Giai pt:\(\sqrt{\left(5-2\sqrt{6}\right)^x}+\sqrt{\left(5+2\sqrt{6}\right)^x}=10\)
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DM
1
TN
15 tháng 2 2018
\(\left(5-2\sqrt{6}\right)^{\frac{x}{2}}+\left(5+2\sqrt{6}\right)^{\frac{x}{2}}=10\)
\(pt\Leftrightarrow\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^{2x}}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^{2x}}=10\)
\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)^x+\left(\sqrt{3}+\sqrt{2}\right)^x=10\)
\(\Leftrightarrow\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^x}+\left(\sqrt{3}+\sqrt{2}\right)^x=10\)
\(\Leftrightarrow\frac{1}{t}+t=10\left(t=\left(\sqrt{3}+\sqrt{2}\right)^x\right)\)
\(\Leftrightarrow t^2-10t+1=0\)\(\Leftrightarrow t=5\pm2\sqrt{6}\)
\(\Rightarrow5\pm2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^x\)
\(\Leftrightarrow\left(\sqrt{3}+\sqrt{2}\right)^{\pm2}=\left(\sqrt{3}+\sqrt{2}\right)^x\)
\(\Rightarrow x=\pm2\). Vậy...
DM
0
NT
2
8 tháng 12 2019
\(\left(\sqrt{2x+3}+2\right)\left(\sqrt{x+6}-\sqrt{x+1}\right)=5\)
\(ĐKXĐ:x\ge-1\).Nhận thấy \(\sqrt{x+6}-\sqrt{x+1}>0\)
\(\Leftrightarrow\left(\sqrt{2x+3}+2\right)\frac{\left(\sqrt{x+6}+\sqrt{x+1}\right)\left(\sqrt{x+6}-\sqrt{x+1}\right)}{\sqrt{x+6}-\sqrt{x+1}}=5\)
\(\Leftrightarrow\left(\sqrt{2x+3}+2\right)\frac{5}{\sqrt{x+6}-\sqrt{x+1}}=5\)
\(\Leftrightarrow\frac{\sqrt{2x+3}+2}{\sqrt{x+6}-\sqrt{x+1}}=1\)
\(\Leftrightarrow\sqrt{2x+3}+2-\sqrt{x+6}+\sqrt{x+1}=0\)
Th1:\(\sqrt{x+1}=2\Leftrightarrow x=3\left(thoaman\right)\)
Th2:\(\sqrt{x+1}-2\ne0\Leftrightarrow x\ne3\)
\(\Leftrightarrow\left(\sqrt{2x+3}-\sqrt{x+6}\right)+\left(2+\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\frac{x-3}{\sqrt{2x+3}+\sqrt{x+6}}+\frac{x-3}{\sqrt{x+1}-2}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{2x+3}+\sqrt{x+6}}+\frac{1}{\sqrt{x+1}-2}\right)=0\)
Tự lm tiếp nha
Y
24 tháng 6 2023
\(\left(2-\sqrt{5}\right)x^2+\left(6-\sqrt{5}\right)x-8+2\sqrt{5}=0\)
\(\Leftrightarrow\left(2-\sqrt{5}\right)x^2-\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)x-(8-2\sqrt{5})=0\)
\(\Leftrightarrow\left(2-\sqrt{5}\right)x\left(x-1\right)+\left(8-2\sqrt{5}\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(2-\sqrt{5}\right)x=-8+2\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-8+2\sqrt{5}}{2-\sqrt{5}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=6+4\sqrt{5}\end{matrix}\right.\)
Vậy \(S=\left\{1;6+4\sqrt{5}\right\}\)
30 tháng 8 2023
6:ĐKXĐ: x>=0; x<>1/25
BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)
=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)
=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)
7:
ĐKXĐ: x>=0
BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)
=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)
=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)
=>\(-\sqrt{x}-2>=0\)(vô lý)
8:
ĐKXĐ: x>=0; x<>9/4
BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)
=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)
=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)
TH1: 9căn x-14>0 và 2căn x-3<0
=>căn x>14/9 và căn x<3/2
=>14/9<căn x<3/2
=>196/81<x<9/4
TH2: 9căn x-14<0 và 2căn x-3>0
=>căn x>3/2 hoặc căn x<14/9
mà 3/2<14/9
nên trường hợp này Loại
9:
ĐKXĐ: x>=0
\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)
=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)
=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)
=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)
10:
ĐKXĐ: x>=0; x<>1/49
\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)
=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)
=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)
=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)
TH1: 6căn x-1>0 và 7căn x-1>0
=>căn x>1/6 và căn x>1/7
=>căn x>1/6
=>x>1/36
TH2: 6căn x-1<0 và 7căn x-1<0
=>căn x<1/6 và căn x<1/7
=>căn x<1/7
=>0<=x<1/49
NK
1
4 tháng 8 2021
Ta có: \(\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^x}=10\)
\(\Leftrightarrow\sqrt{\left(5+2\sqrt{6}\right)^x}=10-5+2\sqrt{6}=5+2\sqrt{6}\)
\(\Leftrightarrow\left(5+2\sqrt{6}\right)^x=\left(5+2\sqrt{6}\right)^2\)
hay x=2
AT
7 tháng 7 2021
\(3\sqrt{9a^6}-6a^3=3\left|3a^3\right|-6a^3\)
Xét \(a\ge0\Rightarrow\) biểu thức \(=9a^3-6a^3=3a^3\)
Xét \(a< 0\Rightarrow\) biểu thức \(=-9a^3-6a^3=-15a^3\)
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(1-3x\right)^2}=\left|x-1\right|+\left|1-3x\right|\)
\(=1-x+3x-1\left(\dfrac{1}{3}< x\le1\right)=2x\)
\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{2-\sqrt{3}}.\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=2\)
\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{6+2\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2=4^2=16\)
\(\sqrt{23-8\sqrt{7}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(2\sqrt{7}-4\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=2\sqrt{7}-4+\sqrt{7}-1=3\sqrt{7}-5\)
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|\)
Xét \(x\ge8\Rightarrow\sqrt{x-4}\ge2\Rightarrow\)biểu thức \(=\sqrt{x-4}+2+\sqrt{x-4}-2\)
\(=2\sqrt{x-4}\)
Xét \(x< 8\Rightarrow\sqrt{x-4}< 2\Rightarrow\) biểu thức \(=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
Nhận xét : \(\sqrt{\left(5-2\sqrt{6}\right)^x}.\sqrt{\left(5+2\sqrt{6}\right)^x}=1\)
Ta đặt \(\sqrt{\left(5-2\sqrt{6}\right)^x}=a\Rightarrow\sqrt{\left(5+2\sqrt{6}\right)^x}=\frac{1}{a}\)
Khi đó phương trình ban đầu trở thành :
\(a+\frac{1}{a}=10\Rightarrow a^2-10a+1=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=5+2\sqrt{6}\\a=5-2\sqrt{6}\end{cases}}\)
+) Với \(a=5+2\sqrt{6}\Rightarrow\sqrt{\left(5-2\sqrt{6}\right)^x}=5+2\sqrt{6}\)
\(\Leftrightarrow\left(5-2\sqrt{6}\right)^x=\left(5+2\sqrt{6}\right)^2=\left(\frac{1}{5-2\sqrt{6}}\right)^2\)
\(\Leftrightarrow x=-2\)
+) Với \(a=5-2\sqrt{6}\Rightarrow\sqrt{\left(5-2\sqrt{6}\right)^x}=5-2\sqrt{6}\)
\(\Leftrightarrow\left(5-2\sqrt{6}\right)^x=\left(5-2\sqrt{6}\right)^2\)
\(\Leftrightarrow x=2\)
Vậy \(x\in\left\{-2,2\right\}\) thỏa mãn đề.