Tìm x nha, mình nhâm nhầm

a,\(\frac{7}{10}\cdot\frac{4}{9}+\frac{3}{10}\cdot\frac{4}{9}-1\frac{7}{9}\)

\(=\frac{14}{45}+\frac{2}{15}-\frac{16}{9}\)

\(=\frac{14}{45}+\frac{6}{45}-\frac{80}{45}\)

\(=\frac{-60}{45}=\frac{-4}{3}\)

b,\(\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{5}{4}-\frac{2}{3}\right)\cdot\left(-3\right)^2+\frac{5}{9}\cdot30\%\)

\(=\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{7}{12}\right)\cdot9+\frac{5}{9}\cdot\frac{3}{10}\)

\(=\frac{-5}{6}+\frac{7}{3}+\frac{1}{6}\)

\(=\frac{-5}{6}+\frac{14}{6}+\frac{1}{6}\)

=\(=\frac{10}{6}=\frac{5}{3}\)

Bài 2: 

Ta có: 2³⁰⁰=2^3x100=(2³)¹⁰⁰=8¹⁰⁰

             3²⁰⁰=3^2x100=(3²)¹⁰⁰=9¹⁰⁰

Vì:8¹⁰⁰<9¹⁰⁰

==> 2³⁰⁰>3²⁰⁰

suy ra: 2^300<3^200

\(\frac{4}{5}\)+ \(\frac{1}{5}\): \(\frac{-2}{15}\)= \(\frac{4}{5}\)+ \(\frac{1}{5}\)* \(\frac{-15}{2}\)=\(\frac{4}{5}\) + \(\frac{-3}{2}\)= \(\frac{-7}{10}\)

b = \(\frac{-7}{12}\). (\(\frac{3}{11}\)+ \(\frac{-14}{11}\)) + \(\frac{5}{6}\)= \(\frac{-7}{12}\). (-1) + \(\frac{5}{6}\)= \(\frac{7}{12}\)+ \(\frac{10}{12}\)= \(\frac{17}{12}\)

b: I=-15+13=-2

Bài 2:

a: \(17-x=3\)

=>\(x=17-3\)

=>x=14(nhận)

b: \(2\cdot\left(x-1\right):3=6\)

=>\(2\left(x-1\right)=6\cdot3=18\)

=>x-1=18/2=9

=>x=9+1=10(nhận)

c: \(x+\left(-2\right)=\left(-11\right)+7\)

=>\(x-2=-4\)

=>\(x=-4+2=-2\left(loại\right)\)

d: \(\left(x-1\right)^2-5=20\)

=>\(\left(x-1\right)^2=25\)

=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)

Câu 3:

a: Đặt *=a

\(\overline{57a3}⋮9\)

=>\(5+7+a+3⋮9\)

=>\(a+15⋮3\)

mà 0<=a<=9

nên a=3

=>*=a

b: \(A=123\cdot7+8+9\)

123*7 là số lẻ

9 là số lẻ

=>123*7+9 chia hết cho 2

mà 8 chia hết cho 2

nên \(A=123\cdot7+9+8⋮2\)

\(123\cdot7⋮3;9⋮3;8⋮̸3\)

=>\(A=123\cdot7+9+8⋮̸3\)

c: \(B=3\cdot5\cdot7+10^{50}\)

\(=5\cdot3\cdot7+5\cdot5^{49}\cdot2^{49}\)

\(=5\left(3\cdot7+5^{49}\cdot2^{49}\right)⋮5\)

=>B là hợp số

con chóa này