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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.
=\(\left|x-3\right|-\left|x+3\right|\)
*x>0
=x-3-x+3
=0
*x<0
=3-x-3+x
=0
\(A=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}\)
\(A=\sqrt{x^2-6x+3^2}-\sqrt{x^2+6x+3^2}\)
\(A=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}\)
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 \(A=1\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
b) Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
d) Để A>0 thì \(\sqrt{x}-2>0\)
hay x>4
a) Ta có:
\(A=2x+\sqrt{x^2-6x+9}\)
\(A=2x+\sqrt{\left(x-3\right)^2}\)
\(A=2x+\left|x-3\right|\)
Nếu \(x< 3\) thì: \(A=2x+3-x=x+3\)
Nếu \(x\ge3\) thì: \(A=2x+x-3=3x-3\)
b) Ta có: \(\left|x\right|=5\Leftrightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
Nếu x = 5: \(A=3\cdot5-3=12\)
Nếu x = -5: \(A=-5+3=-2\)
c) Ta có: \(A=2\Leftrightarrow\orbr{\begin{cases}x+3=2\\3x-3=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\left(ktm\right)\end{cases}}\)
Vậy x = -1
a) \(A=2x+\sqrt{x^2-6x+9}\)
\(=2x+\sqrt{\left(x-3\right)^2}\)
\(=2x+\left|x-3\right|\)
Với x ≥ 3 => A = 2x + x - 3 = 3x - 3
Với x < 3 => A = 2x + 3 - x = x + 3
b) | x | = 5 => x = ±5
Với x = 5 > 3 => A = 3.5 - 3 = 12
Với x = -5 < 3 => A = -5 + 3 = -2
c) A = 2
⇔ 2x + | x - 3 | = 2
⇔ | x - 3 | = 2 - 2x (*)
Với x ≥ 3
(*) ⇔ x - 3 = 2 - 2x
⇔ x + 3x = 2 + 3
⇔ 4x = 5
⇔ x = 5/4 ( ktm )
Với x < 3
(*) ⇔ 3 - x = 2 - 2x
⇔ -x + 2x = 2 - 3
⇔ x = -1 ( tm )
Vậy x = -1
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
b) Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
c) Để \(P< -\dfrac{1}{2}\) thì \(P+\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
\(\Leftrightarrow x< 9\)
Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)
