Bài 6: Biến đối đơn giản biểu thức chứa căn bậc hai

`(\sqrt{x}+2)(x-2\sqrt{x}+4)`
`= (\sqrt{x})^3 +2^3`
`= \sqrt{x^3} +8`

\(=\left(2\sqrt{7}-\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)

\(=7-2\sqrt{21}+2\sqrt{21}=7\)

\(\sqrt{a-2\sqrt{a-1}}\\ =\sqrt{\left(a-1\right)-2\sqrt{a-1}+1}\\ =\sqrt{\left(\sqrt{a-1}\right)^2-2\sqrt{a-1}.1+1^2}\\ =\sqrt{\left(\sqrt{a-1}-1\right)^2}\\ =\left|\sqrt{a-1}-1\right|\)

\(\Leftrightarrow4x-8\sqrt{x}-\sqrt{x}+2=4\sqrt{x}+4\)

\(\Leftrightarrow4x-9\sqrt{x}+2-4\sqrt{x}-4=0\)

\(\Leftrightarrow4x-13\sqrt{x}-2=0\)

Đặt \(\sqrt{x}=a\left(a>=0\right)\)

Pt sẽ là \(4a^2-13a-2=0\)(1)

\(\text{Δ}=\left(-13\right)^2-4\cdot4\cdot\left(-2\right)=169+32=201>0\)

Do đó: Phương trình (1) có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}a_1=\dfrac{13-\sqrt{201}}{8}\left(loại\right)\\a_2=\dfrac{13+\sqrt{201}}{8}\left(nhận\right)\end{matrix}\right.\)

=>\(x=\dfrac{370+26\sqrt{101}}{64}\)

\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\dfrac{2\sqrt{x}}{x-9}\) (ĐKXĐ: \(x>0;x\ne9\))

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\left[\dfrac{x-3\sqrt{x}+x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{2\sqrt{x}}\)

\(=\dfrac{2x}{2\sqrt{x}}\)

\(=\sqrt{x}\)

đk : x khác 9 ; x > 0 

\(=\dfrac{x-3\sqrt{x}+x+3\sqrt{x}}{x-9}:\dfrac{2\sqrt{x}}{x-9}=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)

d: \(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Bài 1

\(1=\left|A\right|\\ 2,=\sqrt{AB}\\ 3,=\dfrac{\sqrt{AB}}{B}\\ 4,=\left|A\right|\sqrt{B}\\ 5,=\dfrac{\sqrt{AB}}{B}\)

Bài 2:

\(a,=3\sqrt{5}-2\sqrt{5}+12\sqrt{5}=13\sqrt{5}\\ b,=\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{3}\)

\(AB=\sin C\cdot BC=\sin36^0\cdot7\approx4,1\left(cm\right)\)

\(ĐK:x\ge0\\ x\sqrt{x}-4x+2\sqrt{x}-8=0\\ \Leftrightarrow x\left(\sqrt{x}-4\right)+2\left(\sqrt{x}-4\right)=0\\ \Leftrightarrow\left(x+2\right)\left(\sqrt{x}-4\right)=0\\ \Leftrightarrow\sqrt{x}-4=0\left(x+2>0\right)\\ \Leftrightarrow x=16\)