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

\(a,3\sqrt{2x}+\sqrt{8x}-\sqrt{18x}=16\left(dk:x\ge0\right)\\ \Leftrightarrow3\sqrt{2x}+2\sqrt{2x}-3\sqrt{2x}=16\\ \Leftrightarrow\sqrt{2x}\left(3+2-3\right)=16\\ \Leftrightarrow2\sqrt{2x}=16\\ \Leftrightarrow\sqrt{2x}=8\\ \Leftrightarrow\left|2x\right|=64\\ \Leftrightarrow2x=64\\ \Leftrightarrow x=32\left(tm\right)\)

Vậy \(S=\left\{32\right\}\)

\(b,\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\left(dk:x\ge-5\right)\)

\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9\left(x+5\right)}=6\\ \Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}\left(2-3+4\right)=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=2\\ \Leftrightarrow\left|x+5\right|=4\\ \Leftrightarrow x+5=4\\ \Leftrightarrow x=-1\left(tm\right)\)

Vậy \(S=\left\{-1\right\}\)

\(\sqrt{6}+\sqrt{15}+\sqrt{21}\)

\(=\sqrt{3}\cdot\sqrt{2}+\sqrt{3}\cdot\sqrt{5}+\sqrt{3}\cdot\sqrt{7}\)

\(=\sqrt{3}\cdot\left(\sqrt{2}+\sqrt{5}+\sqrt{7}\right)\)

A) \(7+4\sqrt{3}\)

\(=4+4\sqrt{3}+3\)

\(=2^2+2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2\)

\(=\left(2+\sqrt{3}\right)^2\)

B) \(3\sqrt{2}+2\sqrt{3}\)

\(=\sqrt{3^2\cdot2}+\sqrt{2^2\cdot3}\)

\(=\sqrt{18}+\sqrt{12}\)

\(=\sqrt{6}\cdot\left(\sqrt{3}+\sqrt{2}\right)\)

C) \(\sqrt{45}-\sqrt{35}\)

\(=\sqrt{5\cdot9}-\sqrt{5\cdot7}\)

\(=\sqrt{5}\cdot\left(\sqrt{9}-\sqrt{7}\right)\)

\(=\sqrt{5}\left(3-\sqrt{7}\right)\)

D) \(x-y^2\)

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

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

\(\sqrt{144a}+3\sqrt{9a^2}+2a=\sqrt{12^2.a}+3\sqrt{\left(3a\right)^2}+2a\)

\(=12\sqrt{a}+3.\left|3a\right|+2a\)

\(=12\sqrt{a}+3.3a+2a\) ( do \(a\ge0\) )

\(=12\sqrt{a}+11a\)

\(\sqrt{\dfrac{1}{9}a}-\sqrt{\dfrac{2}{50}x}+3\sqrt{\dfrac{4}{16}x}\)

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

\(=\dfrac{1}{3}\sqrt{x}-\dfrac{1}{5}\sqrt{x}+3.\dfrac{2}{4}\sqrt{x}=\dfrac{49}{30}\sqrt{x}\)

\(\sqrt{\dfrac{1}{9}x}-\sqrt{\dfrac{2}{50}x}+3\sqrt{\dfrac{4}{16}x}\)

\(=\sqrt{\left(\dfrac{1}{3}\right)^2x}-\sqrt{\dfrac{1}{25}x}+3\sqrt{\dfrac{1}{4}x}\)

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

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

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

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

\(=\dfrac{49}{30}\sqrt{x}\)

\(\sqrt{144a}+3\sqrt{9a^2}+2a\)

\(=\sqrt{12^2\cdot a}+3\cdot\sqrt{3^2\cdot a^2}+2a\)

\(=12\sqrt{a}+3\cdot\left|3a\right|+2a\)

\(=12\sqrt{a}+3\cdot3a+2a\)

\(=12\sqrt{a}+9a+2a\)

\(=12\sqrt{a}+11a\)

3:

a: \(A=5\sqrt{3}-\dfrac{1}{2}\cdot4\sqrt{3}+10\sqrt{3}-7\sqrt{3}\)

=8căn 3-2căn 3

=6căn 3

1:

a: \(=4\sqrt{3}+3\sqrt{3}-3\sqrt{5}+\sqrt{5}\)

=7căn 3-2căn 5

b: \(=\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}\)

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

c: =2căn 5-3căn 5+3*3căn 2+6căn 2

=15căn 2-căn 5

d:

\(=\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}\)

\(=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)

2:

a: Sửa đề: \(\dfrac{a^2+3}{\sqrt{a^2+2}}>2\)

\(A=\dfrac{a^2+3}{\sqrt{a^2+2}}=\dfrac{a^2+2+1}{\sqrt{a^2+2}}=\sqrt{a^2+2}+\dfrac{1}{\sqrt{a^2+2}}\)

=>\(A>=2\cdot\sqrt{\sqrt{a^2+2}\cdot\dfrac{1}{\sqrt{a^2+2}}}=2\)

A=2 thì a^2+2=1

=>a^2=-1(loại)

=>A>2 với mọi a

b: \(\Leftrightarrow\sqrt{a}+\sqrt{b}< =\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}\)

=>\(a\sqrt{a}+b\sqrt{b}>=a\sqrt{b}+b\sqrt{a}\)

=>\(\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)-\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)>=0\)

=>(căn a+căn b)(a-2*căn ab+b)>=0

=>(căn a+căn b)(căn a-căn b)^2>=0(luôn đúng)

1

ĐK: `x>1`

PT trở thành:

\(\sqrt{\dfrac{2x-3}{x-1}}=2\\ \Leftrightarrow\dfrac{2x-3}{x-1}=2^2=4\\ \Leftrightarrow4x-4-2x+3=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow x=\dfrac{1}{2}\left(KTM\right)\)

Vậy PT vô nghiệm.

b

ĐK: \(x\ge2\)

Đặt \(t=\sqrt{x-2}\) (\(t\ge0\))

=> \(x=t^2+2\)

PT trở thành: \(t^2+2-5t+2=0\)

\(\Leftrightarrow t^2-5t+4=0\)

nhẩm nghiệm: `a+b+c=0` (`1+(-5)+4=0`)

\(\Rightarrow\left\{{}\begin{matrix}t=1\left(nhận\right)\\t=4\left(nhận\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{x-2}=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\left(TM\right)\\x=18\left(TM\right)\end{matrix}\right.\)

3:

a: \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)

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

2:

a: 2căn 7=căn 28

3căn 2=căn 18

mà 28>18

nên 2*căn 7>3*căn 2

b: 5=2+3

mà 3>căn 2

nên 2+3>2+căn 2

=>5>2+căn 2

1) a) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)

\(=\sqrt{49.2}-\sqrt{36.2}+0,5\sqrt{4.2}\)

\(=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}\)

\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

b) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49}\)

\(=3\sqrt{a}-4\sqrt{a}+7=7-\sqrt{a}\)

2. a) \(2\sqrt{7}=\sqrt{4.7}=\sqrt{28}\)

\(3\sqrt{2}=\sqrt{9.2}=\sqrt{18}\)

Mà \(\sqrt{28}>\sqrt{18}\Rightarrow2\sqrt{7}>3\sqrt{2}\)

b) \(5=2+3=2+\sqrt{9}\)

Vì \(\sqrt{9}>\sqrt{2}\Rightarrow2+\sqrt{9}>2+\sqrt{2}\Rightarrow5>2+\sqrt{2}\)

3. a) \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)

b) \(\dfrac{x}{y}.\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}.\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)

\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

\(\sqrt{2015}-\sqrt{2014}=\dfrac{1}{\sqrt{2015}+\sqrt{2014}}\)

căn 2016+căn 2015>căn 2015+căn 2014

=>1/(căn 2016+căn 2015)<1/(căn 2015+căn 2014)

=>căn 2016-căn 2015<căn 2015-căn 2014

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

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