\(\sqrt{x+1}\) + \(\sqrt{x+6}\) = 5

\(\sqrt{\left(x+1\right)\left(x-1\right)}\) + \(\sqrt{\left(x+6\right)\left(x-6\right)}\) = 5

x - 1 + x - 6 = 5

2x - 7 = 5

x = 6

\(ĐKXĐ:\left\{{}\begin{matrix}x+1\ge0\\x+6\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge-6\end{matrix}\right.\Leftrightarrow x\ge-1\)

- Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x+6}=b\end{matrix}\right.\left(a\ge0,b\ge\sqrt{5}\right)\)

\(\Rightarrow b^2-a^2=\left(\sqrt{x+6}\right)^2-\left(\sqrt{x+1}\right)=x+6-\left(x+1\right)=5\)

- Ta có hệ phương trình:

\(\left\{{}\begin{matrix}b+a=5\\b^2-a^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+a=5\\\left(b+a\right)\left(b-a\right)=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+a=5\\5\left(b-a\right)=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+a=5\\b-a=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+a=5\\2b=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3+a=5\\b=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{x+6}=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=4\\x+6=9\end{matrix}\right.\)

\(\Leftrightarrow x=3\left(nhận\right)\)

- Vậy tập nghiệm của pt trên là \(S=\left\{3\right\}\)

BH=CH=2,3 (trong tg cân đường cao đồng thời là đường trung trực) \(\Rightarrow BC=2BH=4,6\)

\(\widehat{BAH}=\widehat{CAH}=\dfrac{\widehat{A}}{2}=25^o\) (trong tg cân đường cao đồng thời là đường phân giác)

\(AB=AC\) (cạnh bên tg cân)

Xét tg vuông ABH có

\(\sin\widehat{BAH}=\dfrac{BH}{AC}\Rightarrow AB=AC=\dfrac{BH}{\sin\widehat{BAH}}=\dfrac{2,3}{\sin25^o}\)

\(C_{ABC}=AB+AC+BC=\dfrac{4,6}{\sin25^o}+4,6\)

ai giải hộ mk cái đi!!

Đặt \(A=\sqrt{2003}+\sqrt{2005};B=2\sqrt{2004}\)

\(A^2=2003+2005+2\sqrt{\left(2003.2005\right)}\)

\(=4008+2\sqrt{\left[2004-1\left(2004+1\right)\right]}\)

\(=3008+2\sqrt{\left(2004^2-1\right)}< 2.2004+2\sqrt{\left(2004^2\right)}=4.2004=B^2\\ \Rightarrow A< B\)

\(\dfrac{x-1}{5}+\dfrac{x+1}{7}+\left(x-1\right)\left(x+1\right)=\left(x+1\right)^2\)

\(\Leftrightarrow\dfrac{7\left(x-1\right)+5\left(x+1\right)}{35}+x^2-1=x^2+2x+1\)

\(\Leftrightarrow\dfrac{12x-2}{35}=2x+2\)

\(\Leftrightarrow\dfrac{6x-1}{35}=x+1\)

\(\Leftrightarrow35x+35=6x-1\)

\(\Leftrightarrow x=-\dfrac{36}{29}\)

mik cũng ko chắc đâu nếu sai thì thôi nhé:
 

ĐKXĐ : $-x^2+6x-9\ge 0$

$\iff$$-\left(-x^2+6x-9\right)\le 0$

$\iff$$x^2-6x+9\le 0$

$\iff$${\left(x-3\right)}^2\le 0$

Mà $\left(x-3\right)\ge 0$

Suy ra : ${\left(x-3\right)}^2=0$

$\iff$$x-3=0$

$\iff$$x=3$

ĐKXĐ: \(^{x^2}\)- 6x + 9 ≥0 với mọi x