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

\(\Leftrightarrow A=|x|-\sqrt{\left(x-2\right)^2}\)

\(\Leftrightarrow A=x-|x-2|=x-x+2=2\)

A = \(\sqrt{x^2}-\sqrt{x^2-4x+4}=\sqrt{x^2}-\sqrt{\left(x-2\right)^2}=\left|x\right|-\left|x-2\right|=x-x+2=2\)(vì  \(x\ge2\))

B = \(\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}=\left|x-3\right|-\left|x+3\right|=3-x+x+3=6\)(vì x < 3)

\(B=\sqrt{x^2-6x+3^2}-\sqrt{x^2+6x+3^2}\)

\(\Leftrightarrow B=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}\)

\(\Leftrightarrow B=|x-3|-|x+3|=-x+3-x-3=0\)

Ta có: \(T=\sqrt{x^2-6x+9}+\sqrt{x^2+6x+9}\)

\(=\left|x-3\right|+\left|x+3\right|\)

\(=3-x+x+3\)

\(=6\)

a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x+3}}\)(\(x\ge0,x\ne9\))

b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\sqrt{x}-2\left(x\ge0,x\ne9\right)\)

a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x}+3}\)

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

c) \(6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-\left|3-x\right|\)

mà \(x< 3\Rightarrow3-x>0\Rightarrow6-2x-\left|3-x\right|=6-2x-3+x=3-x\)

a,\(\dfrac{3-\sqrt{x}}{x-9}\)

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

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

a, \(A=\left(\sqrt{12}-2\sqrt{5}\right)\sqrt{3}+\sqrt{60}\)

\(=\left(2\sqrt{3}-2\sqrt{5}\right)\sqrt{3}+2\sqrt{15}\)

\(=2\sqrt{9}-2\sqrt{15}+2\sqrt{15}=2\sqrt{9}\)

b, \(B=\frac{\sqrt{4x}}{x-3}\sqrt{\frac{x^2-6x+9}{x}}=\frac{2\sqrt{x}}{x-3}.\sqrt{\frac{\left(x-3\right)^2}{x}}\)

\(=\frac{2\sqrt{x}}{x-3}.\frac{x-3}{\sqrt{x}}=2\)

em thiếu, giờ mới nhìn lại \(2\sqrt{9}=2.3=6\)

Ta có : \(A=\sqrt{\left(x+3\right)^2}-\frac{\sqrt{x^2-6x+9}}{x+3}\)

\(\Rightarrow A=\left(x+3\right)-\frac{\left(x-3\right)^2}{\left(x+3\right)}\)

\(\Rightarrow A=\frac{\left(x+3\right)^2}{\left(x+3\right)}-\frac{\left(x-3\right)^2}{\left(x+3\right)}\)

\(\Rightarrow A=\frac{\left(x+3\right)^2-\left(x-3\right)^2}{\left(x+3\right)}\)

\(\Rightarrow A=\frac{\left(x+3+x-3\right)\left(x+3-x+3\right)}{\left(x+3\right)}\)

\(\Rightarrow A=\frac{18x}{\left(x+3\right)}\)

Mình quên ghi điều kiện mất là x<-3 nhé

Mấy bạn làm lại giúp mình

Mình cảm ơn  :D

bạn chỉ mình cách ghi dấu căn ik mình làm cho 

Bn ấn vào trả lời

Rồi ấn vào chữ M nằm ngang là xong @@

Nhớ đúng cho mk nha ^^

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

A\(=\left|x-3\right|+\left|x+3\right|\)

\(E=\dfrac{\left|x-3\right|}{\left(x-3\right)\left(x+3\right)}\left(x+3\right)^2=\dfrac{\left|x-3\right|\left(x+3\right)}{x-3}\left(x\ne\pm3\right)\)

Với \(x>3\Leftrightarrow E=x+3\)

Với \(x< 3\Leftrightarrow E=-x-3\)

\(F=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\left(x\ge0;x\ne25\right)\\ F=\dfrac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)

a: \(f\left(x\right)=\sqrt{x^2-6x+9}=\sqrt{\left(x-3\right)^2}=\left|x-3\right|\)

\(f\left(-1\right)=\left|-1-3\right|=4\)

\(f\left(5\right)=\left|5-3\right|=\left|2\right|=2\)

b: f(x)=10

=>\(\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=-7\end{matrix}\right.\)

c: \(A=\dfrac{f\left(x\right)}{x^2-9}=\dfrac{\left|x-3\right|}{\left(x-3\right)\left(x+3\right)}\)

TH1: x<3 và x<>-3

=>\(A=\dfrac{-\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-1}{x+3}\)

TH2: x>3

\(A=\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x+3}\)

\(a,A=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}.\)

\(A=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}.\)

\(A=\left(x-3\right)-\left(x+3\right)\)

\(b,\) Ta có : \(A=1=\left(x-3\right)-\left(x+3\right)\)

                                   \(\Leftrightarrow1=x-3-x-3\Leftrightarrow1=-6\left(ko\right)tm\)

Vậy ko có giá trị của x.

mk ko biết đâu

mk mới hok lớp 5 thui

chúc bạn hok tốt nhé

kb với mk nha

Txđ: x thuộc R

A= \(\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}\)= |x-3| -|x+3|

Th1:  với x>=3 thì A= x-3-x-3= -6

TH2: với x thuộc [-3,3) thì A = -x +3 -x-3= -2x

Th3: với x < -3 thì A = -x+3+x+3 = 6

b. A=1 thuộc TH2 câu a

-2x=1 => x= -1/2 thỏa mãn x thuộc (-3,3) vậy A=1 khi x=-1/2