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

\(A=\sqrt{\left(37-12\right)}.\sqrt{37+12}=\sqrt{25}.\sqrt{49}=5.7=35\)
\(B=\sqrt{21,8+18,2}.\sqrt{21,8-18,2}=\sqrt{40}.\sqrt{3,6}=2\sqrt{10}.\sqrt{3,6}=12\)
\(C=\sqrt{100}.\sqrt{6,5-1,6}.\sqrt{6,5+1,6}=10.\sqrt{4,9}.\sqrt{8,1}=63\)

\(a,\Leftrightarrow\left(2x-2-\sqrt{3}\right)\left(2x+2+\sqrt{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=2+\sqrt{3}\\2x=-2-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2+\sqrt{3}}{2}\\x=\dfrac{-2-\sqrt{3}}{2}\end{matrix}\right.\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\left(\sqrt{x}-\sqrt{13}\right)^2-\left(2\sqrt{13}\right)^2=0\\ \Leftrightarrow\left(\sqrt{x}-3\sqrt{13}\right)\left(\sqrt{x}+\sqrt{13}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\sqrt{13}\\\sqrt{x}=-\sqrt{13}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=117\left(tm\right)\\x\in\varnothing\left(\sqrt{x}\ge0\right)\end{matrix}\right.\\ \Leftrightarrow x=117\)

Chúc bạn học tốt nha !

\(c,=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}=\sqrt{2}+\sqrt{3}\\ d,=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

c) \(\sqrt{5+2\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\left|\sqrt{3}+\sqrt{2}\right|=\sqrt{3}+\sqrt{2}\)

d) \(\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)

Yêu cầu: Tính/ rút gọn

c. $\sqrt{5+2\sqrt{6}}=\sqrt{3+2\sqrt{2.3}+2}=\sqrt{(\sqrt{3}+\sqrt{2})^2}$

$=\sqrt{3}+\sqrt{2}$
d. $\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3.1}+1}=\sqrt{(\sqrt{3}-1)^2}=\sqrt{3}-1$

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

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

\(c,5xy\sqrt{\dfrac{25x^2}{y^6}}=\dfrac{5xy\left|5x\right|}{\left|y^3\right|}=\dfrac{5xy\left(-5x\right)}{y^3}=\dfrac{-25x^2y}{y^3}=\dfrac{-25x^2}{y^2}\)

\(d,=\dfrac{0,2x^3y^3\cdot4}{\left|x^2y^4\right|}=\dfrac{0,8x^3y^3}{x^2y^4}=\dfrac{0,8x}{y}=\dfrac{\dfrac{4}{5}x}{y}=\dfrac{5x}{4y}\)

h) \(\left(\sqrt{2}-3\right)\sqrt{11+6\sqrt{2}}=\left(\sqrt{2}-3\right)\sqrt{\left(\sqrt{2}+3\right)^2}=\left(\sqrt{2}-3\right)\left(\sqrt{2}+3\right)=2-9=-7\)

g) \(\sqrt{49+5\sqrt{96}}-\sqrt{49-5\sqrt{96}}=\sqrt{\left(5+2\sqrt{6}\right)^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}=5+2\sqrt{6}-5+2\sqrt{6}=4\sqrt{6}\)

\(a,=\sqrt{2,25}\cdot\sqrt{400}\cdot\sqrt{\dfrac{1}{4}}=1,5\cdot20\cdot\dfrac{1}{2}=15\\ b,=\sqrt{0,36}\cdot\sqrt{100}\cdot\sqrt{81}=0,6\cdot10\cdot9=54\\ c,=\sqrt{\dfrac{1}{100}\cdot81}=\sqrt{\dfrac{1}{100}}\cdot\sqrt{81}=\dfrac{1}{10}\cdot9=\dfrac{9}{10}\\ d,=\sqrt{0,01\cdot36\cdot3^4}=\sqrt{0,01}\cdot\sqrt{36}\cdot\sqrt{3^4}=0,1\cdot6\cdot9=5,4\)

\(a,=\sqrt{0,25}\cdot\sqrt{0,36}=0,5\cdot0,6=0,3\\ b,=\sqrt{2^4}\cdot\sqrt{\left(-5\right)^2}=2^2\cdot5=20\\ c,=\sqrt{1,44}\cdot\sqrt{100}=1,2\cdot10=12\\ d,=\sqrt{3^4}\cdot\sqrt{5^2}=3^2\cdot5=45\)

a. \(\sqrt{0,25.0,36}=\sqrt{0,25}.\sqrt{0,36}=0,5.0,6=0,3\)

b. \(\sqrt{2^4.\left(-5\right)^2}=\sqrt{2^4}.\sqrt{\left(-5\right)^2}=\sqrt{\left[\left(2\right)^2\right]^2}.\sqrt{5^2}=2^2.5=4.5=20\)

c. \(\sqrt{1,44.100}=\sqrt{1,2^2.10^2}=\sqrt{1,2^2}.\sqrt{10^2}=1,2.10=12\)

d. \(\sqrt{3^4.5^2}=\sqrt{3^4}.\sqrt{5^2}=\sqrt{\left[\left(3\right)^2\right]^2}.\sqrt{5^2}=3^2.5=9.5=45\)

\(\sqrt{5-x}-x=1\left(đk:5\ge x\ge-1\right)\)

\(\Leftrightarrow\sqrt{5-x}=x+1\)

\(\Leftrightarrow5-x=x^2+2x+1\)

\(\Leftrightarrow x^2+3x-4=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-4\left(ktm\right)\end{matrix}\right.\)

a: \(A=x\sqrt{x}-y\sqrt{y}+x\sqrt{y}-y\sqrt{x}\)

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

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

b: \(B=5x^2-7x\sqrt{y}+2y\)

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

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

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