Giair phương trinhf\(\frac{x^2}{2}+\frac{18}{x^2}=13\left(\frac{x}{2}-\frac{3}{x}\right)\)
Giair phương trình: \(2\left(x+\frac{1}{x}\right)^2+\left(x^2+\frac{1}{x^2}\right)^2-\left(x+\frac{1}{x}\right)^2\left(x^2+\frac{1}{x^2}\right)=4-4x+x^2\)
Đặt \(x+\frac{1}{x}=t\Rightarrow\left(x+\frac{1}{x}\right)^2=t^2\Leftrightarrow x^2+\frac{1}{x^2}=t^2-2\)
Khi đó phương trình đã cho
\(\Leftrightarrow2t^2+\left(t^2-2\right)^2-t^2\left(t^2-2\right)=4-4x+x^2\)
\(\Leftrightarrow2t^2+t^4-4t^2+4-t^4+2t^2=x^2-4x+4\)
\(\Leftrightarrow4=x^2-4x+4\)
\(\Leftrightarrow x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Mà ĐKXĐ của phương trình là \(x\ne0\)
Tập nghiệm của pt là \(S=\left\{4\right\}\)
Đặt \(x+\frac{1}{x}=a\)
\(\Rightarrow\left(x+\frac{1}{x}\right)^2=a^2\Leftrightarrow x^2+\frac{1}{x^2}+2=a^2\Leftrightarrow x^2+\frac{1}{x^2}=a^2-2\)
Có \(2a^2+\left(a^2-2\right)^2-a^2\left(a^2-2\right)=\left(2-x\right)^2\)
\(2a^2+a^4-4a^2+4-a^4+2a^2=\left(2-x\right)^2\)
\(\Leftrightarrow4=\left(2-x\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2-x=4\\2-x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=6\end{cases}}\)
Vậy \(S=\left(-2;6\right)\)
Tại sao \(\left(x^2+\frac{1}{x^2}\right)=t^2-2\) thế
Giải phương trình sau
a, \(\frac{3x}{x^2-x+3}-\frac{2x}{x^2-3x+3}=-1\)
b, \(\frac{1}{\left(x^2+2x+2\right)^2}+\frac{1}{\left(x^2+2x+3\right)^2}=\frac{5}{4}\)
c,\(\left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2=\frac{10}{9}\)
d,\(\frac{x^2}{2}+\frac{18}{x^2}=13\left(\frac{x}{2}-\frac{3}{x}\right)\)
a/ Do \(x=0\) không phải nghiệm, pt tương đương:
\(\frac{3}{x+\frac{3}{x}-1}-\frac{2}{x+\frac{3}{x}-3}=-1\)
Đặt \(x+\frac{3}{x}-3=a\) ta được:
\(\frac{3}{a+2}-\frac{2}{a}=-1\)
\(\Leftrightarrow3a-2\left(a+2\right)=-a\left(a+2\right)\)
\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{x}-3=1\\x+\frac{3}{x}-3=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2+x+3=0\end{matrix}\right.\)
b/ Đặt \(x^2+2x+\frac{5}{2}=a>0\)
Phương trình trở thành:
\(\frac{1}{\left(a-\frac{1}{2}\right)^2}+\frac{1}{\left(a+\frac{1}{2}\right)^2}=\frac{5}{4}\)
\(\Leftrightarrow4\left(a+\frac{1}{2}\right)^2+4\left(a-\frac{1}{2}\right)^2=5\left(a^2-\frac{1}{4}\right)^2\)
\(\Leftrightarrow8a^2+2=5\left(a^4-\frac{1}{2}a^2+\frac{1}{16}\right)\)
\(\Leftrightarrow5a^4-\frac{21}{2}a^2-\frac{27}{16}=0\Rightarrow\left[{}\begin{matrix}a^2=\frac{9}{4}\\a^2=-\frac{3}{20}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+2x+\frac{5}{2}=\frac{3}{2}\\x^2+2x+\frac{5}{2}=-\frac{3}{2}\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne\pm1\)
\(\Leftrightarrow\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2+\frac{2x^2}{x^2-1}-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)
\(\Leftrightarrow\left(\frac{x}{x+1}+\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)
\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)
Đặt \(\frac{2x^2}{x^2-1}=a\)
\(\Rightarrow a^2-a-\frac{10}{9}=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{5}{3}\\a=-\frac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2x^2}{x^2-1}=\frac{5}{3}\\\frac{2x^2}{x^2-1}=-\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=-5\left(l\right)\\x^2=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow x=\pm\frac{1}{2}\)
d/ĐKXĐ: ...
\(\Leftrightarrow\left(x^2+\frac{36}{x^2}\right)-13\left(x-\frac{6}{x}\right)=0\)
Đặt \(x-\frac{6}{x}=a\Rightarrow x+\frac{36}{x^2}=a^2+12\)
\(\Rightarrow a^2-13a+12=0\Rightarrow\left[{}\begin{matrix}a=1\\a=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{6}{x}=1\\x-\frac{6}{x}=12\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-x-6=0\\x^2-12x-6=0\end{matrix}\right.\)
giair hệ phương trình:
\(\hept{\begin{cases}6x+\frac{3}{x+y}=13\\12\left(x^2+xy+y^2\right)+\frac{9}{\left(x+y\right)^2}=85\end{cases}}\)
mn giúp mk nha. thak mn nhiều
a, \(\left(12-\frac{7}{18}-10\frac{13}{18}\right):x-1\frac{7}{13}=1\frac{2}{3}\)
b,\(\left[\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right).2,8+1,75\right]:0,05=235\)
c.\(\frac{3-x}{5-x}=\left(\frac{3}{5}\right)^2\)
d,\(\left(1-\frac{3}{10}-x\right):\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=1\)
e,\(\left(12\frac{7}{18}-10\frac{13}{18}\right):x-1\frac{7}{33}:\frac{8}{11}=1\frac{2}{3}\)
các bạn ơi giúp tớ với
Giair phương trình:
\(\left(x^2+\frac{4}{x^2}+1\right)^2=\left(x^2-\frac{4}{x^2}-1\right)^2\)
\(\left(x^2+\frac{4}{x^2}+1\right)^2=\left(x^2-\frac{4}{x^2}-1\right)^2\)
=> \(x^2+\frac{4}{x^2}+1=x^2-\frac{4}{x^2}-1\)
=> \(\left(x^2-x^2\right)+\left(\frac{4}{x^2}+\frac{4}{x^2}\right)=-1-1\)
=> \(\frac{8}{x^2}=-2\)
=> \(8=-2x^2\)
\(=>x^2=-4\)
=> x = rỗng
=> \(x^2=-4\)
Tìm x:
\(\frac{\left(13\frac{2}{9}-15\frac{2}{3}\right)\cdot\left(30^2-5^4\right)}{\left(18\frac{3}{7}-17\frac{1}{4}\right)\cdot\left(25-12\cdot5^2\right)}\cdot x=\frac{\frac{2}{11}+\frac{3}{13}+\frac{4}{15}+\frac{5}{17}}{4\frac{1}{11}+\frac{5}{13}+\frac{9}{15}+\frac{13}{17}}\)
Giải pt"
\(\frac{x^2}{2}+\frac{18}{x^2}=13.\left(\frac{x}{2}-\frac{3}{x}\right)\)
giải phương trình: \(\frac{x^2}{2}+\frac{18}{x^2}=13\left(\frac{x}{2}-\frac{3}{x}\right)\)
Q= \(\frac{\sqrt{a}\left(1-a\right)^2}{1-a^2}:\left[\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
a) Rút gọn biểu thức Q? b) Xét dấu of biểu thức P= a.(Q-\(\frac{1}{2}\))
Tìm x
1)\(\frac{x^4+x^2+1}{x^2}=\frac{x^2+x+1}{x}\)
2)\(\frac{x^2}{2}+\frac{18}{x^2}=13\left(\frac{x}{2}-\frac{3}{x^2}\right)\)
3)\(\frac{x^4+1}{\left(x+1\right)^4}=\frac{1}{2}\)
1/ <=> x2 - x -(x2 - x)/x3 = 0
<=> (x2 - x)(1 - 1/x3) = 0
Phần còn lại bạn làm tiếp nha điều kiện x#0