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

mn ơi dạng này như thế nào vậy

\(a.\dfrac{4}{x+1}>0\)

\(\Leftrightarrow x+1>0\)

\(\Leftrightarrow x>-1\)

\(b.\dfrac{-5}{-x+2}< 0\)

\(\Leftrightarrow-x+2>0\)

\(\Leftrightarrow x< 2\)

\(c.\dfrac{x-3}{-3}< 0\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\)

\(d.\dfrac{x+1}{4}\ge0\)

\(\Leftrightarrow x+1\ge0\)

\(\Leftrightarrow x\ge-1\)

\(e.\dfrac{x+2}{x-3}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2>0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-2\\x>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< -2\\x< 3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\)

\(f.\dfrac{2x-5}{1-3x}< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5>0\\1-3x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5< 0\\1-3x>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{5}{2}\\x>\dfrac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x< \dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{5}{2}\\x< \dfrac{1}{3}\end{matrix}\right.\)

\(\sqrt{5}.\sqrt{45}=\sqrt{5.45}=\sqrt{225}=15\)

\(\sqrt{2,5}.\sqrt{14,4}=\sqrt{2,5.14,4}=\sqrt{36}=6\)

\(\sqrt{45.80}=\sqrt{3600}=60\)

a: \(\sqrt{x+4\sqrt{x-4}}=2\)

\(\Leftrightarrow\sqrt{x-4+2\cdot\sqrt{x-4}\cdot2+4}=2\)

\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)

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

=>x-4=0

hay x=4

b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=1\\x^2-3x+2=x-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=1\\x^2-4x+3=0\end{matrix}\right.\)

=>x=1 hoặc x=3

c: \(\Leftrightarrow\left|2x-3\right|=\left|x-2\right|\)

=>2x-3=x-2 hoặc 2x-3=-x+2

=>x=1 hoặc 3x=5

=>x=1 hoặc x=5/3

D

D

`\sqrt{x^4y^2} = \sqrt{(x^2y)^2} = x^2y`
`-> D`
 

Câu 4: 

a: \(=\dfrac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\)

b: \(=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}\)

c: \(=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\dfrac{x-\sqrt{2}}{x+\sqrt{2}}\)

d: \(=\dfrac{1}{x+\sqrt{5}}\)

Câu 2:

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

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

\(P=\dfrac{\left(2-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}=\dfrac{6+2\sqrt{6}-3\sqrt{6}-6}{3^2-6}=\dfrac{-\sqrt{6}}{3}\)

\(M=\dfrac{\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt[]{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=\dfrac{-2}{\sqrt{2}}=-\sqrt{2}\)\(\sqrt{ }\)

a: \(=\left(\sqrt{2}-\sqrt{5}\right)^2-7\)

\(=7-2\sqrt{10}-7=-2\sqrt{10}\)

b: \(=\left(\sqrt{2}+\sqrt{3}\right)^2-5=5+2\sqrt{6}-5=2\sqrt{6}\)

c: \(=5-\left(\sqrt{2}-\sqrt{3}\right)^2\)

\(=5-5+2\sqrt{6}=2\sqrt{6}\)

b: B=căn x(căn x-4)

a: A=(căn x+căn y)(căn x-căn y-3)

d: D=5x^2-5xcăn y-2x*căn y+2y

=5x(x-căn y)-2căn y(x-căn y)

=(x-căn y)(5x-2căn y)

`b)\sqrt{(x+2)^2}-2=0`

`<=>|x+2|=2`

`<=>x^2+4x+4=4`

`<=>x^2+4x=0`

`<=>x(x+4)=0`

`<=>` $\left[\begin{matrix} x=0\\ x=-4\end{matrix}\right.$

Vậy `S={0;-4}`

=>|x+2|=2

=>x+2=2 hoặc x+2=-2

=>x=0 hoặc x=-4

`=> |x+2| - 2 = 0`

`=> x + 2 - 2=0` hoặc `-x - 2 - 2 =0`

`=> x = 0` hoặc `-x = 4`

`=> x = 0` hoặc `x = -4`

Vậy `S = {0, -4}`

pttđ:

\(x+y+13=4\sqrt{x}+6\sqrt{y}\)

\(x+y+13-4\sqrt{x}-6\sqrt{y}=0\)

\(x+y+4+9-4\sqrt{x}-6\sqrt{y}=0\)

\(\left(\sqrt{x}-2\right)^2+\left(\sqrt{y}-3\right)^2=0\)

=> \(\sqrt{x}-2=0\) ; \(\sqrt{y}-3=0\)

=> x = 2^2  = 4

y = 3^2 = 9