A) -15,5 x 20,8 +2,5 x 9,2 -15,5 x 9,2 + 3,5 x 20,8

=-15,5 x 20,8 + 2,5 x 9,2 + -15,5 x 9,2 +3,5 x 20,8

=[20,8x(-15,5+3,5)]+[9,2x(2,5+-15,5)]

=[20,8x(-12)]+[9,2x(-13)]

=-249,6+-119,6

=-369,2

a) -369,2

b) -7

Ai trả lời nhanh nhất mình tik cho

a) \(\left|2x+1\right|=\left|x+4\right|\Rightarrow\left[{}\begin{matrix}2x+1=x+4\\2x+1=-x-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\3x=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)

b) \(\left|2x-1\right|=x+4\Rightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\3x=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

\(x^{10}=x\)

\(\Rightarrow x^{10}-x=0\)

\(\Rightarrow x\left(x^9-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^9-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^9=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

thank kiu

a) Ta có :

\(27^{27}>27^{26}=\left(27^2\right)^{13}=729^{13}>243^{13}\)

\(\Rightarrow27^{27}>243^{13}\)

\(\Rightarrow-27^{27}< -243^{13}\)

\(\Rightarrow\left(-27\right)^{27}< \left(-243\right)^{13}\)

b) \(\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{8}\right)^{26}=\left(\dfrac{1}{8^2}\right)^{13}=\left(\dfrac{1}{64}\right)^{13}>\left(\dfrac{1}{128}\right)^{13}\)

\(\Rightarrow\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{128}\right)^{13}\)

\(\Rightarrow\left(-\dfrac{1}{8}\right)^{25}< \left(-\dfrac{1}{128}\right)^{13}\)

c) \(4^{50}=\left(4^5\right)^{10}=1024^{10}\)

\(8^{30}=\left(8^3\right)^{10}=512^{10}< 1024^{10}\)

\(\Rightarrow4^{50}>8^{30}\)

d) \(\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{9}\right)^{12}< \left(\dfrac{1}{27}\right)^{12}\)

\(\Rightarrow\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{27}\right)^{12}\)

a) Ta có :

$27^{27}>27^{26}={\left(27^2\right)}^{13}=729^{13}>243^{13}$

$\Rightarrow 27^{27}>243^{13}$

$\Rightarrow -27^{27}<-243^{13}$

$\Rightarrow {\left(-27\right)}^{27}<{\left(-243\right)}^{13}$

b) ${\left(\genfrac{}{}{0ex}{}{1}{8}\right)}^{25}>{\left(\genfrac{}{}{0ex}{}{1}{8}\right)}^{26}={\left(\genfrac{}{}{0ex}{}{1}{8^2}\right)}^{13}={\left(\genfrac{}{}{0ex}{}{1}{64}\right)}^{13}>{\left(\genfrac{}{}{0ex}{}{1}{128}\right)}^{13}$

$\Rightarrow {\left(\genfrac{}{}{0ex}{}{1}{8}\right)}^{25}>{\left(\genfrac{}{}{0ex}{}{1}{128}\right)}^{13}$

$\Rightarrow {\left(-\genfrac{}{}{0ex}{}{1}{8}\right)}^{25}<{\left(-\genfrac{}{}{0ex}{}{1}{128}\right)}^{13}$

c) $4^{50}={\left(4^5\right)}^{10}=1024^{10}$

$8^{30}={\left(8^3\right)}^{10}=512^{10}<1024^{10}$

$\Rightarrow 4^{50}>8^{30}$

d) ${\left(\genfrac{}{}{0ex}{}{1}{9}\right)}^{17}<{\left(\genfrac{}{}{0ex}{}{1}{9}\right)}^{12}<{\left(\genfrac{}{}{0ex}{}{1}{27}\right)}^{12}$

$\Rightarrow {\left(\genfrac{}{}{0ex}{}{1}{9}\right)}^{17}<{\left(\genfrac{}{}{0ex}{}{1}{27}\right)}^{12}$

 Số số hạng của biểu thức đó là:

   (100 - 2) : 2 + 1 = 50 (số hạng)

Tổng của biểu thức đó là:

  (100 + 2) x 50 : 2 = 2550