Giải PT : \(\sqrt{\frac{3x+1}{x+3}}=\sqrt{7}\)
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6 tháng 11 2019 lúc 11:12
giải pt
a) \(\sqrt{x+3}=3-\sqrt{6-x}\)
b) \(\sqrt{3x-2}-\sqrt{x-7}=1\)
c) \(\frac{1-\sqrt{3x+1}}{\sqrt{x-1}-7}=1\)
d) \(\frac{x}{\sqrt{7x-4}-3}=\frac{x}{\sqrt{x+1}}\)
e) \(\sqrt{3x-2}-\sqrt{x-7}=1\)
f) \(2\sqrt{\frac{3x+1}{2x-1}}-\sqrt{\frac{x-1}{2x-1}}=2\)
a)\(ĐK:-3\le x\le6\)
\(PT\Leftrightarrow\sqrt{x+3}+\sqrt{6-x}=3\)
\(\Leftrightarrow x+3+6-x+2\sqrt{\left(x+3\right)\left(6-x\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(6-x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\left(tm\right)\)
b/ ĐKXĐ: \(x\ge7\)
\(\sqrt{3x-2}=1+\sqrt{x-7}\)
\(\Leftrightarrow3x-2=x-6+2\sqrt{x-7}\)
\(\Leftrightarrow x+2=\sqrt{x-7}\)
\(\Leftrightarrow x^2+4x+4=x-7\)
\(\Leftrightarrow x^2+3x+11=0\) (vô nghiệm)
c/ ĐKXĐ: \(x\ge1;x\ne50\)
\(1-\sqrt{3x+1}=\sqrt{x-1}-7\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{3x+1}=8\)
\(\Leftrightarrow4x+2\sqrt{3x^2-2x-1}=64\)
\(\Leftrightarrow\sqrt{3x^2-2x-1}=32-2x\) (\(x\le16\))
\(\Leftrightarrow3x^2-2x-1=\left(32-2x\right)^2\)
d/ ĐKXĐ: \(x\ge\frac{4}{7};x\ne\frac{13}{7}\)
\(\Leftrightarrow\sqrt{x+1}=\sqrt{7x-4}-3\)
\(\Leftrightarrow\sqrt{x+1}+3=\sqrt{7x-4}\)
\(\Leftrightarrow x+10+6\sqrt{x+1}=7x-4\)
\(\Leftrightarrow3\sqrt{x+1}=3x-7\) (\(x\ge\frac{7}{3}\))
\(\Leftrightarrow9\left(x+1\right)=\left(3x-7\right)^2\)
\(\Leftrightarrow...\)
e/ Giống câu b
f/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le-\frac{1}{3}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{\frac{3x+1}{2x-1}}=a\ge0\\\sqrt{\frac{x-1}{2x-1}}=b\ge0\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}2a-b=2\\a^2+5b^2=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=2a-2\\a^2+5b^2=4\end{matrix}\right.\)
\(\Rightarrow a^2+5\left(2a-2\right)^2=4\)
\(\Leftrightarrow a^2+20\left(a^2-2a+1\right)-4=0\)
\(\Leftrightarrow21a^2-40a+16=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{4}{3}\\a=\frac{4}{7}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{\frac{3x+1}{2x-1}}=\frac{4}{3}\\\sqrt{\frac{3x+1}{2x-1}}=\frac{4}{7}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{3x+1}{2x-1}=\frac{16}{9}\\\frac{3x+1}{2x-1}=\frac{16}{49}\end{matrix}\right.\) \(\Leftrightarrow...\)
giải pt
\(\sqrt{5x-1}-\sqrt{3x+13}=\frac{x-7}{3}\)
ĐK: \(x\ge\frac{1}{5}\)
\(\sqrt{5x-1}-\sqrt{3x+13}=\frac{x-7}{3}\Leftrightarrow\frac{\left(\sqrt{5x-1}-\sqrt{3x+13}\right)\left(\sqrt{5x-1}+\sqrt{3x+13}\right)}{\sqrt{5x-1}+\sqrt{3x+13}}=\frac{x-7}{3}\Leftrightarrow\frac{5x-1-3x-13}{\sqrt{5x-1}+\sqrt{3x+13}}=\frac{x-7}{3}\Leftrightarrow\frac{2x-14}{\sqrt{5x-1}+\sqrt{3x+13}}=\frac{x-7}{3}\Leftrightarrow\frac{2\left(x-7\right)}{\sqrt{5x-1}+\sqrt{3x+13}}-\frac{x-7}{3}=0\Leftrightarrow\left(x-7\right)\left(\frac{2}{\sqrt{5x-1}+\sqrt{3x+13}}-\frac{1}{3}\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-7=0\\\frac{2}{\sqrt{5x-1}+\sqrt{3x+13}}-\frac{1}{3}=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=7\left(tm\right)\\\frac{2}{\sqrt{5x-1}+\sqrt{3x+13}}=\frac{1}{3}\left(1\right)\end{matrix}\right.\)
(1)\(\Leftrightarrow6=\sqrt{5x-1}+\sqrt{3x+13}\Leftrightarrow\sqrt{5x-1}=6-\sqrt{3x+13}\Leftrightarrow5x-1=36-12\sqrt{3x+13}+3x+13\Leftrightarrow50-2x=12\sqrt{3x+13}\left(x\le25\right)\Leftrightarrow25-x=6\sqrt{3x+13}\Leftrightarrow625-50x+x^2=108x+468\Leftrightarrow x^2-158x+157=0\Leftrightarrow\left(x-157\right)\left(x-1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-157=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=157\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy S={1;7}
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1/giải pt \(x^2+3x\sqrt[3]{3x+2}-12+\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+8}{x}\)
Giải pt :
a) \(x^2+3x\sqrt[3]{3x+3}-12+\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+8}{x}\)
b) \(\sqrt{\left(x-1\right)\left(3-x\right)}+\sqrt{x+2}=\sqrt{x-1}+\sqrt{3-x}+\frac{x}{2}\)
3 tháng 6 2018 lúc 18:38
B1: giải pt: \(\sqrt{x+3}+\sqrt{2x+4}=12-\sqrt{3x+7}\)
B2: giải pt: \(x^3-3x^2-8x+32=4\sqrt{x+1}\)
giải pt \(10+\sqrt{3}x^3+3x+\frac{\sqrt{3}}{x^3}=5\sqrt{3}x^3+2x+\frac{2\sqrt{3}-1}{x}+\frac{5}{x^2}\)
giải pt
\(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\)
b) \(\sqrt{1-x}+\sqrt{x^2-3x+2}+\left(x-2\right)\sqrt{\frac{x-1}{x-2}}=3\)
a) ĐKXĐ : \(7\le x\le9\)
đặt \(A=\sqrt{x-7}+\sqrt{9-x}\)
\(\Rightarrow A^2=2+2\sqrt{\left(x-7\right)\left(9-x\right)}\le2+\left(x-7\right)+\left(9-x\right)=4\)
\(\Rightarrow A\le2\)
Mà \(x^2-16x+66=\left(x-8\right)^2+2\ge2\)
\(\Rightarrow VT=VP=2\)
do đó : \(x-7=9-x\Leftrightarrow x=8\)( t/m )
b) ĐKXĐ : \(x\le1\)
Ta có : \(\sqrt{1-x}+\sqrt{\left(x-1\right)\left(x-2\right)}-\left|x-2\right|\sqrt{\frac{x-1}{x-2}}=3\)
\(\Leftrightarrow\sqrt{1-x}+\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{\left(x-1\right)\left(x-2\right)}=3\)
\(\Leftrightarrow\sqrt{1-x}=3\Leftrightarrow x=-8\left(tm\right)\)
giải pt \(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}x+\frac{2}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}x\)
Nhân liên hợp rồi rút gọn thì ta sẽ ra. Tôi nghĩ vậy
Giải pt: \(\frac{3+x}{3x}=\sqrt{\frac{1}{9}+\frac{1}{x}\sqrt{\frac{4}{9}+\frac{2}{x^2}}}\)
ĐK: x>0
Đặt a=1/x ta được: a>0
\(a+\frac{1}{3}=\sqrt{\frac{1}{9}+a\sqrt{\frac{4}{9}+2a^2}}\)
\(\Leftrightarrow a^2+\frac{1}{9}+\frac{2}{3}a=\frac{1}{9}+a\sqrt{\frac{4}{9}+2a^2}\)
<=>\(a^2+\frac{2}{3}a=a\sqrt{\frac{4}{9}+2a^2}\)
<=>\(a.\left(a+\frac{2}{3}\right)=a\sqrt{\frac{4}{9}+2a^2}\)
<=>\(a+\frac{2}{3}=\sqrt{\frac{4}{9}+2a^2}\)
<=>\(a^2+\frac{4}{9}+\frac{4}{3}a=\frac{4}{9}+2a^2\)
<=>\(a^2-\frac{4}{3}a=0\Leftrightarrow a=0\left(loại\right);a=\frac{4}{3}\)
<=>\(x=\frac{3}{4}\)(loại -3/2)
Vậy x=3/4
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