giải phương trình:
\(\sqrt{2x^2+16x+8}+\sqrt{x^2-1}=2x+4\)
K
Khách
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22 tháng 8 2021
a, \(16x^2-\left(1+\sqrt{3}\right)^2=0\\ \Rightarrow\left(4x-1-\sqrt{3}\right)\left(4x+1+\sqrt{3}\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x-1-\sqrt{3}=0\\4x+1+\sqrt{3}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{3}}{4}\\x=\dfrac{-1-\sqrt{3}}{4}\end{matrix}\right.\)
b, \(x-2\sqrt{2x}+2=8\\ \Rightarrow x-\sqrt{8x}-6=0\\ \Rightarrow x-6=\sqrt{8x}\\ \Rightarrow\left(x-6\right)^2=\sqrt{8x}^2\\ \Rightarrow x^2-12x+36=8x\\ \Rightarrow x^2-20x+36=0\\ \Rightarrow\left(x^2-2x\right)-\left(18x-36\right)=0\)
\(\Rightarrow x\left(x-2\right)-18\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x-18\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-18=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=18\end{matrix}\right.\)
22 tháng 8 2021
1: Ta có: \(16x^2-\left(\sqrt{3}+1\right)^2=0\)
\(\Leftrightarrow\left(4x-\sqrt{3}-1\right)\left(4x+\sqrt{3}+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}+1}{4}\\x=\dfrac{-\sqrt{3}-1}{4}\end{matrix}\right.\)
2: Ta có: \(x-2\sqrt{2x}+2=8\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=2\sqrt{2}\\\sqrt{x}-2=-2\sqrt{2}\end{matrix}\right.\Leftrightarrow\sqrt{x}=2\sqrt{2}+2\)
\(\Leftrightarrow x=12+8\sqrt{2}\)
UN
0
KN
1
HF
18 tháng 8 2020
Đặt \(a=\sqrt{2x^2+16x+18};b=\sqrt{x^2-1}\left(a,b\ge0\right);\)
Ta có: \(a+b=\sqrt{a^2+2b^2}\Rightarrow a^2+2ab+b^2=a^2+2b^2\)
\(\Leftrightarrow b\left(2a-b\right)=0\)
TH1: \(\sqrt{x^2-1}=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}\left(TM\right)}\)
TH2: \(2\sqrt{2x^2+16x+18}=\sqrt{x^2-1}\Leftrightarrow7x^2+64x+72=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-32+3\sqrt{57}}{7}\left(TM\right)\\x=\frac{-32-3\sqrt{57}}{7}\left(KTM\right)\end{cases}}\)
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)
TH
1
4 tháng 12 2015
ê quân,Bạn làm bài này ở phương pháp gì đây,đặt ẩn phụ hay sử dụng BĐT
14 tháng 8 2021
c: Ta có: \(\sqrt{2x}=\sqrt{5}\)
\(\Leftrightarrow2x=5\)
hay \(x=\dfrac{5}{2}\)
d: Ta có: \(\sqrt{3x-1}=4\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
hay \(x=\dfrac{17}{3}\)
14 tháng 8 2021
Ta có: \(\sqrt{4\cdot\left(1-x\right)^2}=6\)
\(\Leftrightarrow2\left|x-1\right|=6\)
\(\Leftrightarrow\left|x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
ND
Nguyễn Đức Trí
VIP
1 tháng 9 2023
1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
1 tháng 9 2023
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
