x= 1372221/18559

(x-99) (1/30 + 1/32 + 1/34 - 1/36 - 1/38) = 0

SUy ra x - 99 = 0

VẬy x =99

Ta có:

27²⁰ + 3⁶¹ + 9³¹

= (3³)²⁰ + 3⁶¹ + (3²)³¹

= 3⁶⁰ + 3⁶¹ + 3⁶²

= 3⁶⁰.(1 + 3 + 3²)

= 3⁶⁰.(1 + 3 + 9)

= 3⁶⁰.13 chia hết cho 13 (đpcm)

\(\)Bạn viết thiếu 1 vế đúng không?

\(\dfrac{x-69}{30}+\dfrac{x-67}{32}+\dfrac{x-65}{34}=\dfrac{x-63}{36}+\dfrac{x-61}{38}+\dfrac{x-59}{40}\)\(\Rightarrow\left(\dfrac{x-69}{30}-1\right)+\left(\dfrac{x-67}{32}-1\right)+\left(\dfrac{x-65}{34}-1\right)=\left(\dfrac{x-63}{36}-1\right)+\left(\dfrac{x-61}{38}-1\right)+\left(\dfrac{x-59}{40}-1\right)\)

\(\Rightarrow\dfrac{x-99}{30}+\dfrac{x-99}{32}+\dfrac{x-99}{34}=\dfrac{x-99}{36}+\dfrac{x-99}{38}+\dfrac{x-99}{40}\)

\(\Rightarrow\dfrac{x-99}{30}+\dfrac{x-99}{32}+\dfrac{x-99}{34}-\dfrac{x-99}{36}-\dfrac{x-99}{38}-\dfrac{x-99}{40}=0\)\(\Rightarrow\left(x-99\right)\left(\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{34}-\dfrac{1}{36}-\dfrac{1}{38}-\dfrac{1}{40}\right)=0\)

Vì \(\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{34}-\dfrac{1}{36}-\dfrac{1}{38}-\dfrac{1}{40}\ne0\)

Nên:

\(x-99=0\Rightarrow x=99\)

chắc cô giáo mik viết sai đề

Ta có: \(27^{20}+3^{61}+9^{31}\)

\(=\left(3^3\right)^{20}+3^{61}+\left(3^2\right)^{31}\)

\(=3^{60}+3^{61}+3^{62}\)

\(=3^{60}.\left(1+3+3^2\right)\)

\(=3^{60}.13\)

Vì \(13⋮13\) nên \(3^{60}.13⋮13.\)

\(\Rightarrow27^{20}+3^{61}+9^{31}⋮13\left(đpcm\right).\)

Chúc bạn học tốt!

a) 27/35 > 27/40> 21/40 => 27/35>21/40

a) \(\frac{27}{35}=\frac{27\cdot8}{35\cdot8}=\frac{216}{280}\)và\(\frac{21}{40}=\frac{21\cdot7}{40\cdot7}=\frac{147}{280}\)nên\(\frac{27}{35}>\frac{21}{40}\)

b)Ta thấy phân số \(\frac{27}{67}\)và\(\frac{51}{152}\)~\(\frac{1}{2.5}\)và\(\frac{1}{3}\)nên\(\frac{27}{67}>\frac{51}{152}\)

\(a,\frac{\left(-17\right)}{61}:\frac{4}{27}=-\frac{17}{61}\cdot\frac{27}{4}=-\frac{119}{61}\)

\(b,\frac{\left(-1\right)}{31}:\left(-5\right)=-\frac{1}{31}\cdot\left(-\frac{1}{5}\right)=\frac{1}{155}\)

\(\overline{abcd}=100\overline{ab}+\overline{cd}=200\overline{cd}+\overline{cd}=201\overline{cd}=3.67.\overline{cd}⋮67\)

Câu 2 bạn ghi sai đề rồi nhé. 

Ví dụ \(135⋮27\)nhưng \(315⋮̸27\).

Sửa: Cho số \(\overline{abc}\)chia hét cho \(27\). Chứng minh rằng \(\overline{cab}\)cũng chia hết cho \(27\).

Ta có: \(\overline{abc}=100a+10b+c⋮7\Leftrightarrow10000a+1000b+100c⋮27\)

\(\Leftrightarrow10000-370.27a+1000b-37.27b+100c⋮27\)

\(\Leftrightarrow100c+10a+b=\overline{cab}⋮27\).

\(\sqrt{361=19}\)

\(\sqrt{121^2=121}\)

Vaayj \(\sqrt{361}+\sqrt{121^2=}19+121=140\)

\(\sqrt{361}+\sqrt{121^2}\)

\(=19+121\)

\(=140\)

\(k\)\(nha\)