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

Lời giải:

$\sqrt{2022}+\sqrt{2022}=2\sqrt{2022}> 2\sqrt{2021}$ do $2022> 2021$

\(=\sqrt{\sqrt{3}}^2=\sqrt{3}\)

\(\sqrt{3^2.\left(-4\right)^2}=\sqrt{3^2}.\sqrt{\left(-4\right)^2}=3.4=12\)

\(\sqrt{3^2.\left(-4\right)^2}=3.4=12\)

Bạn cần bài nào

ko viết lại đề nhé
a) |x-3|=3-x
\(\Leftrightarrow\left[{}\begin{matrix}x-3=3-x\\x-3=x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x\in R\end{matrix}\right.\)
Vậy...



 

\(\sqrt{3-\sqrt{2}}.\sqrt{3+\sqrt{2}}=\sqrt{3^2-\sqrt{2}^2}=\sqrt{7}\)

`\sqrt{3-\sqrt{2}}.\sqrt{3+\sqrt{2}}`

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

`=\sqrt{9-2}=\sqrt{7}`

`\sqrt{\sqrt{3}}.\sqrt{\sqrt{3}}`

`=\sqrt{\sqrt{3}.\sqrt{3}}`

`=\sqrt{3}`

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

\(\sqrt{\sqrt{3}}.\sqrt{\sqrt{3}}\)

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

\(=\left|\sqrt{3}\right|=\sqrt{3}\)

`a)`\(6\sqrt{6}-27=\sqrt{6}^3-27=\left(\sqrt{6}-3\right)\left(6+3\sqrt{6}+9\right)\)

`b)`\(a-2\sqrt{a}=\sqrt{a}\left(\sqrt{a}-2\right)\)

`c)`\(1-a=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\)

`d)`\(1+a\sqrt{a}=1+\sqrt{a}^3=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)\)

`a)3-2\sqrt{3}=(\sqrt{3})^2-2\sqrt{3}=\sqrt{3}(\sqrt{3}-2)`

`b)2\sqrt{3}-\sqrt{6}=\sqrt{6}.\sqrt{2}-\sqrt{6}=\sqrt{6}(\sqrt{2}-1)`

`a, 3 - 2 sqrt 3 = (sqrt 3)^2 - 2 sqrt 3 = sqrt 3 . (sqrt 3 - 2)`

`b, `2 sqrt 3 - sqrt 6 = sqrt 12 - sqrt 6 = sqrt 6. sqrt 2 - sqrt 6 = sqrt 6(sqrt 2 - 1)` 

\(a)3-2\sqrt{3}\\=3-2\sqrt{3}.1+1\\=(\sqrt{3}-1)^2\\b)2\sqrt{3}-\sqrt{6}\\=2\sqrt{3}-\sqrt{2}\sqrt{3}\\=\sqrt{2}\sqrt{3}(\sqrt{2}-1)\\=\sqrt{6}(\sqrt{2}-1)\)

a: \(A=4\cdot\dfrac{5}{2}\sqrt{x}-\dfrac{8}{3}\cdot\dfrac{3}{2}\sqrt{x}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)

\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}=\dfrac{11}{2}\sqrt{x}\)

b: \(B=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-2y\right)-\dfrac{3}{2}\)

=1/2y+3/4-3/2y-3/2

=-y-3/4