Giai pt:
a. \(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)
b. \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)
giải pt :
a, \(\sqrt{3x^2-7x+3}+\sqrt{x^2-3x+4}=\sqrt{3x^2-5x-1}+\sqrt{x^2-2}\)
b, \(\sqrt{x}+\sqrt{3-x}=x^2-x-2\)
c, \(\sqrt{x+6}+\sqrt{x-1}=x^2-1\)
giải pt :
a,\(\left(\sqrt{5x-1}+\sqrt{x-1}\right)\left(3x-1-\sqrt{5x^2-6x+1}\right)=4x\)
b,\(2\left(\sqrt{x}-\sqrt{x-1}\right)\left(1+\sqrt{x^2-1}\right)=x\sqrt{x}\)
a, ĐK: \(x\ge1\)
Đặt \(\sqrt{5x-1}=a;\sqrt{x-1}=b\left(a,b\ge0\right)\)
\(pt\Leftrightarrow\left(a+b\right)\left(\dfrac{a^2+b^2}{2}-ab\right)=a^2-b^2\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2=2\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)\left(a-b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=b+2\end{matrix}\right.\)
TH1: \(a=b\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}\Leftrightarrow x=0\left(l\right)\)
TH2: \(a=b+2\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}+2\)
\(\Leftrightarrow5x-1=x-1+4+4\sqrt{x-1}\)
\(\Leftrightarrow4x-4-4\sqrt{x-1}=0\)
\(\Leftrightarrow4x-4-4\sqrt{x-1}+1=1\)
\(\Leftrightarrow\left(2\sqrt{x-1}-1\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x-1}-1=1\\2\sqrt{x-1}-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-1}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
giải pt :
a, \(\dfrac{\sqrt{x-3}}{\sqrt{2x-1}-1}=\dfrac{1}{\sqrt{x+3}-\sqrt{x-3}}\)
b, \(\left(\sqrt{x^2+x+1}+\sqrt{4x^2+x+1}\right)\left(\sqrt{5x^2+1}-\sqrt{2x^2+1}\right)=3x^2\)
1.Giai pt bang cach dat an phu :
a, 3x + 14 + 5\(\sqrt{x-2}\) = 7(\(\sqrt{x+1}+\sqrt{x^2-x-2}\) )
b, 7\(\sqrt{3x-7}\) +(4x-7)\(\sqrt{7-x}\) =32
giải pt :
a, \(x^2+5x+2=4\sqrt{x^3+3x^2+x-1}\)
b, \(\sqrt{x+1}+x+3=\sqrt{1-x}+3\sqrt{1-x^2}\)
c,\(\left(2x-3\right)\sqrt{3+x}+2x\sqrt{3-x}=6x-8+\sqrt{9-x^2}\)
a, ĐK: \(\left(x+1\right)\left(x^2+2x-1\right)\ge0\)
\(x^2+5x+2=4\sqrt{x^3+3x^2+x-1}\)
\(\Leftrightarrow x^2+2x-1+3\left(x+1\right)-4\sqrt{\left(x+1\right)\left(x^2+2x-1\right)}=0\)
TH1: \(x\ge-1\)
\(pt\Leftrightarrow\left(\sqrt{x^2+2x-1}-\sqrt{x+1}\right)\left(\sqrt{x^2+2x-1}-3\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=\sqrt{x+1}\\\sqrt{x^2+2x-1}=3\sqrt{x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-1=x+1\\x^2+2x-1=9x+9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2-7x-10=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
TH2: \(x< -1\)
\(pt\Leftrightarrow\left(\sqrt{-x^2-2x+1}-\sqrt{-x-1}\right)\left(\sqrt{-x^2-2x+1}-3\sqrt{-x-1}\right)=0\)
\(\Leftrightarrow...\)
Bài này dài nên ... cho nhanh nha, đoạn sau dễ rồi
giải pt
a) \(\sqrt{x^2+x+1}+\sqrt{3x^2+3x+2}=\sqrt{5x^2+5x-1}\)
b) \(\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\)
c) \(\sqrt{3x^2-5x+7}+\sqrt{3x^2-7x+2}=3\)
d) \(\sqrt{x^2+3x+2}=\sqrt{2x^2+9x+7}-\sqrt{x^2+6x+5}\)
a/ đk: \(\left[{}\begin{matrix}x\le\frac{-5-3\sqrt{5}}{10}\\x\ge\frac{-5+3\sqrt{5}}{10}\end{matrix}\right.\)\(\sqrt{x^2+x+1}+\sqrt{3x^2+3x+2}=\sqrt{5x^2+5x-1}\)
\(\Leftrightarrow\sqrt{x^2+x+1}+\sqrt{3\left(x^2+x+1\right)-1}=\sqrt{5\left(x^2+x+1\right)-6}\)
đặt\(x^2+x+1=t\left(t>0\right)\)
\(\sqrt{t}+\sqrt{3t-1}=\sqrt{5t-6}\)
bình phương 2 vế pt trở thành:
\(t+3t-1+2\sqrt{t\left(3t-1\right)}=5t-6\)
\(\Leftrightarrow2\sqrt{3t^2-t}=t-5\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left(2\sqrt{3t^2-t}\right)^2=\left(t-5\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\11t^2+6t-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left[{}\begin{matrix}t=\frac{-3+2\sqrt{71}}{11}\\t=\frac{-3-2\sqrt{71}}{11}\end{matrix}\right.\end{matrix}\right.\)=> không có gtri t nào t/m
vậy pt vô nghiệm
a/ ĐKXĐ: ...
Đặt \(x^2+x+1=a>0\)
\(\sqrt{a}+\sqrt{3a-1}=\sqrt{5a-6}\)
\(\Leftrightarrow4a-1+2\sqrt{3a^2-a}=5a-6\)
\(\Leftrightarrow2\sqrt{3a^2-a}=a-5\) (\(a\ge5\))
\(\Leftrightarrow4\left(3a^2-a\right)=a^2-10a+25\)
\(\Leftrightarrow11a^2+6a-25=0\)
Nghiệm xấu quá, chắc bạn nhầm lẫn đâu đó
b/
Đặt \(x^2+x+1=a>0\)
\(\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)
\(\Leftrightarrow2a+3+2\sqrt{a^2+3a}=2a+7\)
\(\Leftrightarrow\sqrt{a^2+3a}=2\)
\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{3x^2-5x+7}=3-\sqrt{3x^2-7x+2}\)
\(\Rightarrow3x^2-5x+7=3x^2-7x+11-6\sqrt{3x^2-7x+2}\)
\(\Leftrightarrow3\sqrt{3x^2-7x+2}=2-x\) (\(x\le2\))
\(\Leftrightarrow9\left(3x^2-7x+2\right)=x^2-4x+4\)
\(\Leftrightarrow26x^2-59x+14=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{26}\end{matrix}\right.\)
Do biến đổi ko tương đương nên cần thay lại nghiệm vào pt ban đầu kiểm tra
d/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{x^2+3x+2}+\sqrt{x^2+6x+5}=\sqrt{2x^2+9x+7}\)
\(\Leftrightarrow2x^2+9x+7+2\sqrt{\left(x^2+3x+2\right)\left(x^2+6x+5\right)}=2x^2+9x+7\)
\(\Leftrightarrow\sqrt{\left(x+1\right)^2\left(x+2\right)\left(x+5\right)}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\left(l\right)\\x=-5\end{matrix}\right.\)
giai pt sau
\(\sqrt{3x-1}-\sqrt{x+2}.\sqrt{3x^2+7x+2}+4=4x-2\)
\(x^2-5x+3.\sqrt{2x-1}=2.\sqrt{14-2x}+5\)
\(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
nhiều thế giải ko đổi đâu bạn
đkxđ : \(\frac{1}{2}\le x\le7\)
\(x^2-5x+3\sqrt{2x-1}=2\sqrt{14-2x}+5\)
\(\Leftrightarrow\left(x^2-5x\right)+3\left(\sqrt{2x-1}-3\right)=2\left(\sqrt{14-2x}-2\right)\)
\(\Leftrightarrow x\left(x-5\right)+\frac{3.\left(2x-10\right)}{\sqrt{2x-1}+3}+\frac{2.\left(2x-10\right)}{\sqrt{14-2x}+2}=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+\frac{6}{\sqrt{2x-1}+3}+\frac{4}{\sqrt{14-2x}+2}\right)=0\)
\(\Leftrightarrow x=5\)
còn bài a,c lười đánh lắm
giải pt :
a,\(2x^2-11x+21=3\sqrt[3]{4x-4}\)
b,\(\dfrac{\sqrt{x-3}}{\sqrt{2x-1}-1}=\dfrac{1}{\sqrt{x+3}-\sqrt{x-3}}\)
c,\(\left(\sqrt{x^2+x+1}+\sqrt{4x^2+x+1}\right)\left(\sqrt{5x^2+1}-\sqrt{2x^2+1}\right)=3x^2\)
Giải:
1)a) \(17\sqrt{3x-1}=3x\)
b) \(\sqrt{2+\sqrt{3x-5}}=\sqrt{x+1}\)
c)\(\sqrt{\dfrac{5x+7}{x+3}}=4\)
2)Giai pt :
a) x+y+12=\(4\sqrt{x}+6\sqrt{y}-1\)
b) \(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\)
c)\(\sqrt{x+\sqrt{14x-4y}}+\sqrt{x-\sqrt{14x-4y}}=\sqrt{14}\)
d)x-\(4\sqrt{2x+2}-2\sqrt{2-x+9}=0\)
Bài 1:
b: \(\Leftrightarrow2+\sqrt{3x-5}=x+1\)
\(\Leftrightarrow\sqrt{3x-5}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+1=3x-5\\x>=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+6=0\\x>=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
c: \(\Leftrightarrow5x+7=16\left(x+3\right)\)
=>16x+48=5x+7
=>11x=-41
hay x=-41/11