a: \(x^3+xy^2-y^2-1\)

\(=\left(x^3-1\right)+y^2\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+1\right)+y^2\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+1+y^2\right)\)

b: \(12x^2+4x-6xy-2y\)

\(=4x\left(3x+1\right)-2y\left(3x+1\right)\)

\(=\left(3x+1\right)\left(4x-2y\right)=2\left(2x-y\right)\left(3x+1\right)\)

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

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

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

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

\(250=5^3\cdot2\)

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

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

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

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

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

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

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

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