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

a)\(\sqrt{3x}=4\)

\(\Leftrightarrow3x=16\)

\(\Leftrightarrow x=\dfrac{16}{3}\)

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

\(\Leftrightarrow1-2x=2\)

\(\Leftrightarrow-2x=1\)

\(\Leftrightarrow x=\dfrac{-1}{2}\)

b) \(\sqrt{3x}-\dfrac{1}{2}\sqrt{3x}+\dfrac{3}{4}\sqrt{3x}+5=5\sqrt{3x}\)

\(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{3}{4}-5\right)\sqrt{3x}=-5\)

\(\Leftrightarrow\dfrac{-15}{4}\sqrt{3x}=-5\)

\(\Leftrightarrow\sqrt{3x}=\dfrac{4}{3}\)

\(\Leftrightarrow3x=\dfrac{16}{9}\Leftrightarrow x=\dfrac{16}{27}\)

\(\sqrt{8-2\sqrt{15}}+\sqrt{48+6\sqrt{15}}\\ =\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{45}+\sqrt{3}\right)^2}\\ =\sqrt{5}-\sqrt{3}+3\sqrt{5}+\sqrt{3}\\ =4\sqrt{5}\)

\(\dfrac{x}{x^2+x+1}=\dfrac{1}{4}\)

=>\(x^2+x+1=4x\)

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

\(F=\dfrac{x^5-3x^4+x^3+3x^4-9x^3+3x^2+5x^3-15x^2+5x+12x^2-36x+12+21x}{x^2\left(x^2-3x+1\right)+3x\left(x^2-3x+1\right)+15\left(x^2-3x+1\right)+27x}\)

\(=\dfrac{12x}{27x}=\dfrac{4}{9}\)

a: ĐKXĐ: x>=0; x<>1

b: Khi x=9 thì \(A=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)

c: \(P=A\cdot B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{x+2\sqrt{x}-3+4\sqrt{x}+4+5-x}{x-1}\)

\(=\dfrac{6\sqrt{x}+6}{\left(\sqrt{x}+1\right)^2}=\dfrac{6}{\sqrt{x}+1}\)

1: =>x+1=5

=>x=4

2: \(\Leftrightarrow\left|x-5\right|=2x+2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(2x+2-x+5\right)\left(2x+2+x-5\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(x+7\right)\left(3x-3\right)=0\end{matrix}\right.\Leftrightarrow x=1\)

3: \(\Leftrightarrow\sqrt{3+x}\left(\sqrt{3-x}+1\right)=0\)

=>x+3=0

=>x=-3

\(\dfrac{1}{7}\sqrt{51}với\dfrac{1}{9}\sqrt{150}\)

<=> \(\dfrac{\sqrt{51}}{7}với\dfrac{\sqrt{150}}{9}\)

<=> \(9\sqrt{51}với7\sqrt{150}\)

<=> \(\sqrt{4131}với\sqrt{7350}\)

=> \(\sqrt{4131}< \sqrt{7350}\)

=> \(\dfrac{1}{7}\sqrt{51}< \dfrac{1}{9}\sqrt{150}\)

a: \(=3-2\sqrt{2}-\sqrt{2}+1+1+\dfrac{1}{2}\sqrt{2}\)

\(=-\dfrac{5}{2}\sqrt{2}+5\)

b: \(=\dfrac{x-4+10-x}{\sqrt{x}+2}=\dfrac{6}{\sqrt{x}+2}\)

c: \(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)

Câu 1:

ĐK: \(x\geq \frac{-3}{2}\)

\(\sqrt{2x+3}=3-\sqrt{5}\)

\(\Rightarrow 2x+3=(3-\sqrt{5})^2=14-6\sqrt{5}\)

\(\Rightarrow x=\frac{11-6\sqrt{5}}{2}\)

Câu 2: ĐK: \(x\geq 0\)

\(\sqrt{5+\sqrt{7x}}=2+\sqrt{7}\)

\(\Rightarrow 5+\sqrt{7x}=(2+\sqrt{7})^2=11+4\sqrt{7}\)

\(\Rightarrow \sqrt{7x}=6+4\sqrt{7}\)

\(\Rightarrow 7x=(6+4\sqrt{7})^2\Rightarrow x=\frac{(6+4\sqrt{7})^2}{7}\)

Câu 3: ĐK: \(x\geq 0\)

\((\sqrt{x}-2)(5-\sqrt{x})=4-x\)

\(\Leftrightarrow 5\sqrt{x}-x-10+2\sqrt{x}=4-x\)

\(\Leftrightarrow 7\sqrt{x}=14\Rightarrow \sqrt{x}=2\Rightarrow x=4\)

Câu 4: ĐK: \(x\ge 1\)

Sửa đề \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)

\(\Leftrightarrow \frac{\sqrt{x-1}}{2}-\frac{3}{2}\sqrt{9}.\sqrt{x-1}+24\sqrt{\frac{1}{64}}\sqrt{x-1}=-17\)

\(\Leftrightarrow \frac{\sqrt{x-1}}{2}-\frac{9\sqrt{x-1}}{2}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow \sqrt{x-1}(\frac{1}{2}-\frac{9}{2}+3)=-17\)

\(\Leftrightarrow -\sqrt{x-1}=-17\Rightarrow \sqrt{x-1}=17\Rightarrow x=17^2+1=290\)

a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)

b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)

c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)

\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)