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24 tháng 9 2021
1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
25 tháng 9 2021
1) ĐKXĐ: \(16x^2-25\ge0\)
\(\Leftrightarrow x^2\ge\dfrac{25}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{4}\\x\le-\dfrac{5}{4}\end{matrix}\right.\)
2) ĐKXĐ: \(4x^2-49\ge0\Leftrightarrow x^2\ge\dfrac{49}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{7}{2}\\x\le-\dfrac{7}{2}\end{matrix}\right.\)
3) ĐKXĐ: \(8-x^2\ge0\Leftrightarrow x^2\le8\)
\(\Leftrightarrow-2\sqrt{2}\le x\le2\sqrt{2}\)
4) ĐKXĐ: \(x^2-12\ge0\Leftrightarrow x^2\ge12\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\sqrt{3}\\x\le-2\sqrt{3}\end{matrix}\right.\)
5) ĐKXĐ: \(x^2+4\ge0\left(đúng\forall x\right)\)
26 tháng 8 2021
`a)sqrt{1-4x+4x^2}+5=x-2`
`<=>\sqrt{(2x-1)^2}=x-2-5`
`<=>|2x-1|=x-7(x>=7)`
`<=>[(2x-1=x-7),(2x-1=7-x):}`
`<=>[(x=-6(ktm)),(3x=8):}`
`<=>x=8/3(ktm)`
Vậy PTVN
`b)3sqrt{12+4x}+4/7sqrt{147+49x}=3/2sqrt{48+16x}+4(x>=-3)`
`<=>6sqrt{x+3}+4sqrt{x+3}=6sqrt{x+3}+4`
`<=>4sqrt{x+3}=4`
`<=>sqrt{x+3}=1<=>x+3=1`
`<=>x=-2(tm)`
Vậy `S={-2}`
26 tháng 8 2021
a) \(\sqrt{1-4x+4x^2}+5=x-2\Leftrightarrow\sqrt{\left(1-2x\right)^2}+5=x-2\Leftrightarrow\left|1-2x\right|=x-7\left(1\right)\)TH1: \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow1-2x=x-7\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)(không thỏa đk)
TH2: \(1-2x< 0\Leftrightarrow x>\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow2x-1=x-7\Leftrightarrow x=-6\)(không thỏa đk)
Vậy \(S=\varnothing\)
b) \(3\sqrt{12+4x}+\dfrac{4}{7}\sqrt{147+49x}=\dfrac{3}{2}\sqrt{48+16x}+4\Leftrightarrow6\sqrt{3+x}+4\sqrt{3+x}=6\sqrt{3+x}+4\Leftrightarrow4\sqrt{3+x}=4\Leftrightarrow\sqrt{3+x}=1\Leftrightarrow3+x=1\Leftrightarrow x=-2\)
22 tháng 7 2023
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
22 tháng 7 2023
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
25 tháng 10 2020
a) \(\frac{3}{4}\sqrt{x}-\sqrt{9x}+5=\frac{1}{4}\sqrt{9x}\)
ĐK : x ≥ 0
⇔ \(\frac{3}{4}\sqrt{x}-\sqrt{3^2x}-\frac{1}{4}\sqrt{3^2x}=-5\)
⇔ \(\frac{3}{4}\sqrt{x}-3\sqrt{x}-\frac{1}{4}\cdot3\sqrt{x}=-5\)
⇔ \(-\frac{9}{4}\sqrt{x}-\frac{3}{4}\sqrt{x}=-5\)
⇔ \(-3\sqrt{x}=-5\)
⇔ \(\sqrt{x}=15\)
⇔ \(x=225\)( tm )
b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)
ĐK : x ≤ 3
⇔ \(\sqrt{3-x}-\sqrt{3^2\left(3-x\right)}+\frac{5}{4}\sqrt{4^2\left(3-x\right)}=6\)
⇔ \(\sqrt{3-x}-3\sqrt{3-x}+\frac{5}{4}\cdot4\sqrt{3-x}=6\)
⇔ \(-2\sqrt{3-x}+5\sqrt{3-x}=6\)
⇔ \(3\sqrt{3-x}=6\)
⇔ \(\sqrt{3-x}=2\)
⇔ \(3-x=4\)
⇔ \(x=-1\)( tm )
c) \(\sqrt{9x^2+12x+4}=4\)
⇔ \(\sqrt{\left(3x+2\right)^2}=4\)
⇔ \(\left|3x+2\right|=4\)
⇔ \(\orbr{\begin{cases}3x+2=4\\3x+2=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-2\end{cases}}\)
d) \(\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}\)
ĐK : x ≥ 1
⇔ \(\frac{1}{3}\sqrt{x-1}+2\sqrt{2^2\left(x-1\right)}-12\sqrt{\left(\frac{1}{5}\right)^2\cdot\left(x-1\right)}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+2\cdot2\sqrt{x-1}-12\cdot\frac{1}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+4\sqrt{x-1}-\frac{12}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{29}{15}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\sqrt{x-1}=1\)
⇔ \(x-1=1\)
⇔ \(x=2\)( tm )
tại sao
cô mình bảo max A=1 tại x=4