\(\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\frac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}=\frac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\frac{x^3+y^3}{x\left(x^3-y^3\right)}\)

uầy

so amazing 

ai lm đc là legendary đó

viết cái đề nhìn zô đã ko mún làm r

`bạn tự kẻ hình nhé
ta đễ dàng cm dk DM=CM
Từ đó ta có SAMDAMD=1/2 SDACDAC=1/3 SABCABC
SBDMBDM = 1/2SBDCBDC= 1/6 SABCABC
Suy ra SABMABM=(1/3+1/6)SABCABC= 1/2SABCABC= 15m^2

cho hỏi là làm sao dm=cm

a) P xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x+5\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}}\)

Vậy P xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{\left(x-5\right)\left(x+5\right)2}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)

Có: \(P=0\)

\(\Rightarrow P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}=0\Leftrightarrow x\left(x^2+4x-5\right)=0\Leftrightarrow x^2+4x-5=0\)

\(\Leftrightarrow\left(x^2-x\right)+\left(5x-5\right)=0\)

\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

Vậy \(P=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

a(a + 2) + b(b - 2) - 2ab

= a² + 2a + b² - 2b - 2ab

= (a² - 2ab + b²) +(2a - 2b)

= (a - b)² + 2(a - b)

= 7² + 2.7

= 49 + 14 =63

\(a\left(a+2\right)+b\left(b-2\right)-2ab=a^2+2a+b^2-2b-2ab\)

\(=\left(a^2-2ab+b^2\right)+\left(2a-2b\right)=\left(a-b\right)^2+2\left(a-b\right)\)

Với \(a-b=7\)thì biểu thức có giá trị là: \(7^2-7=49-7=42\)

a,ĐK:  \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)

\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)

c, Với x = 4 thỏa mãn ĐKXĐ thì

\(A=\frac{-3}{4-3}=-3\)

d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)

\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)

Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)

DDK : \(x\ge1\)

\(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)

\(\Leftrightarrow\sqrt{x-1}=\sqrt{3x-2}+\sqrt{5x-1}\)

\(\Rightarrow x-1=3x-2+5x-2+2\sqrt{\left(3x-2\right)\left(5x-1\right)}\)

\(\Leftrightarrow x-1-3x+2-5x+2=2\sqrt{15x^2-3x-10x+2}\)

\(\Leftrightarrow3-7x=2\sqrt{15x^2-13x+2}\)

\(\Rightarrow9-42x+49x^2=4\left(15x^2-13x+2\right)\)

\(\Leftrightarrow9-42x+49x^2=60x^2-52x+8\)

\(\Leftrightarrow11x^2-10x-1=0\)

\(\Leftrightarrow11x^2-11x+x-1=0\)

\(\Leftrightarrow\left(11x+1\right)\left(x-1\right)=0\)

Giải nốt nha .