Giải pt sau
\(\frac{X^2}{\sqrt{5}}-\sqrt{20=0}\)
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BC
1
28 tháng 9 2021
\(\sqrt{x+6-4\sqrt{x+2}}-\sqrt{9-4\sqrt{5}}=0\left(đk:x\ge-2\right)\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+2}-2\right)^2}=\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(\Leftrightarrow\left|\sqrt{x+2}-2\right|=\left|\sqrt{5}-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}-2=\sqrt{5}-2\\\sqrt{x+2}-2=2-\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=5\\x+2=21-8\sqrt{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=19-8\sqrt{5}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{3;19-8\sqrt{5}\right\}\)
14 tháng 7 2019
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
14 tháng 7 2019
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
3 tháng 5 2019
1)
a) \(\left\{{}\begin{matrix}2x-y=5\\x+y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2x-y+x+y=5+4\\x+y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x=9\\x+y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vậy (x;y)=(3;1)
b) \(16x^5-8x^3+x=0\Leftrightarrow x\left(16x^4-8x^2+1\right)=0\Leftrightarrow x\left[\left(4x^2\right)^2-2.4x^2.1+1^2\right]=0\Leftrightarrow x\left(4x^2-1\right)^2=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\4x^2-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=\frac{\pm1}{2}\end{matrix}\right.\)
Vậy S={\(-\frac{1}{2};0;\frac{1}{2}\)}
2)
A=\(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{4}+\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{5}+1}{5-1}=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{5}+1}{4}=\frac{\sqrt{5}-1+\sqrt{5}+1}{4}=\frac{2\sqrt{5}}{4}=\frac{\sqrt{5}}{2}\)
B=\(\frac{4}{3+\sqrt{5}}-\frac{8}{1+\sqrt{5}}+\frac{15}{\sqrt{5}}=\frac{4\left(3-\sqrt{5}\right)}{9-5}-\frac{8\left(1-\sqrt{5}\right)}{1-5}+3\sqrt{5}=\frac{4\left(3-\sqrt{5}\right)}{4}-\frac{8\left(\sqrt{5}-1\right)}{4}+3\sqrt{5}=3-\sqrt{5}-2\sqrt{5}+2+3\sqrt{5}=5\)
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 \(S=\left\{1;6+4\sqrt{5}\right\}\)
QT
1
25 tháng 5 2015
\(\sqrt{\frac{1}{4}x^2+x+1}=\sqrt{\left(\frac{1}{2}x\right)^2+2.\frac{1}{2}x.1+1^2}=\sqrt{\left(\frac{1}{2}x+1\right)^2}=\left|\frac{1}{2}x+1\right|\)
\(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
phương trình <=> \(\left|\frac{1}{2}x+1\right|=\sqrt{5}-1\)
<=> \(\frac{1}{2}x+1=\sqrt{5}-1\) hoặc \(\frac{1}{2}x+1=-\sqrt{5}+1\)
+) \(\frac{1}{2}x+1=\sqrt{5}-1\)<=> \(x=2\sqrt{5}+4\)
+) \(\frac{1}{2}x+1=-\sqrt{5}+1\) <=> \(x=-2\sqrt{5}\)
Vậy pt có 2 nghiệm \(x=2\sqrt{5}+4\); \(x=-2\sqrt{5}\)
NT
1
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 8 2020
ĐKXĐ: \(x>0\)
\(\Leftrightarrow x+\frac{1}{x}-4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+5=0\)
Đặt \(\sqrt{x}+\frac{1}{\sqrt{x}}=t>0\Rightarrow t^2=x+\frac{1}{x}+2\Rightarrow x+\frac{1}{x}=t^2-2\)
Pt trở thành:
\(t^2-2-4t+5=0\Leftrightarrow t^2-4t+3=0\) \(\Rightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{\sqrt{x}}=1\\\sqrt{x}+\frac{1}{\sqrt{x}}=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x}+1=0\left(vn\right)\\x-3\sqrt{x}+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}=\frac{3\pm\sqrt{5}}{2}\Rightarrow x=\frac{7\pm3\sqrt{5}}{2}\)
TG
28 tháng 11 2021
b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)
\(\Rightarrow a^2+3-4a=0\)
=> (a - 3).(a - 1) = 0
=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)
Bình phương lên giải tiếp nhé!
c) Tương tư câu b nhé
De bai sai ha ban ?
\(\Leftrightarrow\left|x-1,5\right|=-\left|2,5-x\right|.\)(1)
VT >=0; VP <=0. Để đẳng thức 1 xảy ra thì VT = VP = 0.
Nhưng vì VP = 0 =>x= 2,5 thì VT = 1 nên PT vô nghiệm.