Ta có: AB//CD

=>\(\widehat{BAD}+\widehat{ADC}=180^0\)

=>\(2\cdot\left(\widehat{KAD}+\widehat{KDA}\right)=180^0\)

=>\(\widehat{KAD}+\widehat{KDA}=90^0\)

=>ΔKAD vuông tại K

=>\(\widehat{AKD}=90^0\)

Độ dài cạnh là \(\sqrt{64}=8\left(cm\right)\)

Chu vi hình vuông là 8x4=32(cm)

=>Chọn B

b nha

\(\dfrac{x}{9}-\dfrac{3}{y}=\dfrac{1}{18}\)

=>\(\dfrac{xy-27}{9y}=\dfrac{1}{18}\)

=>\(18\cdot\left(xy-27\right)=9y\)

=>2(xy-27)=y

=>2xy-y=54

=>y(2x-1)=54

mà x,y là các số tự nhiên

nên \(\left(2x-1;y\right)\in\left\{\left(1;54\right);\left(3;18\right);\left(9;6\right);\left(27;2\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(1;54\right);\left(2;18\right);\left(5;6\right);\left(14;2\right)\right\}\)

=>A max=1+54=55

\(\dfrac{1}{4}=\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{9}=\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{100}=\dfrac{1}{10^2}< \dfrac{1}{9\cdot10}=\dfrac{1}{9}-\dfrac{1}{10}\)

Do đó: \(A=\dfrac{1}{4}+\dfrac{1}{9}+..+\dfrac{1}{100}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

=>\(A< 1-\dfrac{1}{10}=\dfrac{9}{10}\)

=>A<B

\(\left(7x+2\right)^{-1}=3^{-2}\)

=>\(\dfrac{1}{7x+2}=\dfrac{1}{3^2}=\dfrac{1}{9}\)

=>7x+2=9

=>7x=7

=>x=1

\(\dfrac{-7}{13}+2+\dfrac{6}{13}\\ =\left(\dfrac{-7}{13}+\dfrac{6}{13}\right)+2\\ =\dfrac{-1}{13}+\dfrac{26}{13}\\ =\dfrac{-1+26}{13}\\ =\dfrac{25}{13}\)

1: (2x-1)*x>0

TH1: \(\left\{{}\begin{matrix}2x-1>0\\x>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x>0\end{matrix}\right.\Leftrightarrow x>\dfrac{1}{2}\)

TH2: \(\left\{{}\begin{matrix}2x-1< 0\\x< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x< 0\end{matrix}\right.\)

=>x<0

2: 
ĐKXĐ: x<>1

\(\dfrac{x+3}{x-1}< 0\)

TH1: \(\left\{{}\begin{matrix}x+3>0\\x-1< 0\end{matrix}\right.\)

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

=>-3<x<1

TH2: \(\left\{{}\begin{matrix}x+3< 0\\x-1>0\end{matrix}\right.\)

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

=>Loại

c: 

ĐKXĐ: x<>0

\(\dfrac{x^2-2}{5x}< 0\)

TH1: \(\left\{{}\begin{matrix}x^2-2< 0\\5x>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2< 2\\x>0\end{matrix}\right.\Leftrightarrow0< x< \sqrt{2}\)

TH2: \(\left\{{}\begin{matrix}x^2-2>0\\5x< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2>2\\x< 0\end{matrix}\right.\Leftrightarrow-\sqrt{2}< x< 0\)

d: (x-3)(x+7)>0

TH1: \(\left\{{}\begin{matrix}x-3>0\\x+7>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>3\\x>-7\end{matrix}\right.\Leftrightarrow x>3\)

TH2: \(\left\{{}\begin{matrix}x-3< 0\\x+7< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\x< -7\end{matrix}\right.\)

=>x<-7

\(x^2=x^4\)

=>\(x^2\left(1-x^2\right)=0\)

=>\(\left[{}\begin{matrix}x^2=0\\1-x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

\(x^2=x^4\\ x^4-x^2=0\\ x^2\left(x^2-1\right)=0\)

TH1: `x^2=0`

`=> x=0`

TH2: `x^2-1=0`

`=>x^2=1^2`

`=>x=1` hoặc `x=-1` 

Tuổi con là (28-8):(3-1)=20:2=10(tuổi)

Tuổi mẹ là 3x10+8=38(tuổi)

2 lần tuổi conlà:

28 - 8 = 20 (tuổi) 

Tuổi con là:

20 : 2 = 10 (tuổi)

Tuổi mẹ là: 

28 + 10 = 38 (tuổi)

ĐS: ...