a, CM: AD//AB=AE//AC

Xét tam giác ABC có:

AD//AB vì đề bài cho cạnh BC lấy D ( lấy sao cho AD=AB)

AE//AC vì đề bài cho cạnh AC lấy E  ( lấy sao cho AE=AC)

VÌ ĐỀU CHUNG MỘT TAM GIÁC NÊN 3 CẠNH = NHAU 

\(\Rightarrow\) AD/AB=AE/AC.

b, AB = 2cm vì AD= 2cm( AD//AB \(\Rightarrow=\)nhau và = 2 cm)

đêm hôm khuya khoắt đăng lên lm j :v 

\(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left(a+b+c-a\right)\left(3x^2+b^2+c^2+3ab+2bc+3ac\right)-\left(b^3+c^3\right)\)

\(=\left(3a^2+b^2+c^2+3ab+2ac-b^2+bc-c^2\right)\left(b+c\right)\)

\(=\left(3a^2+3ab+3ac+3bc\right)\left(b+c\right)\)

\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)

\(\text{ (a+b+c)^3−a^3−b^3−c^3 =a^3+3a^2(b+c)+3a(b+c)^2+(b+c)^2−a^3−b^3−c^3 =3(b+c)(a^2+ab+ac)+b^3+3b^2c+3bc^2+c^3−b^3−c^3 =3(b+c)(a^2+ab+ac+bc) =3(b+c)[a(a+b)+c(a+b)] =3(b+c)(a+b)(a+c)}\)

2X hay 2x ?

\(\frac{2x-1}{x^2+1}=\frac{2}{x^2-x+1}-\frac{1}{x+1}\)đề như này hả , hay thế nào == 

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

\(2x^4-x^3-2x-1=3x^3+3x+1-x^4\)

\(2x^4-x^3-2x-1-3x^3-3x-1+x^4=0\)

\(3x^4-4x^3-x-2=0\)

\(x=-0.618034;1.618034\)( đù -.- ra lắm :v ) 

Ta có : \(x.\left(x+1\right)-2x=0\)

\(\Leftrightarrow2x+x-2x=0\)

\(\Leftrightarrow x=0\)

Vậy x=0 

hok tốt!!

x(x-1)-2x=0

x^2-x-2x=0

x^2-3x=0

x(x-3)=0

\(\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Trả lời:

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

\(\Leftrightarrow x-2+2=x^2+2x\)

\(\Leftrightarrow x=x^2+2x\)

\(\Leftrightarrow x-x^2-2x=0\)

\(\Leftrightarrow x^2+x=0\)

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

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

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

Vậy\(x\in\left\{0;-1\right\}\)

Hok tốt!

Good girl

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

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

\(\Leftrightarrow x\left(x+2-1\right)=0\)

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

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

KL

Câu 1 :

Đặt \(n^2+2n+4=a^2\)

\(\Leftrightarrow\left(n+1\right)^2+3=a^2\)

\(\Leftrightarrow a^2-\left(n+1\right)^2=3\)

\(\Leftrightarrow\left(a-n-1\right)\left(a+n+1\right)=3\)

TH1 \(\hept{\begin{cases}a+n+1=3\\a-n-1=1\end{cases}}\)

TH2 : \(\hept{\begin{cases}a+n+1=-3\\a-n-1=-1\end{cases}}\)

TH3 : \(\hept{\begin{cases}a+n+1=-1\\a-n-1=-3\end{cases}}\)

TH4 : \(\hept{\begin{cases}a+n+1=1\\a-n-1=3\end{cases}}\)

Bạn tính ra trong từng TH nhé !

Câu 1 :

Giả sử : \(n^2+2n+4=k^2\left(k\inℤ\right)\)

\(\Rightarrow k^2-\left(n^2+2n+1\right)=3\)

\(\Rightarrow k^2-\left(n+1\right)^2=3\)

\(\Rightarrow\left(k+n+1\right)\left(k-n-1\right)=3\)

Do k + n + 1 > k - n - 1 ( với k;n thuộc Z )

\(\Rightarrow\hept{\begin{cases}k+n+1=3\\k-n-1=1\end{cases}}\Rightarrow\hept{\begin{cases}k+n=2\\k-n=2\end{cases}}\Rightarrow\hept{\begin{cases}k=2\\n=0\end{cases}}\) 

Vậy n = 0