Giải pt : \(\sqrt{x^2+91}=\sqrt{x-2}+x^2\)
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24 tháng 9 2021
\(ĐK:0< x\le4\)
Đặt \(\sqrt{2+\sqrt{x}}=a>0;\sqrt{2-\sqrt{x}}=b>0\)
\(\Rightarrow a^2+b^2=2+\sqrt{x}+2-\sqrt{x}=4\)
\(PT\Leftrightarrow\dfrac{a^2}{\sqrt{2}+a}+\dfrac{b^2}{\sqrt{2}-b}=\sqrt{2}\\ \Leftrightarrow\dfrac{a^2\sqrt{2}-a^2b+b^2\sqrt{2}+ab^2}{2+\sqrt{2}\left(a-b\right)-ab}=\sqrt{2}\\ \Leftrightarrow\sqrt{2}\left(a^2+b^2\right)+ab\left(b-a\right)=2\sqrt{2}+2\left(a-b\right)-\sqrt{2}ab\\ \Leftrightarrow4\sqrt{2}-ab\left(a-b\right)=2\sqrt{2}+2\left(a-b\right)-\sqrt{2}ab\\ \Leftrightarrow\left(2+ab\right)\left(a-b\right)=2\sqrt{2}+\sqrt{2}ab\\ \Leftrightarrow\left(2+ab\right)\left(a-b\right)-\sqrt{2}\left(2+ab\right)=0\\ \Leftrightarrow\left(a-b-\sqrt{2}\right)\left(2+ab\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}ab=-2\\a-b=\sqrt{2}\end{matrix}\right.\)
Xét \(\left\{{}\begin{matrix}ab=-2\\a^2+b^2=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=8\\\left(a+b\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=\pm2\sqrt{2}\\a+b=0\end{matrix}\right.\left(loại.vì.a>0;b\ge0\right)\)
Xét \(\left\{{}\begin{matrix}a-b=\sqrt{2}\\a^2+b^2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\b^2+2\sqrt{2}b+2+b^2=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\2b^2+2\sqrt{2}b-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\b^2+b\sqrt{2}-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\\left[{}\begin{matrix}b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\\b=\dfrac{-\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\left(sd.\Delta\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\left(b\ge0\right)\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2+\sqrt{x}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\\sqrt{2-\sqrt{x}}=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)
Tới đây dễ r nha
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2021
a.
ĐKXĐ: \(1\le x\le7\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2021
b. ĐKXĐ: ...
Biến đổi pt đầu:
\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a^2b^2-b^4=b-a\)
\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)
Thế vào pt dưới:
\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)
\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)
\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)
\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)
\(\Leftrightarrow...\)
V
1
21 tháng 6 2021
`ĐK:x>=2`
`pt<=>sqrt{(x-1)(x-2)}+sqrt{x+3}=sqrt{x-2}+sqrt{(x-1)(x+3)}`
`<=>sqrt{x-1}(sqrt{x-2}-sqrt{x+3})-(sqrt{x-2}-sqrt{x+3})=0`
`<=>(sqrt{x-2}-sqrt{x+3})(sqrt{x-1}-1)=0`
`+)sqrt{x-2}=sqrt{x+3}`
`<=>x-2=x+3`
`<=>0=5` vô lý
`+)sqrt{x-1}-1=0`
`<=>x-1=1`
`<=>x=2(tm)`.
Vậy `x=2`.
BC
0
Lời giải:
ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow (\sqrt{x^2+91}-10)=(\sqrt{x-2}-1)+(x^2-9)$
$\Leftrightarrow \frac{x^2-9}{\sqrt{x^2+91}+10}=\frac{x-3}{\sqrt{x-2}+1}+(x-3)(x+3)$
$\Leftrightarrow (x-3)\left[\frac{x+3}{\sqrt{x^2+91}+10}-1-(x+3)\right]=0$
Dễ thấy với $x\geq 2$ thì:
$x+3> \frac{x+3}{\sqrt{x^2+91}+10}$
$\Rightarrow 1+(x+3)> \frac{x+3}{\sqrt{x^2+91}+10}$
Vậy biểu thức trong ngoặc vuông lớn hơn $0$
$\Rightarrow x-3=0$
$\Leftrightarrow x=3$ (tm)