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a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
`a)sqrt{x^2-6x+9}=2`
`<=>sqrt{(x-3)^2}=2`
`<=>|x-3|=2`
`**x-3=2`
`<=>x=5`
`**x-3=-2`
`<=>x=1`
Vậy `S={1,5}`
`b)sqrt{4x-20}+sqrt{x-5}-1/3sqrt{9x-45}=4`
đk:`x>=5`
`pt<=>2sqrt{x-5}+sqrt{x-5}-1/3*3*sqrt{x-5}=4`
`<=>2sqrt{x-5}=4`
`<=>sqrt{x-5}=2`
`<=>x-5=4<=>x=9`
Vậy `S={9}`
Lời giải:
a.
PT $\Leftrightarrow \sqrt{(x-3)^2}=2$
$\Leftrightarrow |x-3|=2$
$\Leftrightarrow x-3=\pm 2$
$\Leftrightarrow x=1$ hoặc $x=5$
b. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4(x-5)}+\sqrt{x-5}-\frac{1}{3}\sqrt{9(x-5)}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x=2^2+5=9$ (thỏa mãn)
a) ĐKXĐ: \(x\ge0\)
Ta có: \(3\sqrt{18x}-5\sqrt{8x}+4\sqrt{50x}=38\)
\(\Leftrightarrow9\sqrt{2x}-10\sqrt{2x}+20\sqrt{2x}=38\)
\(\Leftrightarrow19\sqrt{2x}=38\)
\(\Leftrightarrow\sqrt{2x}=2\)
\(\Leftrightarrow2x=4\)
hay x=2(thỏa ĐK)
b) ĐKXĐ: \(x\ge0\)
Ta có: \(3\sqrt{12x}-2\sqrt{27x}+4\sqrt{3x}=8\)
\(\Leftrightarrow6\sqrt{3x}-6\sqrt{3x}+4\sqrt{3x}=8\)
\(\Leftrightarrow\sqrt{3x}=2\)
\(\Leftrightarrow3x=4\)
hay \(x=\dfrac{4}{3}\)
c) ĐKXĐ: \(x\ge5\)
Ta có: \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
hay x=9
a)
\(3.3\sqrt{2x}-5.2\sqrt{2x}+4.5.\sqrt{2x}=38\\ \Leftrightarrow19\sqrt{2x}=38\\ \Leftrightarrow\sqrt{2x}=2\\ \Leftrightarrow x=2\)
b)
\(3.2.\sqrt{3x}-2.3.\sqrt{3x}+4.\sqrt{3x}=8\\ \Leftrightarrow4\sqrt{3x}=8\\ \Leftrightarrow\sqrt{3x}=2\\\Leftrightarrow x=\dfrac{2^2}{3}=\dfrac{4}{3} \)
c)
\(\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\)
`sqrt{x-5}+2sqrt{4x-20}-1/2sqrt{9x-45}=12`
Điều kiện:`x>=5`
`pt<=>sqrt{x-5}+2sqrt{4(x-5)}-1/2sqrt{9(x-5)}=12`
`<=>sqrt{x-5}+4sqrt{x-5}-3/2sqrt{x-5}=12`
`<=>7/2sqrt{x-5}=12`
`<=>sqrt{x-5}=24/7`
`<=>x-5=576/49`
`<=>x=821/49(Tmđk)`
Vậy `S={821/49}.`
Ta có: \(\sqrt{x-5}+2\sqrt{4x-20}-\dfrac{1}{3}\sqrt{9x-45}=12\)
\(\Leftrightarrow4\sqrt{x-5}=12\)
\(\Leftrightarrow x-5=9\)
hay x=14