\(3x^2+6x-3=\sqrt{\frac{x+7}{3}}\)
Những câu hỏi liên quan
tìm điều kiện xác định của biểu thức:
\(a)\frac{6x}{-\sqrt{x+7}}-\frac{3}{-5x-4}+\frac{\sqrt{x}}{-3x+2}\)
\(b)\frac{5-\sqrt{x}}{x+4}+\frac{\sqrt{x-2}-3}{-2x-10}\)
\(c)\frac{\sqrt{6x}}{-x-3}-\frac{4x}{2x+3}\)
\(d)\frac{\sqrt{2x-7}}{3x-4}-\frac{\sqrt{6x}}{x-3}+3x-1\)
a) \(\left\{{}\begin{matrix}x\ge0\\-\sqrt{x+7}< 0\\-5x-4\ne0\\-3x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x+7>0\\-5x\ne4\\-3x\ne-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>-7\\x\ne\frac{-4}{5}\\x\ne\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne\frac{2}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x\ge0\\x+4\ne0\\x-2\ge0\\-2x-10\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-4\\x\ge2\\-2x\ne10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne-5\end{matrix}\right.\Leftrightarrow x\ge2\)
c) \(\left\{{}\begin{matrix}x\ge0\\-x-3\ne0\\2x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-3\\x\ne-\frac{3}{2}\end{matrix}\right.\Leftrightarrow x\ge0\)
d) \(\left\{{}\begin{matrix}2x-7\ge0\\x\ge0\\3x-4\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{7}{2}\\x\ge0\\x\ne\frac{4}{3}\\x\ne3\end{matrix}\right.\Leftrightarrow x\ge\frac{7}{2}\)
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1) Rút gọn biểu thức: a, sqrt{2+sqrt{3}}-sqrt{2-sqrt{3}} b, sqrt{4-sqrt{7}}+sqrt{4+sqrt{7}}2) Giải phương trình: a, left(xsqrt{frac{6}{x}}+sqrt{frac{2x}{3}}+sqrt{6x}right).sqrt{6x}2 b, left(sqrt{frac{3}{x}}+sqrt{frac{x}{3}}+sqrt{3x}right).sqrt{3x}3 c, sqrt{x^2+2x+1}-sqrt{x^2-1}0 d, sqrt{x}+sqrt{x+1}frac{1}{sqrt{x}}
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1) Rút gọn biểu thức:
a, \(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
b, \(\sqrt{4-\sqrt{7}}+\sqrt{4+\sqrt{7}}\)
2) Giải phương trình:
a, \(\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right).\sqrt{6x}=2\)
b, \(\left(\sqrt{\frac{3}{x}}+\sqrt{\frac{x}{3}}+\sqrt{3x}\right).\sqrt{3x}=3\)
c, \(\sqrt{x^2+2x+1}-\sqrt{x^2-1}=0\)
d, \(\sqrt{x}+\sqrt{x+1}=\frac{1}{\sqrt{x}}\)
Giải phương trình:
a)\(6\sqrt{x+8}=16+3x-x^2\)
b)\(\sqrt{3x+1}-\sqrt{x+3}+7x^2-7=0\)
c) \(6\left(x+\frac{1}{x}\right)+2=7\sqrt{x+3}\)
d) \(3x+10=\frac{1}{x}+4\sqrt{6x+3}\)
Giải phương trình
\(3x^2+6x-3=\sqrt{\frac{x+7}{3}}\)
\(3x^2+6x-3=\sqrt{\frac{x+7}{3}}\)
\(\Leftrightarrow\left(3x^2+6x-3\right)^2=\left(\sqrt{\frac{x+7}{3}}\right)^2\)
\(\Leftrightarrow9x^4+36x^3+18x^2-36x+9=\frac{x+7}{3}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{69}+7}{6}\\x=\frac{\sqrt{73}-5}{6}\end{cases}}\)
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1. giải các phương trình :
a/sqrt{6x^2-12x+7}x^2-2x
frac{2}{sqrt{3+x}}frac{sqrt{3+x}}{x-1}
c/x^2+sqrt{-x-1}4+sqrt{-x-1}
d/frac{3x^2+1}{sqrt{x-1}}frac{4}{sqrt{x-1}}
e/sqrt{-x^2+3x+4}2x^2-6x+2
f/frac{sqrt{4x^2+7x-2}}{x+2}sqrt{2}
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1. giải các phương trình :
a/\(\sqrt{6x^2-12x+7}=x^2-2x\)
\(\frac{2}{\sqrt{3+x}}=\frac{\sqrt{3+x}}{x-1}\)
c/\(x^2+\sqrt{-x-1}=4+\sqrt{-x-1}\)
d/\(\frac{3x^2+1}{\sqrt{x-1}}=\frac{4}{\sqrt{x-1}}\)
e/\(\sqrt{-x^2+3x+4}=2x^2-6x+2\)
f/\(\frac{\sqrt{4x^2+7x-2}}{x+2}=\sqrt{2}\)
a, ĐK: \(6x^2-12x+7\ge0\) (*)
\(PT\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\6x^2-12x+7=x^4-4x^3+4x^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\x^4-4x^3-2x^2+12x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x\ge0\\\left(x-1\right)^2\left(x^2-2x-7\right)=0\end{matrix}\right.\) \(\Rightarrow x=1\pm2\sqrt{2}\) (thỏa mãn ĐK)
Vậy...
22 tháng 10 2019 lúc 17:12
Giải pt
a) 2x^2+sqrt{x^2-5x-6}10x+15
b) 5sqrt{3x^2-4x-2}-6x^2+8x+70
c) x^2+sqrt{2x^2+4x+3}6-2x
d) 2sqrt{frac{3x-1}{x}}frac{x}{3x-1}+1
e) sqrt{frac{24x-4}{x}}frac{x}{6x-1}+1
f) sqrt{frac{2x-1}{x}}+1+sqrt{frac{x}{2x-1}}frac{3x}{2x-1}
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Giải pt
a) \(2x^2+\sqrt{x^2-5x-6}=10x+15\)
b) \(5\sqrt{3x^2-4x-2}-6x^2+8x+7=0\)
c) \(x^2+\sqrt{2x^2+4x+3}=6-2x\)
d) \(2\sqrt{\frac{3x-1}{x}}=\frac{x}{3x-1}+1\)
e) \(\sqrt{\frac{24x-4}{x}}=\frac{x}{6x-1}+1\)
f) \(\sqrt{\frac{2x-1}{x}}+1+\sqrt{\frac{x}{2x-1}}=\frac{3x}{2x-1}\)
a/ ĐKXĐ: ...
\(\Leftrightarrow2\left(x^2-5x-6\right)+\sqrt{x^2-5x-6}-3=0\)
Đặt \(\sqrt{x^2-5x-6}=a\ge0\)
\(2a^2+a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-5x-6}=1\Leftrightarrow x^2-5x-7=0\)
b/ ĐKXĐ: ...
\(\Leftrightarrow5\sqrt{3x^2-4x-2}-2\left(3x^2-4x-2\right)+3=0\)
Đặt \(\sqrt{3x^2-4x-2}=a\ge0\)
\(-2a^2+5a+3=0\) \(\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{3x^2-4x-2}=3\Leftrightarrow3x^2-4x-11=0\)
c/ \(\Leftrightarrow x^2+2x-6+\sqrt{2x^2+4x+3}=0\)
Đặt \(\sqrt{2x^2+4x+3}=a>0\Rightarrow x^2+2x=\frac{a^2-3}{2}\)
\(\frac{a^2-3}{2}-6+a=0\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x-6=0\)
d/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{3x-1}{x}}=a>0\)
\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\)
\(\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{\frac{3x-1}{x}}=1\Leftrightarrow3x-1=x\)
e/ĐKXĐ: ...
\(\Leftrightarrow2\sqrt{\frac{6x-1}{x}}=\frac{x}{6x-1}+1\)
Đặt \(\sqrt{\frac{6x-1}{x}}=a>0\)
\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{\frac{6x-1}{x}}=1\Rightarrow6x-1=x\)
f/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{x}{2x-1}}=a>0\)
\(\frac{1}{a}+1+a=3a^2\)
\(\Leftrightarrow3a^3-a^2-a-1=0\)
\(\Leftrightarrow\left(a-1\right)\left(3a^2+2a+1\right)=0\)
\(\Leftrightarrow a=1\Rightarrow\sqrt{\frac{x}{2x-1}}=1\Rightarrow x=2x-1\)
Giải phương trình : 3x2 + 6x - 3 = \(\sqrt{\frac{x+7}{3}}\)
Đặt : \(\sqrt{\frac{x+7}{3}}\)= t + 1
=> x+7/3 = t^2+2t+1
<=> x+7 = 3t^2+6t+3
<=> 3t^2+6t+3-x-7 = 0
<=> 3t^2+6t-x = 4
pt <=> 3x^2+6x-3 = t+1
<=>3x^2+6x-t = 1+3
<=> 3x^2+6x-t = 4
Từ đó ta có hệ pt đối xứng loại 2 :
3t^2+6t-x = 4
3x^2+6x-1 = 4
Đến đó bạn tự giải nha
Tk mk nha
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1)7sqrt{3x-7}+left(4x-7right)sqrt{7-x}322)4x^2-11x+6left(x-1right)sqrt{2x^2-6x+6}3)9+3sqrt{xleft(3-2xright)}7sqrt{x}+5sqrt{3-2x}4)sqrt{2x^2+4x+7}x^4+4x^3+3x^2-2x-75)frac{6-2x}{sqrt{5-x}}+frac{6+2x}{sqrt{5+x}}frac{8}{3}6)2left(5x-3right)sqrt{x+1}+left(x+1right)sqrt{3-x}3left(5x+1right)7)sqrt{7x+7}+sqrt{7x-6}+2sqrt{49x^2+7x-42}181-14x
Đọc tiếp
1)\(7\sqrt{3x-7}+\left(4x-7\right)\sqrt{7-x}=32\)
2)\(4x^2-11x+6=\left(x-1\right)\sqrt{2x^2-6x+6}\)
3)\(9+3\sqrt{x\left(3-2x\right)}=7\sqrt{x}+5\sqrt{3-2x}\)
4)\(\sqrt{2x^2+4x+7}=x^4+4x^3+3x^2-2x-7\)
5)\(\frac{6-2x}{\sqrt{5-x}}+\frac{6+2x}{\sqrt{5+x}}=\frac{8}{3}\)
6)\(2\left(5x-3\right)\sqrt{x+1}+\left(x+1\right)\sqrt{3-x}=3\left(5x+1\right)\)
7)\(\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{49x^2+7x-42}=181-14x\)
\(Tính:\)
\(a.\frac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}=2\)
\(b.\frac{6x-3}{\sqrt{x}-\sqrt{-x}}=3+2\sqrt{x-x^2}\)
\(c.\sqrt{4-3\sqrt{10-3x}}=x-2\)
\(d.x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)
\(e.\sqrt{2059-x}+\sqrt{2035-x}-\sqrt{2154-x}=24\)