Bài 8: Rút gọn biểu thức chứa căn bậc hai

Bài 1:

a: \(x^2=9\)

=>\(x^2=3^2=\left(-3\right)^2\)

=>\(x=\pm3\)

b: \(9x^2=16\)

=>\(\left[{}\begin{matrix}3x=4\\3x=-4\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

c: \(\left(x+2\right)^2=16\)

=>\(\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

d: \(x^2-9+\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(x+3\right)+\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(x+3+1\right)=0\)

=>\(\left(x-3\right)\left(x+4\right)=0\)

=>\(\left[{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

Bài 2:

a: ĐKXĐ: x>0

\(\sqrt{x}=21\)

=>\(x=21^2\)

=>\(x=441\)

b: ĐKXĐ: x>=0

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

=>\(\sqrt{x}=-\dfrac{1}{2}\)(vô lý)

Vậy: \(x\in\varnothing\)

c: ĐKXĐ: x>=0

\(\left(\sqrt{x}+1\right)^2=4\)

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

=>\(\sqrt{x}=1\)

=>x=1(nhận)

d: ĐKXĐ: x>=0

\(\left|\sqrt{x}-1\right|=2\)

=>\(\left[{}\begin{matrix}\sqrt{x}-1=2\\\sqrt{x}-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-1\left(loại\right)\end{matrix}\right.\)

=>\(\sqrt{x}=3\)

=>x=9(nhận)

e: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-6x+9}=5x-2\)

=>\(\sqrt{\left(x-3\right)^2}=5x-2\)

=>\(\left\{{}\begin{matrix}5x-2>=0\\\left(5x-2\right)^2=\left(x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{5}\\\left(5x-2\right)^2-\left(x-3\right)^2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{5}\\\left(5x-2-x+3\right)\left(5x-2+x-3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{5}\\\left(4x+1\right)\left(6x-5\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{5}\\\left[{}\begin{matrix}x=-\dfrac{1}{4}\left(loại\right)\\x=\dfrac{5}{6}\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy: x=5/6

f: \(\sqrt{\left(x-1\right)^2}=3-2x\)

=>\(\left\{{}\begin{matrix}3-2x>=0\\\left(3-2x\right)^2=\left(x-1\right)^2\end{matrix}\right.\)

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

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

=>\(\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x-2\right)\left(3x-4\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left[{}\begin{matrix}x=2\left(loại\right)\\x=\dfrac{4}{3}\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)

Lời giải:
Xét tử thức:

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

\(\Rightarrow C=\frac{x-1}{\sqrt{x}}: \frac{\sqrt{x}+1}{x}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}}.\frac{x}{\sqrt{x}+1}=\sqrt{x}(\sqrt{x}-1)\)

\(P=\dfrac{x+3}{\sqrt{x}+3}\) (ĐK: \(x\ge0\))

Mà: \(x\ge0\Rightarrow\left\{{}\begin{matrix}x+3\ge3\\\sqrt{x}+3\ge3\end{matrix}\right.\) nên:

\(P=\dfrac{x+3}{\sqrt{x}+3}\ge\dfrac{3}{3}=1\)

Dấu "=" xảy ra:

\(\dfrac{x+3}{\sqrt{x}+3}=1\)

\(\Leftrightarrow x=\sqrt{x}\)

\(\Leftrightarrow x=0\left(tm\right)\)

Vậy: \(P_{min}=1\) khi \(x=0\) 

Lời giải:
a.  ĐKXĐ: $x>0; x\neq 1$

\(P=\left[\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}+\frac{\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\frac{x+\sqrt{x}}{\sqrt{x}+2}\)

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

b.

$P>2 \Leftrightarrow \frac{x}{\sqrt{x}-1}-2>0$

$\Leftrightarrow \frac{x-2\sqrt{x}+2}{\sqrt{x}-1}>0$

$\Leftrightarrow \frac{(\sqrt{x}-1)^2+1}{\sqrt{x}-1}>0$

$\Leftrightarrow \sqrt{x}-1>0$ (do $(\sqrt{x}-1)^2+1>0$)

$\Leftrightarrow x>1$

Kết hợp đkxđ suy ra $x>1$
c. 

$\frac{1}{P}=\frac{\sqrt{x}-1}{x}$

Áp dụng BĐT Cô-si:

$x+4\geq 4\sqrt{x}\Rightarrow x\geq 4(\sqrt{x}-1)$
$\Rightarrow \frac{\sqrt{x}-1}{x}\leq \frac{\sqrt{x}-1}{4(\sqrt{x}-1)}=\frac{1}{4}$

Vậy $\frac{1}{P}$ max $=\frac{1}{4}$ khi $x=4$

a: ΔABC vuông tại A

=>\(BC^2=AB^2+AC^2\)

=>\(BC=\sqrt{6^2+8^2}=10\left(cm\right)\)

Xét ΔABC vuông tại A có AH là đường cao

nên \(AH\cdot BC=AB\cdot AC\)

=>\(AH\cdot10=6\cdot8=48\)

=>AH=4,8cm

Xét ΔABC vuông tại A có \(sinACB=\dfrac{AB}{BC}=\dfrac{3}{5}\)

=>\(\widehat{ACB}\simeq36^052'\)

b: ΔHAB vuông tại H có HE là đường cao

nên \(AE\cdot AB=AH^2\left(1\right)\)

ΔHAC vuông tại H có HF là đường cao

nên \(AF\cdot AC=AH^2\left(2\right)\)

Từ (1),(2) suy ra \(AE\cdot AB=AF\cdot AC\)

=>\(\dfrac{AE}{AC}=\dfrac{AF}{AB}\)

Xét ΔAEF vuông tại A và ΔACB vuông tại A có

\(\dfrac{AE}{AC}=\dfrac{AF}{AB}\)

Do đó: ΔAEF đồng dạng với ΔACB

=>\(\widehat{AFE}=\widehat{ABC}\)

1: Khi x=36 thì \(A=\dfrac{6}{2\cdot6-4}=\dfrac{6}{12-4}=\dfrac{6}{8}=\dfrac{3}{4}\)

2: 

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >4\end{matrix}\right.\)

\(C=B:A\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{3\sqrt{x}-x}{x-4}\right):\dfrac{\sqrt{x}}{2\sqrt{x}-4}\)

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

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

3: \(C\cdot\sqrt{x}< \dfrac{4}{3}\)

=>\(\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{3}< 0\)

=>\(\dfrac{2\sqrt{x}\cdot3-4\left(\sqrt{x}+2\right)}{3\left(\sqrt{x}+2\right)}< 0\)

=>\(6\sqrt{x}-4\sqrt{x}-8< 0\)

=>\(2\sqrt{x}-8< 0\)

=>\(\sqrt{x}< 4\)

=>\(0< =x< 16\)

Kết hợp ĐKXĐ của C, ta được: \(\left\{{}\begin{matrix}0< x< 16\\x< >4\end{matrix}\right.\)

\(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)

Điều kiện: \(x\ge5\).

Phương trình tương đương với:

\(\sqrt{4\left(x-5\right)}+\dfrac{3\sqrt{x-5}}{\sqrt{9}}=3\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}=3\)

\(\Leftrightarrow\sqrt{x-5}=1\Rightarrow x-5=1\Leftrightarrow x=6\left(TM\right)\)

Vậy: Phương trình có tập nghiệm \(S=\left\{6\right\}\).