giải các pt sau
a) \(\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\)
b) \(x^2+\sqrt[3]{x^4-x^2}=2x+1\)
c) \(2x+\frac{x-1}{x}=\sqrt{1-\frac{1}{x}}+3\sqrt{x-\frac{1}{x}}\)
mn giúp mk vs ạ
giải pt
a) \(3\sqrt{x}+\frac{3}{2\sqrt{x}}=2x+\frac{1}{2x}-7\)
b) \(5\sqrt{x}+\frac{5}{2\sqrt{x}}=2x+\frac{1}{2x}+4\)
c) \(\sqrt{2x^2+8x+5}+\sqrt{2x^2-4x+5}=6\sqrt{x}\)
d) \(x+1+\sqrt{x^2-4x+1}=3\sqrt{x}\)
e) \(x^2+2x\sqrt{x-\frac{1}{x}}=3x+1\)
f) \(x^2-6x+x\sqrt{\frac{x^2-6}{x}}-6=0\)
g) \(\frac{3x^2}{3+\sqrt{x}}+6+2\sqrt{x}=5x\)
h) \(\frac{x^2}{4-3\sqrt{x}}+8=3\left(x+2\sqrt{x}\right)\)
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
e/ ĐKXĐ: ...
\(\Leftrightarrow x^2-1+2x\sqrt{\frac{x^2-1}{x}}=3x\)
Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:
\(\frac{x^2-1}{x}+2\sqrt{\frac{x^2-1}{x}}=3\)
Đặt \(\sqrt{\frac{x^2-1}{x}}=a\ge0\)
\(a^2+2a=3\Leftrightarrow a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\frac{x^2-1}{x}}=1\Leftrightarrow x^2-1=x\Leftrightarrow x^2-x-1=0\)
f/ ĐKXĐ: ...
\(\Leftrightarrow x^2-6+x\sqrt{\frac{x^2-6}{x}}-6x=0\)
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{x^2-6}{x}+\sqrt{\frac{x^2-6}{x}}-6=0\)
Đặt \(\sqrt{\frac{x^2-6}{x}}=a\ge0\)
\(a^2+a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\frac{x^2-6}{x}}=2\Leftrightarrow x^2-4x-6=0\)
giải pt
a) \(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(\sqrt{2x+3}-\sqrt{4-x}\right)^2-10\)
b) \(\sqrt{4x+1}+2\sqrt{1-x}+10\sqrt{-4x^2+3x+1}=13\)
c) \(\left(x^2+1\right)^2=13-x\sqrt{2x^2+4}\)
d) \(\left(\sqrt{x+1}+\sqrt{x-1}\right)^2-3=\frac{1}{\sqrt{x+1}-\sqrt{x-1}}\)
e) \(\left(\frac{2x-3}{\sqrt{x^2-1}}+2\right)\left(\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{x+1}}\right)=\frac{1}{x^2-1}\)
a/ ĐKXĐ: \(-\frac{3}{2}\le x\le4\)
\(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(x+7-2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-10\)
\(\Leftrightarrow\sqrt{2x+3}+\sqrt{4-x}=3\left(x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-52\)
Đặt \(\sqrt{2x+3}+\sqrt{4-x}=a>0\Rightarrow a^2=x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\)
Phương trình trở thành:
\(a=3a^2-52\Leftrightarrow3a^2-a-52=0\Rightarrow\left[{}\begin{matrix}a=-4\left(l\right)\\a=\frac{13}{3}\end{matrix}\right.\)
\(\sqrt{2x+3}+\sqrt{4-x}=\frac{13}{3}\)
Phương trình này vô nghiệm nên ko muốn giải tiếp, bạn bình phương lên và chuyển vế thôi :(
b/ ĐKXĐ: \(-\frac{1}{4}\le x\le1\)
Đặt \(\sqrt{4x+1}+2\sqrt{1-x}=a>0\Rightarrow a^2=5+4\sqrt{-4x^2+3x+1}\)
\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}\)
Pt trở thành:
\(a+10\left(\frac{a^2-5}{4}\right)=13\)
\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}=1\)
\(\Leftrightarrow-4x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{4}\end{matrix}\right.\)
c/ \(\Leftrightarrow x^2\left(x^2+2\right)=12-x\sqrt{2x^2+4}\)
\(\Leftrightarrow x^2\left(2x^2+4\right)=24-2x\sqrt{2x^2+4}\)
Đặt \(x\sqrt{2x^2+4}=a\) ta được:
\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)
d/ ĐKXĐ: \(x\ge1\)
Nhân cả tử và mẫu của vế phải với liên hợp của nó ta được:
\(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{x-1}\right)^2-3=\frac{\sqrt{x+1}+\sqrt{x+1}}{2}\)
Đặt \(\sqrt{x+1}+\sqrt{x-1}=a>0\)
\(\Rightarrow a^2-3=\frac{a}{2}\Rightarrow2a^2-a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}+\sqrt{x-1}=2\)
\(\Leftrightarrow x+\sqrt{x^2-1}=2\)
\(\Leftrightarrow\sqrt{x^2-1}=2-x\) (\(x\le2\))
\(\Leftrightarrow x^2-1=x^2-4x+4\)
\(\Rightarrow x=\frac{5}{4}\)
giải pt sau
a, \(x^2+3-\sqrt{2x^2-3x+2}=\frac{3}{2}\left(x+4\right)\)
b, 2(1-x) \(\sqrt{x^2+2x-1}=x^2-2x-1\)
c, \(x^3+1=2\sqrt[3]{2x-1}\)
d, \(x^2-3x+1=\frac{-\sqrt{3}}{3}\sqrt{x^4+x^2+1}\)
TXĐ: D=R
\(\Leftrightarrow2x^2+6-2\sqrt{2x^2-3x+2}=3\left(x+4\right)\)
\(\Leftrightarrow\frac{2x^2-3x-6}{2}-4=\sqrt{2x^2-3x+2}-4\)
\(\Leftrightarrow\frac{2x^2-3x-14}{2}=\frac{2x^2-3x-14}{\sqrt{2x^2-3x+2}+4}\)
\(\left[{}\begin{matrix}2x^2-3x-14=0\\\frac{1}{2}=\frac{1}{\sqrt{2x^2-3x+2}+4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-2\\x=\frac{7}{2}\end{matrix}\right.\\\text{ pt vô nghiệm}\end{matrix}\right.\)
Vậy ....
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 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
a) \(2\sqrt{\frac{x}{x-1}}-\sqrt{\frac{x-1}{x}}=\frac{5x-2}{x}\)
b) \(3\sqrt{\frac{2x}{x-1}}+4\sqrt{\frac{x-1}{2x}}=\frac{5x-3}{2x}+9\)
c) \(\sqrt{\frac{x}{3-2x}}+5\sqrt{\frac{3-2x}{x}}=\frac{12-9x}{x}+6\)
d) \(\frac{x-1}{x}-2\sqrt{\frac{x-1}{x}}=3\)
e) \(\sqrt{\frac{x}{x-1}}+\sqrt{\frac{x-1}{x}}=\frac{3}{\sqrt{2}}\)
f) \(\sqrt{x-\frac{1}{x}}=\frac{1}{\sqrt{x}}-\sqrt{x}\)
a/ ĐKXĐ: ...
\(\Leftrightarrow2\sqrt{\frac{x}{x-1}}-\sqrt{\frac{x-1}{x}}=\frac{2\left(x-1\right)}{x}+3\)
Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)
\(\frac{2}{a}-a=2a^2+3\Leftrightarrow2a^3+a^2+3a-2=0\)
\(\Leftrightarrow\left(2a-1\right)\left(a^2+a+2\right)=0\Leftrightarrow a=\frac{1}{2}\)
\(\Rightarrow\sqrt{\frac{x-1}{x}}=\frac{1}{2}\Leftrightarrow4\left(x-1\right)=x\)
b/ ĐKXĐ: ...
\(\Leftrightarrow3\sqrt{\frac{2x}{x-1}}+4\sqrt{\frac{x-1}{2x}}=\frac{3\left(x-1\right)}{2x}+10\)
Đặt \(\sqrt{\frac{x-1}{2x}}=a>0\)
\(\frac{3}{a}+4a=3a^2+10\Leftrightarrow3a^3-4a^2+10a-3=0\)
\(\Leftrightarrow\left(3a-1\right)\left(a^2-a+3\right)=0\Leftrightarrow a=\frac{1}{3}\)
\(\Leftrightarrow\sqrt{\frac{x-1}{2x}}=\frac{1}{3}\Leftrightarrow9\left(x-1\right)=2x\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{\frac{x}{3-2x}}+5\sqrt{\frac{3-2x}{x}}=\frac{4\left(3-2x\right)}{x}+5\)
Đặt \(\sqrt{\frac{3-2x}{x}}=a>0\)
\(\frac{1}{a}+5a=4a^2+5\Leftrightarrow4a^3-5a^2+5a-1=0\)
\(\Leftrightarrow\left(4a-1\right)\left(a^2-a+1\right)=0\Leftrightarrow a=\frac{1}{4}\)
\(\Leftrightarrow\sqrt{\frac{3-2x}{x}}=\frac{1}{4}\Leftrightarrow16\left(3-2x\right)=x\)
d/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)
\(a^2-2a=3\Leftrightarrow a^2-2a-3=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=3\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\frac{x-1}{x}}=3\Leftrightarrow x-1=9x\)
e/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{x}{x-1}}=a>0\)
\(a+\frac{1}{a}=\frac{3}{\sqrt{2}}\Leftrightarrow a^2-\frac{3}{\sqrt{2}}a+1=0\)
\(\Rightarrow\left[{}\begin{matrix}a=\sqrt{2}\\a=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{\frac{x}{x-1}}=\sqrt{2}\\\sqrt{\frac{x}{x-1}}=\frac{\sqrt{2}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(x-1\right)\\2x=x-1\end{matrix}\right.\)
f/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{\frac{x^2-1}{x}}=\frac{1-x}{\sqrt{x}}\)
Bình phương 2 vế:
\(\frac{x^2-1}{x}=\frac{\left(1-x\right)^2}{x}\Leftrightarrow x^2-1=x^2-2x+1\)
\(\Rightarrow x=1\)
a)Giải các phương trình sau bằng phương pháp đặt ẩn phụ:
1) \(x^2-3x-3=\frac{3\left(\sqrt[3]{x^3-4x^2+4}-1\right)}{1-x}\) ;2)\(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)
b) Giải các phương trình sau(không giới hạn phương pháp):
1)\(2\left(1-x\right)\sqrt{x^2+2x-1}=x^2-2x-1\) ; 2)\(\sqrt{2x+4}-2\sqrt{2-x}=\frac{12x-8}{\sqrt{9x^2+16}}\)
3)\(\frac{3x^2+3x-1}{3x+1}=\sqrt{x^2+2x-1}\) ; 4) \(\frac{2x^3+3x^2+11x-8}{3x^2+4x+1}=\sqrt{\frac{10x-8}{x+1}}\)
5)\(13x-17+4\sqrt{x+1}=6\sqrt{x-2}\left(1+2\sqrt{x+1}\right)\);
6)\(x^2+8x+2\left(x+1\right)\sqrt{x+6}=6\sqrt{x+1}\left(\sqrt{x+6}+1\right)+9\)
7)\(x^2+9x+2+4\left(x+1\right)\sqrt{x+4}=\frac{5}{2}\sqrt{x+1}\left(2+\sqrt{x+4}\right)\)
8)\(8x^2-26x-2+5\sqrt{2x^4+5x^3+2x^2+7}\)
À do nãy máy lag sr :) Chứ bài đặt ẩn phụ mệt lắm :)
giải pt:
1) \(4\sqrt{\frac{x^2}{3}+4}=1+\frac{3x}{2}+\sqrt{6x}\)
2) \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
3) \(\sqrt{1+x}+\sqrt{1-x}+\frac{x^2}{4}=2\)
ĐKXĐ : x\(\ge0\)
ADBĐT BCS ta được
\(\left(\frac{x^2}{3}+4\right)\left(3+1\right)\ge\left(x+2\right)^2\)
\(\Rightarrow4\sqrt{\frac{x^2}{3}+4}\ge2x+4\)(do x\(\ge0\)) (1)
Do x\(\ge0\)nên ADBĐT Cauchy ta được:
\(\sqrt{6x}\le\frac{x+6}{2}\)\(\Rightarrow1+\frac{3x}{2}+\sqrt{6x}\le1+\frac{3x}{2}+\frac{x+6}{2}=1+\frac{4x+6}{2}=2x+4\)(2)
Từ (1) và (2) \(\Rightarrow4\sqrt{\frac{x^2}{3}+4}\ge1+\frac{3x}{2}+\sqrt{6x}\)
Dấu = xảy ra \(\Leftrightarrow x=6\)(thỏa mãn ĐKXĐ)
3) ĐKXĐ \(-1\le x\le1\)
Khi đó phương trình đã cho \(\Leftrightarrow4\left(\sqrt{1+x}+\sqrt{1-x}\right)=8-x^2\)
\(\Leftrightarrow\hept{\begin{cases}16\left(2+2\sqrt{1-x^2}\right)=\left(7+1-x^2\right)\left(2\right)\\8-x^2\ge0\end{cases}}\)
Đặt \(\sqrt{1-x^2}=a\ge0\)
Khi đó phương trình (2) trở thành:
\(\hept{\begin{cases}16\left(2+2a\right)=\left(7+a^2\right)\\x^2\le8\end{cases}}\)
\(\Leftrightarrow a^4+14a^2+49=32+32a\)
\(\Leftrightarrow a^4+14a^2-32a+17=0\)
\(\Leftrightarrow a^4-2a^2+1+16a^2-32a+16=0\)
\(\Leftrightarrow\left(a^2-1\right)^2+16\left(a-1\right)^2=0\)
\(\Leftrightarrow a=1\)
hay \(\sqrt{1-x^2}=1\)
\(\Leftrightarrow x=0\)(thỏa mãn)
Bài 1: Tìm điều kiện để các phân thức sau có nghĩa
a)\(\frac{x-1}{x+1}b)\frac{2x+1}{-3x+5}c)\frac{3x-1}{x^2-4}d)\frac{x-1}{x^2+4}e)\frac{x-1}{\left(x-2\right)\left(x+3\right)}g)\frac{x-1}{x+2}:\frac{x}{x+1}\)
Bài 2 :Tìm điều kiện để các căn thức sau có nghĩa:\(1)\sqrt{3x}|2)\sqrt{-x}|3)\sqrt{3x+2}|4)\sqrt{5-2x}|5)\sqrt{x^2}|6)\sqrt{-4x^2}|7)\sqrt{x-3}+\sqrt{2x+2}|8)\sqrt{\frac{-3}{x+2}}|9)\frac{3}{2x-4}\)