Giải PT
\(\sqrt{2x+1}-\sqrt{18x+9}=\sqrt{32+16x}-18\)
K
Khách
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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)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 6 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\-4+\sqrt{7}\le x\le-1\end{matrix}\right.\)
Khi x thỏa ĐKXĐ, vế phải luôn dương, bình phương 2 vế ta được:
\(\Leftrightarrow3x^2+16x+17+2\sqrt{\left(x^2-1\right)\left(2x^2+16x+18\right)}=4x^2+16x+16\)
\(\Leftrightarrow2\sqrt{\left(x^2-1\right)\left(2x^2+16x+18\right)}=x^2-1\)
\(\Leftrightarrow4\left(x^2-1\right)\left(2x^2+16x+18\right)=\left(x^2-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\4\left(2x^2+16x+18\right)=x^2-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\7x^2+64x+73=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\x=\dfrac{-32+3\sqrt{57}}{7}\\x=\dfrac{-32-3\sqrt{57}}{7}\left(loại\right)\end{matrix}\right.\)
23 tháng 6 2018
\(\sqrt{2x^2+16x+18}+\sqrt{x^2-1}=2x+4\)
\(\Leftrightarrow\sqrt{2x^2+16x+18}-\left(2x+4\right)+\sqrt{x^2-1}=0\)
\(\Leftrightarrow\dfrac{2x^2+16x+18-\left(4x^2+16x+16\right)}{\sqrt{2x^2+16x+18}+\left(2x+4\right)}+\sqrt{x^2-1}=0\)
\(\Leftrightarrow\dfrac{2x^2+16x+18-4x^2-16x-16}{\sqrt{2x^2+16x+18}+\left(2x+4\right)}+\sqrt{x^2-1}=0\)
\(\Leftrightarrow\dfrac{-2x^2+2}{\sqrt{2x^2+16x+18}+\left(2x+4\right)}+\sqrt{x^2-1}=0\)
\(\Leftrightarrow\dfrac{-2\left(x^2-1\right)}{\sqrt{2x^2+16x+18}+\left(2x+4\right)}+\sqrt{x^2-1}=0\)
\(\Leftrightarrow\sqrt{x^2-1}\left(1-\dfrac{2\sqrt{x^2-1}}{\sqrt{2x^2+16x+18}+\left(2x+4\right)}\right)=0\)
Tới đây đơn giản rồi
AY
2
AT
25 tháng 7 2018
đkxđ: x≥\(-\dfrac{1}{2}\)
\(\sqrt{18x+9}-\sqrt{8x+4}+\dfrac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\dfrac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\left(3-2+\dfrac{1}{3}\right)\sqrt{2x+1}=4\)
\(\Leftrightarrow\dfrac{4}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\sqrt{2x+1}=3\Leftrightarrow2x+1=9\Leftrightarrow x=4\)
vậy x = 4
25 tháng 7 2018
Bình phương 2 vế ,ta có:
\(26x+13+\dfrac{1}{9}\left(2x+1\right)-2\sqrt{9.4\left(2x+1\right)^2}-2.\dfrac{1}{3}\sqrt{4\left(2x+1\right)^2}+2.\dfrac{1}{3}\sqrt{9\left(2x+1\right)^2}=16\) \(\dfrac{236}{9}x+\dfrac{118}{9}-2.6.\left(2x+1\right)-\dfrac{2}{3}.2.\left(2x+1\right)+\dfrac{2}{3}.3.\left(2x+1\right)=16\)
\(\dfrac{236}{9}x+\dfrac{118}{9}-24x-12-\dfrac{8}{3}x-\dfrac{4}{3}+4x+2=16\)
\(\dfrac{32}{9}x+\dfrac{16}{9}=16\)
\(\dfrac{16}{9}\left(2x+1\right)=16\)
\(2x+1=9\Rightarrow2x=8\Rightarrow x=4\)
Vậy x=4
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: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
NM
1
KN
29 tháng 1 2020
\(ĐKXĐ:2x^2+16x+18\ge0;x^2-1\ge0\)
\(pt\Leftrightarrow\sqrt{x^2-1}=2x+4-\sqrt{2x^2+16x+18}\)(1)
\(\Leftrightarrow\sqrt{x^2-1}\left(\frac{2\sqrt{x^2-1}}{2x+4+\sqrt{2x^2+16x+18}}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-1}=0\\2\sqrt{x^2-1}=2x+4+\sqrt{2x^2+16x+18}\left(2\right)\end{cases}}\)
Lấy(1) + (2), ta được: \(3\sqrt{x^2-1}=4x+8\Leftrightarrow x=\frac{3\sqrt{57}-32}{7}\)
PT
15
12 tháng 1 2016
\(pt\Rightarrow\sqrt{x^2-1}=2x+4-\sqrt{2x^2+16x+18}\)
\(\Rightarrow\sqrt{\frac{1}{2}.\left(2x+4\right)^2-\frac{1}{2}.\left(2x^2+16x+18\right)}=2x+4-\sqrt{2x^2+16x+18}\)
Chia 2 vế cho \(\sqrt{2x^2+16x+18}\)
\(\Rightarrow\sqrt{\frac{\left(2x+4\right)^2}{2.\left(2x^2+16x+18\right)}-\frac{1}{2}}=\frac{2x+4}{\sqrt{2x^2+16x+18}}-1\)
Đặt \(\frac{2x+4}{\sqrt{2x^2+16x+18}}=a\)
\(\Rightarrow\sqrt{\frac{1}{2}a^2-\frac{1}{2}}=a-1\left(a\ge1\right)\)
Kết quả x = 1 nha , chính xác r nek
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\sqrt{1-8x+16x^2}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{1^2-2\cdot4x\cdot1+\left(4x\right)^2}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{\left(4x-1\right)^2}=\dfrac{1}{3}\)
\(\Leftrightarrow\left|4x-1\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=\dfrac{1}{3}\left(ĐK:x\ge\dfrac{1}{4}\right)\\4x-1=\dfrac{1}{3}\left(ĐK:x< \dfrac{1}{4}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=\dfrac{1}{6}\left(tm\right)\end{matrix}\right.\)
b) \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\) (ĐK: \(x\ge2\))
\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)
\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)
\(\Leftrightarrow6\sqrt{x-2}=18\)
\(\Leftrightarrow\sqrt{x-2}=3\)
\(\Leftrightarrow x-2=9\)
\(\Leftrightarrow x=9+2\)
\(\Leftrightarrow x=11\left(tm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 6 2021
ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(16x^2-48x+35+\left(\sqrt{6x-9}-\sqrt{2x-2}\right)=0\)
\(\Leftrightarrow\left(4x-7\right)\left(4x-5\right)+\dfrac{4x-7}{\sqrt{6x-9}+\sqrt{2x-2}}=0\)
\(\Leftrightarrow\left(4x-7\right)\left(4x-5+\dfrac{1}{\sqrt{6x-9}+\sqrt{2x-2}}\right)=0\)
\(\Leftrightarrow4x-7=0\)
\(\sqrt{2x+1}-\sqrt{18x+9}=\sqrt{32x+16}-18\left(đk:x\ge-\dfrac{1}{2}\right)\)
\(\Leftrightarrow\sqrt{2x+1}-3\sqrt{2x+1}-4\sqrt{2x+1}=-18\)
\(\Leftrightarrow6\sqrt{2x+1}=18\)
\(\Leftrightarrow\sqrt{2x+1}=3\)
\(\Leftrightarrow2x+1=9\)
\(\Leftrightarrow x=4\left(tm\right)\)
\(\sqrt{2x+1}-9\sqrt{2x+1}-16\sqrt{2x+1}=-18\)
\(-24\sqrt{2x+1}=-18\)
\(\sqrt{2x+1}=\dfrac{3}{4}\)
\(\sqrt{\left(2x+1\right)^2}=\dfrac{9}{16}\)
\(2x+1=\dfrac{9}{16}\)
\(x=\dfrac{-7}{32}\)