chu vi là 2.3,14.1062.8
diện tích là 3,14.10.10=314

a)
\(=\dfrac{4.4.25+4.11.4.25}{29.2.48+71.2.48}=\dfrac{400+400.11}{96.29+71.96}=\dfrac{400\left(1+11\right)}{96\left(29+71\right)}=\dfrac{400.12}{96.100}=\dfrac{2.2.100.12}{12.2.2.2.100}=\dfrac{1}{2}\)

có AB//CD,AD//BC
=> ABCD là hình thoi
=> B=D

F=35.1.8+32=280+32=312

a)
a//b => B2=A4=40⁰(so le trong)
b)
B1=A1 ( a//b và ở vị trí đồng vị)
c)
B1=180⁰-B2
B1=180⁰-40⁰=140⁰

1A
a) \(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(3+x+y\right)\left(x-y\right)\)
b) \(=\left(x^2+2xy+y^2\right)-4x^2y^2=\left(x+y\right)^2-\left(2xy\right)^2=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c)
\(=x^4\left(x^2-1\right)+2x^2\left(x-1\right)=x^4\left(x-1\right)\left(x+1\right)+2x^2\left(x-1\right)=\left(x^4\left(x+1\right)+2x^2\right)\left(x-1\right)\)
 

có \(2^{332}< 2^{333}\)
\(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
 mà \(8^{111}< 9^{111}\)
=>\(2^{332}< 2^{333}< 3^{222}< 3^{223}\)

tam giác ABC có 
AC²=BC²+AB²
AC²=18²+70²=5224
AC\(\approx72,28m\)

thay x=1 vào M ta có
\(M=\left(7-2\right)\left(4+14+49\right)-\left(64-8\right)\)
\(M=5.67-56=335-56=279\)
\(P=\left(2x-1\right)\left(4.x^2-2.x+1\right)+\left(2x-1\right)\left(1+2x+4x^2\right)\)
\(P=\left(2x-1\right)\left(4x^2-2x+1+1+2x+4x^2\right)\)
\(P=\left(2x-1\right)\left(8x^2+2\right)\)
thay x=10 vào P ta có
\(P=\left(2.10-1\right)\left(8.10^2+2\right)=\left(20-1\right)\left(800+2\right)=19.802=15238\)

a)
\(\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{3}\right)^3\)=>n=3
b)\(\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^4=>n=4\)
c)
\(\left(3.2\right)^n=36\)
6ⁿ=6²
n=2