`[27^10+9^5]/[9^13+27^2]`

`=[(3^3)^10+(3^2)^5]/[(3^2)^13+(3^3)^2]`

`=[3^30+3^10]/[3^26+3^6]`

`=[3^6(3^24+3^4)]/[3^6(3^20+1)]`

`=[3^24+3^4]/[3^20+1]`

`=81`

Có :  \(\widehat{ACB}+\widehat{A'CB'}=60^0\)

Mà \(\widehat{ACB}=\widehat{A'CB'}\) (2 góc đối đỉnh )

=>\(\widehat{ACB}=\widehat{A'CB'}=\dfrac{60}{2}=30^0\) 

Ta có \(\widehat{ACB}+\widehat{BCA'}=180^0\) (2 góc kề bù )

=> \(\widehat{BCA'}=180^0-\widehat{ACB}=180^0-30^0=150^0\) 

\(\dfrac{x}{3}=\dfrac{y}{6}\Rightarrow x=\dfrac{1}{2}y\)

Ta có: \(x.y=62\)

\(\Rightarrow\dfrac{1}{2}y.y=62\)

\(\Rightarrow\dfrac{1}{2}y^2=62\)

\(\Rightarrow y^2=124\)

\(\Rightarrow y=2\sqrt{31}\)

\(\Rightarrow x=\sqrt{31}\)

\(\dfrac{x}{3}=\dfrac{y}{3}\Rightarrow\dfrac{4x}{12}=\dfrac{y}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\dfrac{4x}{12}=\dfrac{y}{3}=\dfrac{4x-y}{12-3}=\dfrac{42}{9}=\dfrac{14}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{14}{3}\Rightarrow x=14\\\dfrac{y}{3}=\dfrac{14}{3}\Rightarrow y=14\end{matrix}\right.\)

\(b,\)

\(27^2:25^3=\left(3^3\right)^2:\left(5^2\right)^3=3^6:5^6=\left(\dfrac{3}{5}\right)^6\)

\(c,\)

\(25^4.2^8=\left(5^2\right)^4.2^8=5^8.2^8=\left(5.2\right)^8.10^8\)

\(d,\)

\(32^4.64^3=\left(2^5\right)^4.\left(2^6\right)^3=2^{20}.2^{18}=2^{38}\)

\(25^3:5^2=\left(5^2\right)^3:5^2=\left(5^2\right)^2=5^4\)

\(e,\)

\(64^2.16^4=\left(2^6\right)^2.\left(2^4\right)^4=2^{12}.2^{16}=2^{28}\)

\(625^5.5^3=\left(5^4\right)^5:5^3=5^{20}:5^3=5^{17}\)

\(\left|x-3,5\right|=3x\)  (*)

ĐK \(x\ge0\)

Từ (*) => \(\left[{}\begin{matrix}x-3,5=3x\\x-3,5=-3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=3,5\\4x=3,5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{4}\left(loại\right)\\x=\dfrac{7}{8}\left(tm\right)\end{matrix}\right.\)

Vậy x=7/8

Do |x-3,5| >= 0 suy ra x >=0

Trường hợp 1: x - 3,5 =3x => 2x = -3,5 (Loại gì x < 0)

Trường hợp 2: 3,5 -x = 3x => 4x = 3,5 => x = 3,5/4 = 0,875

Kết luận : X = 0,875

Mọi thắc mắc tới toán có thể liên hệ để được giải đáp thêm

Ta có :

\(\widehat{xoy}+\widehat{yoz}=180^0\) (vì kề bù)`

Mà : \(\widehat{yoz}=75^0\)

`=>`\(\widehat{xoy}=180^0-\widehat{yoz}\)

\(=180^0-75^0\)

`= 105^0`

Vậy : \(\widehat{xoy}=105^0\)

\(\left(-\dfrac{6}{13}+\dfrac{5}{13}-\dfrac{47}{13}-\dfrac{1}{8}\right):\left(99\dfrac{17}{65}-100\dfrac{5}{52}+\dfrac{1}{130}\right)\)

\(=\left(\dfrac{-6+5-47}{13}-\dfrac{1}{8}\right):\left(99\dfrac{17}{65}-99\dfrac{5}{52}-1+\dfrac{1}{130}\right)\)

\(=\left(\dfrac{-48}{13}-\dfrac{1}{8}\right):\left(\dfrac{17}{65}-\dfrac{5}{52}-1+\dfrac{1}{130}\right)\)

\(=\left(\dfrac{-384}{104}-\dfrac{13}{104}\right):\left(\dfrac{68}{260}-\dfrac{25}{260}-\dfrac{260}{260}+\dfrac{2}{260}\right)\)

\(=\dfrac{-397}{104}:\dfrac{-215}{260}\)

\(=\dfrac{-397}{104}:-\dfrac{43}{52}\)

\(=\dfrac{-397}{104}\cdot\dfrac{52}{-43}=\dfrac{397}{86}\)