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\(a,A=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}.\)
\(A=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}.\)
\(A=\left(x-3\right)-\left(x+3\right)\)
\(b,\) 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.
\(A=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}\)
\(A=\sqrt{x^2-6x+3^2}-\sqrt{x^2+6x+3^2}\)
\(A=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}\)
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 \(A=1\)
=\(\left|x-3\right|-\left|x+3\right|\)
*x>0
=x-3-x+3
=0
*x<0
=3-x-3+x
=0
\(A=\sqrt{x^2}-\sqrt{x^2-4x+4}\)
\(\Leftrightarrow A=|x|-\sqrt{\left(x-2\right)^2}\)
\(\Leftrightarrow A=x-|x-2|=x-x+2=2\)
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)
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}\)
\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)
\(=\frac{\left(x-3\right)\left(x-2\right)+3\sqrt{\left(x-4\right)\left(x-2\right)}}{3\left(x-4\right)+\left(x-3\right)\sqrt{\left(x-4\right)\left(x-2\right)}}\)
\(=\frac{\sqrt{x-2}\left(\left(x-3\right)\sqrt{x-2}+3\sqrt{x-4}\right)}{\sqrt{x-4}\left(3\sqrt{x-4}+\left(x-3\right)\sqrt{x-2}\right)}\)
\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)
a)\(5x+\sqrt{x^2+6x+9}=5x+\sqrt{\left(x+3\right)^2}\)
\(=5x+x+3\)
\(=6x+3\)
b)Với B=-9 suy ra
\(6x+3=-9\)
\(\Rightarrow6x=-12\)
\(\Rightarrow x=-2\)