Bài 3: Liên hệ giữa phép nhân và phép khai phương

Bài 2:

a: \(\sqrt{\dfrac{130}{200}}:\dfrac{\sqrt{5}}{\sqrt{8}}=\dfrac{\sqrt{130}}{10\sqrt{2}}\cdot\dfrac{2\sqrt{2}}{\sqrt{5}}=\dfrac{\sqrt{26}}{5}\)

b: \(=\sqrt{\dfrac{1}{5}:\dfrac{4}{5}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

c: \(=\left(3\sqrt{5}+3\sqrt{5}-5\sqrt{5}\right):3\sqrt{5}\)

\(=\dfrac{\sqrt{5}}{3\sqrt{5}}=\dfrac{1}{3}\)

\(a.\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}=\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}-\sqrt{45+2.2\sqrt{2}.3\sqrt{5}+8}=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}=-4\sqrt{5}\) \(b.\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}.\sqrt{3}+3}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\left(16-15\right)=2\)

\(\left(\sqrt{x^2+5}+x\right)\left(\sqrt{y^2+5}+y\right)=5\)

⇔ \(\left(\sqrt{x^2+5}+x\right)\left(\sqrt{x^2+5}-x\right)\left(\sqrt{y^2+5}+y\right)=5\left(\sqrt{x^2+5}-x\right)\) ⇔ \(5\left(\sqrt{y^2+5}+y\right)=5\left(\sqrt{x^2+5}-x\right)\)

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

Tương tự : \(x+y=\sqrt{y^2+5}-\sqrt{x^2+5}\left(2\right)\)

Cộng từng vế của ( 1 ; 2 ) , ta có : x + y = 0

Câu a :

\(\dfrac{\sqrt{15}-10\sqrt{2}-\sqrt{10}}{\sqrt{5}}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{3}-\sqrt{40}-\sqrt{2}\right)}{\sqrt{5}}\)

\(=\sqrt{3}-\sqrt{40}-\sqrt{2}\)

Câu b :

\(\dfrac{\sqrt{8}-4\sqrt{3}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{4}-\sqrt{24}\right)}{\sqrt{2}}\)

\(=\sqrt{4}-\sqrt{24}\)

Wish you study well !!

\(a.\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3=11-4\sqrt{6}\)

\(b.\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=\left(1+\sqrt{3}\right)^2-2=4+2\sqrt{3}-2=2+2\sqrt{3}\) \(c.\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{9-5}=6+4=10\) \(d.\left(\sqrt{\sqrt{11}+\sqrt{7}}-\sqrt{\sqrt{11}-\sqrt{7}}\right)^2=\sqrt{11}+\sqrt{7}+\sqrt{11}-\sqrt{7}-2\sqrt{11-7}=2\sqrt{11}-4\) \(e.\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}=\sqrt{2}\) \(f.\sqrt{21-12\sqrt{3}}-\sqrt{3}=\sqrt{12-2.2\sqrt{3}.3+9}-\sqrt{3}=2\sqrt{3}-3-\sqrt{3}=\sqrt{3}-3\)

\(g.\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{3+2\sqrt{3}+1}=\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)=\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)=2\left(2+\sqrt{3}\right)\left(\sqrt{3}-2\right)=2\left(3-4\right)=-2\)

\(h.\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{16-2.4\sqrt{2}+2}}}=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}}=\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\)

DK B>0

bieu thuc thanh b.[b-1]

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

Bài 1:

a, \(\sqrt{2x-1}=5\Rightarrow2x-1=25\Rightarrow2x=25+1=26\) \(\Rightarrow x=26:2=13\)

b,\(\sqrt{4\left(x-1\right)}=12\Rightarrow4\left(x-1\right)=12^2=144\)\(\Rightarrow x-1=144:4=36\Rightarrow x=36+1=37\)

c,\(\sqrt{x^2-6x+9}=5\Rightarrow\sqrt{\left(x-3\right)^2}=5\)\(\Rightarrow\left|x-3\right|=5\Rightarrow\left[{}\begin{matrix}x-3=5\\3-x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

Bài 2:

a, Để căn thức trên có nghĩa\(\Rightarrow2x-\dfrac{1}{3}\ge0\Rightarrow2x\ge\dfrac{1}{3}\Rightarrow x\ge\dfrac{1}{3}:2=\dfrac{1}{6}\)

Vậy để căn thức trên có nghĩa thì x>= 1/6

b, x<= 5/3

c, -1<=x<5

d, x>=6; x<=-1

Mình k chắc có đúng ko đâu

Biết đâu làm đó , sai thôi đừngg chửi nhé

1, Rút gọn

a) A = \(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}\) = \(\dfrac{\left(\sqrt{x}\right)^2+\sqrt{xy}}{\left(\sqrt{y}\right)^2+\sqrt{xy}}\) = \(\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{y}+\sqrt{x}\right)}\) = \(\dfrac{\sqrt{x}}{\sqrt{y}}\)

b) B = \(\sqrt{\dfrac{\left(a-b\right)^3.b^3}{c}}\) . \(\sqrt{\dfrac{bc^3}{\left(a-b\right)}}\)

= \(\sqrt{\dfrac{\left(a-b\right)^3.b^3}{c}.\dfrac{bc^3}{\left(a-b\right)}}\) = \(\sqrt{\left(a-b\right)^2.b^4.c^2}\)

= \(\left|a-b\right|\) . \(\left|b^2\right|\) . \(\left|c\right|\)

= -(a -b) .b². c

bài 2:

a/ \(\sqrt{x^2-4}-\sqrt{x-2}=0\) đk: x≥2

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

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

vậy pt có 1 nghiệm x = 2

b/ \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=5\)

Ta có: \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=\sqrt{3\left(x^2+4x+4\right)+4}+\sqrt{\left(y^2-4y+4\right)+9}=\sqrt{3\left(x+2\right)^2+4}+\sqrt{\left(y-2\right)^2+9}\ge\sqrt{4}+\sqrt{9}=2+3=5\)

=> Dấu ''='' xảy ra khi x = -2; y = 2

Vậy pt có nghiệm x=-2; y = 2

a) \(\dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}=\dfrac{\sqrt{2}\sqrt{3}+\sqrt{2}\sqrt{5}}{\sqrt{7}\sqrt{3}+\sqrt{7}\sqrt{5}}\)

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

b) \(\dfrac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}=\dfrac{9\sqrt{5}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}=3\dfrac{3\sqrt{3}+3\sqrt{5}}{3\sqrt{3}+3\sqrt{5}}=3.1=3\)

c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

= \(1-\sqrt{3}\)

P/s: bạn làm thêm bước nữa nha, mình lười, hehe

d) \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}.1+1^2}}{\sqrt{5}-1}\)

\(=\dfrac{\sqrt{\left(\sqrt{5-1}\right)^2}}{\sqrt{5}-1}=\dfrac{\left|\sqrt{5}-1\right|}{\sqrt{5}-1}=\dfrac{\sqrt{5}-1}{\sqrt{5}-1}=1\)

c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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