cho phuong trinh:\(\dfrac{2+\sqrt{x}}{\sqrt{2}+\sqrt{2+\sqrt{x}}}+\dfrac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)
a/tim dieu kien cua x de phuong trinh co nghia
b/giai phuong trinh
Giai phuong trinh: \(\dfrac{2+\sqrt{x}}{\sqrt{2}+\sqrt{2+\sqrt{x}}}+\dfrac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)
ĐK: \(0< x\le4\)
Đặt \(\sqrt{2+\sqrt{x}}=a\left(a>0\right)\) ; \(\sqrt{2-\sqrt{x}}=b\left(b\ge0\right)\)
=> \(a^2+b^2=2+\sqrt{x}+2-\sqrt{x}=4\) (1)
Ta có: \(\dfrac{a^2}{\sqrt{2}+a}+\dfrac{b^2}{\sqrt{2}-b}=\sqrt{2}\)
<=> \(\dfrac{a^2.\sqrt{2}-a^2b+b^2.\sqrt{2}+ab^2}{2+\sqrt{2}\left(a-b\right)-ab}=\sqrt{2}\)
<=> \(\left(a^2+b^2\right)\sqrt{2}+ab\left(b-a\right)=2\sqrt{2}+2\left(a-b\right)-ab.\sqrt{2}\)
<=> \(4\sqrt{2}+ab\left(b-a\right)=2\sqrt{2}+2\left(a-b\right)-ab.\sqrt{2}\) ( Theo 1)
<=> \(\left(a-b\right)\left(2+ab\right)=2\sqrt{2}+ab.\sqrt{2}\)
<=> \(\left(a-b-\sqrt{2}\right)\left(ab+2\right)=0\)
<=> \(\left[{}\begin{matrix}ab+2=0\\a-b-\sqrt{2}=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}ab=-2\\a-b=\sqrt{2}\end{matrix}\right.\) mà a2 + b2 = 4
Xét \(\left\{{}\begin{matrix}ab=-2\\a^2+b^2=4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left(a-b\right)^2=8\\\left(a+b\right)^2=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}a-b=\pm\sqrt{8}\\a+b=0\end{matrix}\right.\) ( Loại vì \(a>0;b\ge0\) )
Xét \(\left\{{}\begin{matrix}a-b=\sqrt{2}\\a^2+b^2=4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\\left(b+\sqrt{2}\right)^2+b^2=4\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\2b^2+2b.\sqrt{2}-2=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\b^2+b.\sqrt{2}-1=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\\left[{}\begin{matrix}b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\\b=\dfrac{-\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)
#Lề: Bn lấy cái đề ở đâu hay v?
v cac bac e giai xong lau roi cac bac a voi lai co cach giai ko can dai nhu the dau nhe
1) Tim dieu kien cua bat phuong trinh
a) (2x-1)\(\sqrt{x-2018}\ge\sqrt{x-2018}\)
1) phuong trinh \(\dfrac{4}{\sqrt{x+2}}+\sqrt{x-2}=\sqrt{2-x}\) co bao nhieu nghiem:
A. 1 B. 2 C. Vo so D. 0
phương trình này vô số nghiệm á bạn
đáp án là câu C nha
giai phuong trinh :
\(\dfrac{\sqrt{x+3}+\sqrt{x-1}}{\sqrt{x+3}-\sqrt{x-1}}=\dfrac{13-x^2}{4}\)
\(\Leftrightarrow\dfrac{x+3+x-1+2\sqrt{\left(x+3\right)\left(x-1\right)}}{x+3-x+1}=\dfrac{13-x^2}{4}\)
\(\Leftrightarrow2x+2+2\sqrt{\left(x+3\right)\left(x-1\right)}=13-x^2\)
\(\Leftrightarrow\sqrt{4\left(x+3\right)\left(x-1\right)}=13-x^2-2x-2=-x^2-2x+11\)
=>\(x\simeq1,37\)
giai phuong trinh \(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}}+\dfrac{2005}{2006}\)
\(\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}}=\dfrac{1}{2006}\sqrt{2006^2+2005^2+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{\left(2006-2005\right)^2+2.2005.2006+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{1+2.2005.2006+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{\left(2005.2006+1\right)^2}=\dfrac{2005.2006+1}{2006}=2005+\dfrac{1}{2006}\)
Phương trình tương đương:
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2005+\dfrac{1}{2006}+\dfrac{2005}{2006}\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2006\)
TH1: \(x\ge2\): \(x-1+x-2=2006\Rightarrow2x=2009\Rightarrow x=\dfrac{2009}{2}\)
TH2: \(x\le1\) : \(1-x+2-x=2006\Rightarrow-2x=2003\Rightarrow x=\dfrac{-2003}{2}\)
TH3: \(1< x< 2:\) \(x-1+2-x=2006\Rightarrow3=2006\) (vô nghiệm)
Vậy \(\left[{}\begin{matrix}x=\dfrac{2009}{2}\\x=\dfrac{-2003}{2}\end{matrix}\right.\)
tim m de he phuong trinh va phuong trinh co nghiem
\(a,\sqrt{x^2+3x+2m}=\sqrt{4x-x^2}\)
b, \(\left\{{}\begin{matrix}x+y+1=x\\x^2+y^2=m\end{matrix}\right.\)
Giai phuong trinh: \(\sqrt{3x+x^2+\dfrac{9}{4}}+\sqrt{x^2+3x+1}=0\)
Lời giải:
Với mọi $x$ thuộc ĐKXĐ, ta luôn có:
\(\left\{\begin{matrix} \sqrt{3x+x^2+\frac{9}{4}}\geq 0\\ \sqrt{x^2+3x+1}\geq 0\end{matrix}\right.\)
Do đó, để \(\sqrt{3x+x^2+\frac{9}{4}}+\sqrt{x^2+3x+1}=0\) thì:
\(\left\{\begin{matrix} \sqrt{3x+x^2+\frac{9}{4}}= 0\\ \sqrt{x^2+3x+1}=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x=\frac{-3}{2}\\ x=\frac{3\pm \sqrt{5}}{2}\end{matrix}\right.\) (vô lý)
Do đó pt vô nghiệm.
Giai phuong trinh : \(\frac{2+\sqrt{x}}{\sqrt{2}+\sqrt{\sqrt{x}+2}}+\frac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)
Xin lỗi bạn nhiều nhiều lắm mình không biết làm bài này vì mình chưa học
không cần đâu bạn à bài này mình giải được rồi
A=(\(\frac{1}{\sqrt{x}-1}\)+\(\frac{1}{\sqrt{x}+1}\))2.\(\frac{x^2-1}{2}\)-\(\sqrt{1-x^2}\)
1)tim dk cua x de bieu thuc A co nghia
2)rut gon bieu thuc A
3) giai phuong trinh theo x khi A=-2