\(A=\sqrt[]{50}+\sqrt[]{65}\Rightarrow A^2=50+65+2\sqrt[]{50.65}=115+2\sqrt[]{5.10.5.13=}115+10\sqrt[]{130}\left(1\right)\)

\(B=\sqrt[]{15}+\sqrt[]{115}\Rightarrow B^2=15+115+2\sqrt[]{15.115}=15+115+2\sqrt[]{3.5.5.23}=15+115+10\sqrt[]{69}\left(2\right)\)Ta có  \(10\sqrt[]{130}< 10\sqrt[]{69.2}=10\sqrt[]{2}\sqrt[]{69}< 15+10\sqrt[]{69}\left(3\right)\)

\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow A^2< B^2\Rightarrow A< B\)

\(\Rightarrow\sqrt[]{50}+\sqrt[]{65}< \sqrt[]{15}+\sqrt[]{115}\)

So sánh gì thế em, em nhập đủ đề vào hi

N = -3\(x\)(4\(x^2\) +5) - 2\(x^2\).(4 -6\(x\)) + 9\(x^2\) 

Vì |\(x\)| = 1;  ⇔ (|\(x\)|)² =  \(x^2\) = 1

Thay \(x^2\) = 1 vào N ta có:

N = -3\(x\)(4\(x^2\) + 5) - 2\(x^2\).(4 -6\(x\)) + 9\(x^2\) 

N = -3\(x\)( 4 + 5) - 2(4 - 6\(x\)) + 9

N = -3\(x\).9 - 8 + 12\(x\) + 9

N = - 27\(x\) + 12\(x\) + 1

N = -15\(x\) + 1

|\(x\)| =1 ⇒ \(x\) = 1; -1

thay \(x\) = 1 vào N = -15\(x\) + 1 = -15 + 1 = - 14

Thay \(x\) = -1 vào N = -15\(x\) + 1 = (-15).(-1) + 1 = 16

a) \(3^{54}\)

\(2^{200}=4^{100}>3^{54}\)

\(\Rightarrow3^{54}< 2^{200}\)

b) \(15^{12}=3^{12}.5^{12}\)

\(1^3.125^3=\left(5^3\right)^3=5^9< 3^{12}.5^{12}\)

\(\Rightarrow15^{12}>1^3.125^3\)

c) \(78^{12}-78^{11}=78^{11}.\left(7-1\right)=78^{11}.6\)

\(78^{11}-78^{10}=78^{10}.\left(7-6\right)=78^{10}.6< 78^{11}.6\)

\(\Rightarrow78^{12}-78^{11}>78^{11}-78^{10}\)

d) \(72^{45}-72^{44}=72^{44}.\left(72-1\right)=72^{44}.72>27^{44}\)

\(\Rightarrow72^{45}-72^{44}>27^{44}\)

e) \(3^{39}=\left(3^3\right)^{13}=27^{13}>11^{11}\)

\(\Rightarrow3^{39}>11^{11}\)

Giải chi tiết dùm mik nhé.

a, (\(\dfrac{2x}{7}-5\)) : (-8) = 0,75

    \(\dfrac{2}{7}x\) - 5              = 0,75 \(\times\) (-8)

     \(\dfrac{2}{7}x\) - 5          = -6

      \(\dfrac{2}{7}\)\(x\)               = -1

         \(x=-\dfrac{7}{2}\)

b, (3\(x-1\))² + 1 = 5

     (3\(x-1\))²      = 4

      \(\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\)

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

Vậy \(x\in\) { \(-\dfrac{1}{3}\); 1}

c, 7 - (1 - 2\(x\))² = 6

          (1 - 2\(x\))² = 1

           \(\left[{}\begin{matrix}1-2x=1\\1-2x=-1\end{matrix}\right.\)

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

Vậy \(x\in\){ 0; 1}

a, \(A=x^2\left(2x-1\right)+x\left(x+8\right)=2x^3-x^2+x^2+8x=2x^3+8x\)

Thay x = -2, ta có:

\(2\cdot\left(-2\right)^3+8\cdot\left(-2\right)=-32\)

b, \(A=2x^3+8x=0\\ \Leftrightarrow2x\left(x^2+4\right)=0\\ \Leftrightarrow x=0\)

Vậy A=0 khi x=0

a,A = \(x^2\).( 2\(x\) - 1) + \(x\)(\(x+8\))

A = 2\(x^3\) - \(x^2\) + \(x^2\) + 8\(x\)

A = 2\(x^3\) + 8\(x\)

b, \(x=-2\) ⇒ A = 2.(-2)³ + 8.(-2) = - 32 

A = 0 ⇔ 2\(x^3\) + 8\(x\) = 0

             2\(x\left(x^2+4\right)\) = 0 

              vì \(x^2\) + 4 > 0 ∀ \(x\) ⇒ \(x\)  =0 

\(\left(2x^2-3\right)\left(x^2+1\right)\left(\sqrt{x}-5\right)=0\left(ĐKXĐ:x\ge0\right)\\ \Leftrightarrow2\left(x^2-\dfrac{3}{2}\right)\left(x^2+1\right)\left(\sqrt{x}-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-\dfrac{3}{2}=0\\x^2+1=0\\\sqrt{x}-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{3}{2}\\x^2=-1\left(vô.lí\right)\\\sqrt{x}=5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{3}{2}}\left(loại:-\sqrt{\dfrac{3}{2}}\right)\\x=5^2=25\end{matrix}\right.\\ Vậy:x\in\left\{\sqrt{\dfrac{3}{2}};25\right\}\)

x ϵ {\(\sqrt{\dfrac{3}{2}}\);25}

Vì góc AOD là góc bẹt nên ta có:

    180 - 30 = 150 độ

Vì góc AOD và góc COB ; AOC và DOB là các cặp góc đối đỉnh nên góc AOC = 150 độ

góc BOC  = 30 độ

góc AOD = 30độ

góc BOD = 150 độ.

\(\widehat{BOD}\) = 150^o.

\(x=\dfrac{a-3}{9}\left(a\ne0\right)\)

a) Để \(x>0\Rightarrow\dfrac{a-3}{9}>0\Rightarrow a-3>0\Rightarrow a>3\)

b) Để \(x< 0\Rightarrow\dfrac{a-3}{9}< 0\Rightarrow a-3< 0\Rightarrow a< 3\)

c) Để \(x\in Z\Rightarrow a-3\in\left\{0;\pm9;\pm18;\pm27...\right\}\)

\(\Rightarrow a\in\left\{3;9;-3;1;-5;-24;27...\right\}\)

a) x = \(\dfrac{24}{-54}\) = \(\dfrac{-4}{9}\) = \(\dfrac{-20}{45}\)

y = \(\dfrac{-33}{55}\) = \(\dfrac{-3}{5}\) = \(\dfrac{-27}{45}\)

Vì \(\dfrac{-20}{45}\) > \(\dfrac{-27}{45}\) ⇒ \(\dfrac{24}{-54}\) > \(\dfrac{-33}{55}\) ⇒ x > y 

b) x = \(\dfrac{-19}{23}\) = \(\dfrac{4}{23}\) - 1

y = \(\dfrac{-25}{29}\) = \(\dfrac{4}{29}\) - 1

Vì \(\dfrac{4}{23}\) > \(\dfrac{4}{29}\) ⇒ \(\dfrac{4}{23}\) - 1 > \(\dfrac{4}{29}\) - 1 ⇒ \(\dfrac{-19}{23}\) > \(\dfrac{-25}{29}\) ⇒ x > y