\(=\left(\frac{-5}{28}+\frac{7}{4}+\frac{8}{35}\right):\frac{-69}{20}\)
\(=\frac{14}{5}:\frac{-69}{20}\)
\(=\frac{-56}{69}\)
\(\left(1900-2x\right):35-32=16\)
\(\left(1900-2x\right):35=48\)
\(1900-2x=1680\)
\(2x=220\)
\(x=110\)
\(\left(1900-2x\right):35=16+32\)
\(\left(1900-2x\right):35=48\)
\(1900-2x=48.35\)
\(1900-2x=1680\)
\(2x=1900-1680\)
\(2x=220\)
\(x=220:2\)
\(x=110\)
Vậy x=110
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) 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
\(=\frac{\left(x+1\right)\left(x+2\right)\left(x-5\right)\left(x+5\right)}{\left(x+2\right)\left(x+5\right)}=\left(x+1\right)\left(x-5\right)=x^2-4x-5\)
\(\frac{\left(-27\right)^{10}.16^{25}}{-3.\left(-32\right)^{15}}=\frac{\left(-3\right)^{30}.\left(-2\right)^{100}}{\left(-3\right).\left(-2\right)^{75}}=\frac{\left(-3\right).\left(-3\right)^{29}.\left(-2\right)^{75}.\left(-2\right)^{25}}{\left(-3\right).\left(-2\right)^{75}}\)
= \(\left(-3\right)^{29}.\left(-2\right)^{25}\)