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\(A=3\sqrt{8}-\sqrt{50}-\sqrt{\sqrt{2}-1}\)
\(\Leftrightarrow6\sqrt{2}-5\sqrt{2}-\sqrt{\sqrt{2}-1}\)
\(\Leftrightarrow\sqrt{2}-\sqrt{\sqrt{2}-1}\)
\(B=2.\dfrac{2}{x-1}.\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{\sqrt{x^2-2x+1}}{2x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{\sqrt{\left(x-1\right)^2}}{x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{x-1}{x}\)
\(\Leftrightarrow\)\(2.\dfrac{1}{x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x}\)
giải pt
a)\(\dfrac{1}{x+1}+\dfrac{3}{2x+1}=\dfrac{8}{x-2}\)
b)\(\sqrt{2x+1}+\sqrt{3-x}=\sqrt{3x+5}\)
a. Pt đã cho tương đương với:
\(\sqrt{3x-2}=\sqrt{x+7}+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\3x-2=x+7+1+2\sqrt{x+7}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\2x-10=2\sqrt{x+7}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\x-5=\sqrt{x+7}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\x^2-10x+25=x+7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\x^2-11x+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\\left(x-2\right)\left(x-9\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\\left[{}\begin{matrix}x=2\\x=9\end{matrix}\right.\end{matrix}\right.\)(Loại )
\(\Leftrightarrow x=9\)
Vậy pt có nghiệm x =9
b. Đk: \(x\ne1;y\ne2\)
Đặt \(\dfrac{1}{x-1}=a;\dfrac{1}{y-2}=b\)
Khi đó hệ đã cho trở thành:
\(\left\{{}\begin{matrix}a+b=2\\-3a+2b=1\end{matrix}\right.\)
Giải hệ trên tìm a,b rồi từ đó tìm được x;y. Nhớ đối chiếu với Đk trước khi kết luận.
a) ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\0>x\ge-1\end{matrix}\right.\). Để pt có nghiệm => x>0=> \(x\ge1\) pt<=> \(x-\sqrt{1-\dfrac{1}{x}}=\sqrt{x-\dfrac{1}{x}}.Bìnhphương2vetaco\left(x-\sqrt{1-\dfrac{1}{x}}\right)^2=x-\dfrac{1}{x}\)\(\Leftrightarrow x^2+1-\dfrac{1}{x}-2x\sqrt{1-\dfrac{1}{x}}=x-\dfrac{1}{x}\Leftrightarrow x^2-x+1=2\sqrt{x^2-x}\Leftrightarrow\left(\sqrt{x^2-x}-1\right)^2=0\Leftrightarrow x^2-x=1\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)
b) ĐKXĐ\(0\le x\le1\) pt \(\Leftrightarrow\left(\sqrt{x^2+x}+\sqrt{x-x^2}\right)^2=\left(x+1\right)^2\Leftrightarrow2x+2x.\sqrt{1-x^2}=x^2+2x+1\Leftrightarrow x^2-2x\sqrt{1-x^2}+1-x^2+x^2=0\Leftrightarrow\left(x-\sqrt{1-x^2}\right)^2+x^2=0\)
Ta có :
\(\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{\left(\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x+1}-\sqrt{x+2}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}=-\sqrt{x+1}+\sqrt{x+2}\)
Tương tự :
\(\dfrac{1}{\sqrt{x+2}+\sqrt{x+3}}=-\sqrt{x+2}+\sqrt{x+3}\)
\(\dfrac{1}{\sqrt{x+3}+\sqrt{x+4}}=-\sqrt{x+3}+\sqrt{x+4}\)
....
\(\dfrac{1}{\sqrt{x+2019}+\sqrt{x+2010}}=-\sqrt{x+2019}+\sqrt{x+2010}\)
Từ những ý trên , pt trở thành :
\(-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+4}-.....-\sqrt{x+2019}+\sqrt{x+2020}=11\)
\(\Leftrightarrow\sqrt{x+2020}-\sqrt{x+1}=11\)
\(\Leftrightarrow x+2020-2\sqrt{\left(x+2020\right)\left(x+1\right)}+x+1=121\)
\(\Leftrightarrow2x+1900=2\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x+950=\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x^2+1900x+902500=x^2+2021x+2020\)
\(\Leftrightarrow121x-900480=0\)
\(\Leftrightarrow x=\dfrac{900480}{121}\)
\(x^2-3x+1=\dfrac{\sqrt{3}}{3}\sqrt{x^4+x^2+1}=0\\ \Leftrightarrow x^2-3x+1=-\dfrac{\sqrt{3}}{3}\sqrt{x^4+x^2+1}\\ \Leftrightarrow\left(x^2-3x+1\right)^2=\dfrac{1}{3}\left(x^4+x^2+1\right)\\ \Leftrightarrow x^4+9x^2+1-6x^3-6x+2x^2=\dfrac{1}{3}x^4+\dfrac{1}{3}x^2+\dfrac{1}{3}\\ \Leftrightarrow3x^4+27x^2+3-18x^3-18x+6x^2=x^4+x^2+1\\ \Leftrightarrow2x^4-18x^3+32x^2-18x+2=0\\ \Leftrightarrow x^4-9x^3+16x^2-9x+1=0\\ \Leftrightarrow\left(x^4-2x^3+x^2\right)-\left(7x^3-14x^2+7x\right)+\left(x^2-2x+1\right)=0\\ \Leftrightarrow x^2\left(x^2-2x+1\right)-7x\left(x^2-2x+1\right)+\left(x^2-2x+1\right)=0\\ \Leftrightarrow\left(x^2-7x+1\right)\left(x^2-2x+1\right)=0\\ \Leftrightarrow\left(x^2-7x+\dfrac{49}{4}-\dfrac{45}{4}\right)\left(x^2-2x+1\right)=0\\ \Leftrightarrow\left[\left(x-\dfrac{7}{2}\right)^2-\dfrac{45}{4}\right]\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{2}-\dfrac{3\sqrt{5}}{2}\right)\left(x-\dfrac{7}{2}+\dfrac{3\sqrt{5}}{2}\right)\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{2}-\dfrac{3\sqrt{5}}{2}=0\\x-\dfrac{7}{2}+\dfrac{3\sqrt{5}}{2}=0\\\left(x-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+3\sqrt{5}}{2}\\x=\dfrac{7-3\sqrt{5}}{2}\\x=1\end{matrix}\right.\)
Vậy.....................