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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 \(x< 3\Rightarrow3-x>0\Rightarrow6-2x-\left|3-x\right|=6-2x-3+x=3-x\)
\(E=\dfrac{\left|x-3\right|}{\left(x-3\right)\left(x+3\right)}\left(x+3\right)^2=\dfrac{\left|x-3\right|\left(x+3\right)}{x-3}\left(x\ne\pm3\right)\)
Với \(x>3\Leftrightarrow E=x+3\)
Với \(x< 3\Leftrightarrow E=-x-3\)
\(F=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\left(x\ge0;x\ne25\right)\\ F=\dfrac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
\(A=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}\)
\(A=\sqrt{x^2-6x+3^2}-\sqrt{x^2+6x+3^2}\)
\(A=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}\)
b)\(\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}=1\)
\(TH1:x-3>=0\)
\(< =>x+3>=0\)
\(\left|x-3\right|-\left|x+3\right|=1\)
\(x-3-x-3=1\)
\(-6=1\)(loại)
\(TH2:x-3< =0\)
\(x+3>=0\)
\(< =>\left|x-3\right|-\left|x+3\right|=1\)
\(3-x-x-3\)
\(-2x=1\)
\(x=-\frac{1}{2}\left(TM\right)\)
\(TH3:x-3< =0\)
\(x+3< =0\)
\(< =>\left|x-3\right|-\left|x+3\right|=1\)
\(3-x+X+3=1\)
\(6=1\)(loại)
\(< =>x=\left\{\frac{1}{2}\right\}\)để \(A=1\)
Bn ấn vào trả lời
Rồi ấn vào chữ M nằm ngang là xong @@
Nhớ đúng cho mk nha ^^
\(a,A=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}.\)
\(A=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}.\)
\(A=\left(x-3\right)-\left(x+3\right)\)
\(b,\) 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.
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, \(B=\frac{\sqrt{4x}}{x-3}\sqrt{\frac{x^2-6x+9}{x}}=\frac{2\sqrt{x}}{x-3}.\sqrt{\frac{\left(x-3\right)^2}{x}}\)
\(=\frac{2\sqrt{x}}{x-3}.\frac{x-3}{\sqrt{x}}=2\)
1.
$x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{(x-3)^2}=x+3+|x-3|$
$=x+3+(3-x)=6$
2.
$\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{(x+2)^2}-\sqrt{x^2}$
$=|x+2|-|x|=x+2-(-x)=2x+2$
3.
$\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}$
$=\sqrt{(\sqrt{x^2-1}+1)^2}-\sqrt{(\sqrt{x^2-1}-1)^2}$
$=|\sqrt{x^2-1}+1|+|\sqrt{x^2-1}-1|$
$=\sqrt{x^2-1}+1+|\sqrt{x^2-1}-1|$
4.
$\frac{\sqrt{x^2-2x+1}}{x-1}=\frac{\sqrt{(x-1)^2}}{x-1}$
$=\frac{|x-1|}{x-1}=\frac{x-1}{x-1}=1$
5.
$|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=2-x+\frac{\sqrt{(x-2)^2}}{x-2}$
$=2-x+\frac{|x-2|}{x-2}|=2-x+\frac{2-x}{x-2}=2-x+(-1)=1-x$
6.
$2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\frac{\sqrt{(x-5)^2}}{x-5}$
$=2x-1-\frac{|x-5|}{x-5}$
\(B=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}\)
\(B=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}\)
\(B=\left|x-3\right|-\left|x+3\right|\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -3\\B=-x+3+x+3=6\end{matrix}\right.\\\left\{{}\begin{matrix}-3\le x< 3\\B=-x+3-x-3=-2x\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge3\\B=x-3-x-3=-6\end{matrix}\right.\end{matrix}\right.\)
b)
\(B=1\Leftrightarrow-3\le x< 3\Rightarrow B=-2x=1\Rightarrow x=-\dfrac{1}{2}̸\)
Ta có: \(T=\sqrt{x^2-6x+9}+\sqrt{x^2+6x+9}\)
\(=\left|x-3\right|+\left|x+3\right|\)
\(=3-x+x+3\)
\(=6\)