\(42\cdot40+6\cdot8\cdot21-22\cdot96\)

\(=42\cdot40+24\cdot42-22\cdot96\)

\(=42\left(40+24\right)-22\cdot1,5\cdot64\)

\(=64\cdot42-64\cdot33=64\cdot9=576\)

\(250=5^3\cdot2\)

=>\(Ư\left(250\right)=\left\{1;-1;2;-2;5;-5;10;-10;25;-25;50;-50;125;-125;250;-250\right\}\)

\(Ư\left(250\right)\)\(=\left\{1;2;5;10;25;50;125;250\right\}\)

@Chu Gia Hân

bạn hỏi gì vậy???

Ta có: \(\widehat{BDA}+\widehat{DBA}=90^0\)(ΔBAD vuông tại A)

\(\widehat{CEB}+\widehat{CBE}=90^0\)(ΔCBE vuông tại C)

mà \(\widehat{DBA}=\widehat{CBE}\)

nên \(\widehat{BDA}=\widehat{CEB}\)

=>\(\widehat{CED}=\widehat{CDE}\)

=>ΔCDE cân tại C

ΔCDE cân tại C

mà CH là đường cao

nên CH là phân giác của góc ECD

a: \(x-\dfrac{3}{4}=6\cdot\dfrac{3}{8}\)

=>\(x-\dfrac{3}{4}=\dfrac{3}{4}\cdot3\)

=>\(x=\dfrac{9}{4}+\dfrac{3}{4}=\dfrac{12}{4}=3\)

b: \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)

=>\(\dfrac{7}{8}:x=\dfrac{5}{2}\)

=>\(x=\dfrac{7}{8}:\dfrac{5}{2}=\dfrac{7}{8}\cdot\dfrac{2}{5}=\dfrac{7}{20}\)

c: \(x+\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{3}{4}\)

=>\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

=>\(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{9}{12}-\dfrac{2}{12}=\dfrac{7}{12}\)

\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2004}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2003}{2004}\)

\(=\dfrac{1}{2004}\)

a: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

=>\(x\left(x-2\right)\left(x+2\right)=0\)

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

b: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

=>\(\left(x+2\right)\left(x+2-x+2\right)=0\)

=>4(x+2)=0

=>x+2=0

=>x=-2

c: \(6x^3+7x^2+2x=0\)

=>\(x\left(6x^2+7x+2\right)=0\)

=>\(x\left(6x^2+4x+3x+2\right)=0\)

=>\(x\left(3x+2\right)\left(2x+1\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\3x+2=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d: \(x^2+4x=7\)

=>\(x^2+4x+4=11\)

=>\(\left(x+2\right)^2=11\)

=>\(\left[{}\begin{matrix}x+2=\sqrt{11}\\x+2=-\sqrt{11}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{11}-2\\x=-\sqrt{11}-2\end{matrix}\right.\)

1: ta có: AB//CD
mà E\(\in\)AB; F\(\in\)CD

nên AE//DF; BE//CF

2: Ta có: AE+EB=AB

DF+FC=DC

mà AE=DF và AB=DC

nên EB=FC

3: Xét tứ giác AEFD có

AE//FD

AE=FD

Do đó: AEFD là hình bình hành

4: Xét tứ giác BEFC có

BE//FC
BE=FC

Do đó BEFC là hình bình hành