1. Phạm Anh Sơn

2. Hoàng Thị Thu Phúc 

a: AB=AC=5/2=2,5m

BC=căn AB^2+AC^2

=>\(BC=\sqrt{2.5^2\cdot2}\simeq3,5\left(m\right)\)

b: \(AD=\dfrac{2.5\cdot2.5}{3.5}\simeq1,8\left(m\right)\)

Ví dụ 9:

a.

$x-\sqrt{x}-\sqrt{xy}+\sqrt{y}=(x-\sqrt{xy})-(\sqrt{x}-\sqrt{y})$

$=\sqrt{x}(\sqrt{x}-\sqrt{y})-(\sqrt{x}-\sqrt{y})$

$=(\sqrt{x}-\sqrt{y})(\sqrt{x}-1)$

b.

$=\sqrt{(x-3)(x+3)}-2\sqrt{x-3}$

$=\sqrt{x-3}(\sqrt{x+3}-2)$
c.

$x-\sqrt{x}-6=(x+2\sqrt{x})-(3\sqrt{x}+6)$

$=\sqrt{x}(\sqrt{x}+2)-3(\sqrt{x}+2)=(\sqrt{x}+2)(\sqrt{x}-3)$

Bài 3: 

ĐKXĐ: $x>0; x\neq 1$

a. 

\(E=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}:\left[\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}+\frac{2-x}{\sqrt{x}(\sqrt{x}-1)}\right]\)

\(=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}:\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}(\sqrt{x}-1)}\)

\(=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}:\frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-1)}=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}.\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}+1}=\frac{x}{\sqrt{x}-1}\)

b.

$E=\frac{x}{\sqrt{x}-1}>1$
$\Leftrightarrow \frac{x}{\sqrt{x}-1}-1>0$

$\Leftrightarrow \frac{x-\sqrt{x}+1}{\sqrt{x}-1}>0(*)$

Do $x-\sqrt{x}+1=(\sqrt{x}-\frac{1}{2})^2+\frac{3}{4}>0$ với mọi $x$ thuộc đkxđ nên $(*)\Leftrightarrow \sqrt{x}-1>0$

$\Leftrightarrow x>1$

Kết hợp đkxđ suy ra $x>1$

\(P=\sqrt{x}+1+\dfrac{1}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}}+1=3\)

Dấu = xảy ra khi x=1

Lời giải:

ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow x+4\sqrt{x-4}=4$

$\Leftrightarrow (x-4)+4\sqrt{x-4}=0$

$\Leftrightarrow \sqrt{x-4}(\sqrt{x-4}+4)=0$

Hiển nhiên $\sqrt{x-4}+4>0$ với mọi $x\geq 4$

$\Rightarrow \sqrt{x-4}=0$

$\Leftrightarrow x=4$ (tm)

(\(x\) - 2)(\(\sqrt{3x+1}\) ) - 1 = 3\(x\)  Đk : 3\(x\) + 1 ≥ 0;  \(x\) ≥ - \(\dfrac{1}{3}\)

(\(x\) - 2)(\(\sqrt{3x+1}\)) - (3\(x\) + 1) = 0

\(\sqrt{3x+1}\).(\(x\) - 2 - \(\sqrt{3x+1}\)) = 0

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

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

\(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x^2-4x+4=3x+1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x^2-7x+3=0\end{matrix}\right.\)

\(x^2\) - 7\(x\) + 3 = 0

△ = 49 -12 = 37

\(x_1\) = \(\dfrac{7+\sqrt{37}}{2}\)

\(x_{_{ }2}\) = \(\dfrac{-7-\sqrt{37}}{2}\) (loại)

a) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)

K=\(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)

\(=\left[\dfrac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\dfrac{\left(\sqrt{a}-1\right)+2}{a-1}\right]\)

\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\):\(\dfrac{1}{\sqrt{a}-1}\)

=\(\dfrac{a-1}{\sqrt{a}}\)

b) Khi a=3+\(2\sqrt{2}\), ta có:

K=\(\dfrac{2+2\sqrt{2}}{\sqrt{\left(\sqrt{2}+1\right)^2}}=\dfrac{\sqrt{2}\left(\sqrt{2+1}\right)}{\sqrt{2}+1}=\sqrt{2}\)

c) Vì\(\sqrt{a}>0\) nên để K<0 thì a-1<0 =>a<1

\(Q=\left(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right):\dfrac{2}{\sqrt{5}-\sqrt{3}}\)

\(Q=\left(\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\cdot\dfrac{\sqrt{5}-\sqrt{3}}{2}\)

\(Q=\left(\dfrac{\sqrt{3}}{1}+\dfrac{\sqrt{5}}{1}\right)\cdot\dfrac{\sqrt{5}-\sqrt{3}}{2}\)

\(Q=\left(\sqrt{5}+\sqrt{3}\right)\cdot\dfrac{\sqrt{5}-\sqrt{3}}{2}\)

\(Q=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}\)

\(Q=\dfrac{\left[\left(\sqrt{5}\right)^2-\left(\sqrt{3}\right)^2\right]}{2}\)

\(Q=\dfrac{5-3}{2}\)

\(Q=\dfrac{2}{2}\)

\(Q=1\)