2/3√9x-9 -1/4√16x-16 +3√x-1=4
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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 \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
GC
1
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
25 tháng 11 2023
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)
LT
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 7 2023
\(\sqrt{x-1}\) - \(\sqrt{9x-9}\) + \(\sqrt{16x-16}\) = 4 (đk \(x\ge\)1)
\(\sqrt{x-1}-\) \(\sqrt{9\left(x-1\right)}\) + \(\sqrt{16\left(x-1\right)}\) = 4
\(\sqrt{x-1}\) - 3\(\sqrt{x-1}\) + \(4\sqrt{x-1}\) = 4
\(\sqrt{x-1}\)( 1 - 3 + 4 ) = 4
\(\sqrt{x-1}\) . 2 = 4
\(\sqrt{x-1}\) = 4 : 2
\(\sqrt{x-1}\) = 2
\(x-1\) =4
\(x=4+1\)
\(x=5\) (thỏa mãn)
Vậy \(x\) = 5
TT
1
16 tháng 7 2021
Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}-2\sqrt{16x+16}=\sqrt{x+1}-8\)
\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}-8\sqrt{x+1}-\sqrt{x+1}=-8\)
\(\Leftrightarrow\sqrt{x+1}=2\)
\(\Leftrightarrow x+1=4\)
hay x=3
30 tháng 9 2021
b: Ta có: \(\sqrt{9x^2-9}+\sqrt{4x^2-4}=\sqrt{16x^2-16}+2\)
\(\Leftrightarrow\sqrt{x^2-1}=2\)
\(\Leftrightarrow x^2-1=4\)
hay \(x\in\left\{\sqrt{5};-\sqrt{5}\right\}\)
HP
30 tháng 9 2021
a. \(x+\sqrt{x^2-4x+4}=\dfrac{1}{2}\)
<=> \(x+\sqrt{\left(x-2\right)^2}=\dfrac{1}{2}\)
<=> \(x+\left|x-2\right|=\dfrac{1}{2}\)
<=> \(\left[{}\begin{matrix}x+x-2=\dfrac{1}{2}\\x+\left[-\left(x-2\right)\right]=\dfrac{1}{2}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}2x=\dfrac{5}{2}\\x-x+2=\dfrac{1}{2}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\0=\dfrac{-3}{2}\left(VLí\right)\end{matrix}\right.\)
Vậy nghiệm của PT là \(S=\left\{\dfrac{5}{4}\right\}\)
b. \(\sqrt{9x^2-9}+\sqrt{4x^2-4}=\sqrt{16x^2-16}+2\)
<=> \(\sqrt{9\left(x^2-1\right)}+\sqrt{4\left(x^2-1\right)}=\sqrt{16\left(x^2-1\right)}+2\)
<=> \(3\sqrt{x^2-1}+2\sqrt{x^2-1}-4\sqrt{x^2-1}=2\)
<=> \(\left(3+2-4\right)\sqrt{x^2-1}=2\)
<=> \(\sqrt{x^2-1}=2\)
<=> x2 - 1 = 4
<=> x2 = 5
<=> x = \(\sqrt{5}\)
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