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\(ĐK:-1\le x\le1\\ PT\Leftrightarrow13\left(1-2x^2\right)\sqrt{\left(1-x^2\right)\left(1+x^2\right)}+9\left(1+2x^2\right)\sqrt{\left(1+x^2\right)\left(1-x^2\right)}=0\\ \Leftrightarrow\sqrt{1-x^4}\left(13-26x^2+9+18x^2\right)=0\\ \Leftrightarrow\sqrt{1-x^4}\left(22-8x^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1-x^4=0\\22-8x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(1+x^2\right)\left(1-x\right)\left(1+x\right)=0\\x^2=\dfrac{22}{8}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{\sqrt{11}}{2}\left(ktm\right)\\x=-\dfrac{\sqrt{11}}{2}\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
mình nhầm mẫu nhé :v mình làm lại
\(=\left(\dfrac{x-\sqrt{x}-2x+4\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\right):\dfrac{2-\sqrt{x}}{x-1}\)
\(=\dfrac{-x+3\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{2-\sqrt{x}}=\dfrac{\left(2-\sqrt{x}\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(2-\sqrt{x}\right)\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)
\(\Leftrightarrow x^2+3x-12-10x-30=0\)
\(\Leftrightarrow x^2-7x-42=0\)
\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)
b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};1\right\}\)
Lời giải:
Đặt $\frac{1}{x-y+2}=a;\frac{1}{x+y-1}=b$ thì HPT trở thành cơ bản:
\(\left\{\begin{matrix}
14a-10b=9\\
3a+2b=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
14a-10b=9\\
15a+10b=20\end{matrix}\right.\)
$\Rightarrow (14a-10b)+(15a+10b)=9+20$
$\Leftrightarrow 29a=29\Leftrightarrow a=1$.
$b=\frac{4-3a}{2}=\frac{1}{2}$
Vậy: \(\left\{\begin{matrix} \frac{1}{x-y+2}=1\\ \frac{1}{x+y-1}=\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-y+2=1\\ x+y-1=2\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x-y=-1\\ x+y=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\end{matrix}\right.\)
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y+1=4xy\\\left(\dfrac{x}{y+1}\right)^2+\left(\dfrac{y}{x+1}\right)^2=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=4xy\\\left(\dfrac{x}{y+1}\right)^2+\left(\dfrac{y}{x+1}\right)^2=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{x}{y+1}\right)\left(\dfrac{y}{x+1}\right)=\dfrac{1}{4}\\\left(\dfrac{x}{y+1}\right)^2+\left(\dfrac{y}{x+1}\right)^2=\dfrac{1}{2}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x}{y+1}=u\\\dfrac{y}{x+1}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^2+v^2=\dfrac{1}{2}\\uv=\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow u^2-2uv+v^2=0\Leftrightarrow u=v=\pm\dfrac{1}{2}\)
TH1: \(u=v=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y+1}=\dfrac{1}{2}\\\dfrac{y}{x+1}=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=1\\x-2y=-1\end{matrix}\right.\) \(\Leftrightarrow...\)
Th2: \(u=v=-\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y+1}=-\dfrac{1}{2}\\\dfrac{y}{x+1}=-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x+y=-1\\x+2y=-1\end{matrix}\right.\) \(\Leftrightarrow...\)
\(\left(x\ne-y;x>\dfrac{y}{2}\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7-\left(x+y\right)}{x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7}{x+y}=2\end{matrix}\right.\)
\(đặt:\dfrac{1}{\sqrt{2x-y}}=a>0;\dfrac{1}{x+y}=b\)
\(\Rightarrow\left\{{}\begin{matrix}4a-21b=\dfrac{1}{2}\\3a+7b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\left(tm\right)\\b=\dfrac{1}{14}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{2x-y}}=\dfrac{1}{2}\\\dfrac{1}{x+y}=\dfrac{1}{14}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\x+y=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=8\end{matrix}\right.\)(thỏa)
(1) + rút y từ pt (2) thay vào pt (1), ta được pt bậc hai 1 ẩn x, dễ rồi, tìm x rồi suy ra y
(2) + (3)
+ pt nào có nhân tử chung thì đặt nhân tử chung (thật ra chỉ có pt (2) của câu 2 là có nhân từ chung)
+ trong hệ, thấy biểu thức nào giống nhau thì đặt cho nó 1 ẩn phụ
VD hệ phương trình 3: đặt a= x+y ; b= căn (x+1)
+ khi đó ta nhận được một hệ phương trình bậc nhất hai ẩn, giải hpt đó rồi suy ra x và y
ĐKXĐ: \(x\ne0;\pm1\)
\(\left(\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)^2}-\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)\dfrac{1}{x}=2\)
\(\Leftrightarrow\dfrac{x^2+3x+2}{\left(x-1\right)^2}-\dfrac{x-2}{x-1}=2x\)
\(\Leftrightarrow\dfrac{x^2-2x+1+5x-5+6}{\left(x-1\right)^2}-\dfrac{x-1-1}{x-1}=2x\)
\(\Leftrightarrow1+\dfrac{5}{x-1}+\dfrac{6}{\left(x-1\right)^2}-1+\dfrac{1}{x-1}=2x\)
\(\Leftrightarrow\dfrac{3}{x-1}+\dfrac{3}{\left(x-1\right)^2}=x=x-1+1\)
Đặt \(\dfrac{1}{x-1}=a\) phương trình trở thành:
\(3a+3a^2=\dfrac{1}{a}+1=\dfrac{a+1}{a}\)
\(\Leftrightarrow3a\left(a+1\right)-\dfrac{a+1}{a}=0\)
\(\Leftrightarrow\left(a+1\right)\left(3a-\dfrac{1}{a}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\\3a=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=-1\\a=\dfrac{\sqrt{3}}{3}\\a=\dfrac{-\sqrt{3}}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{x-1}=-1\\\dfrac{1}{x-1}=\dfrac{\sqrt{3}}{3}\\\dfrac{1}{x-1}=\dfrac{-\sqrt{3}}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=1+\sqrt{3}\\x=1-\sqrt{3}\end{matrix}\right.\)