Bài toán :
Cho \(\left(x+\sqrt{y^2+1}\right)\left(y+\sqrt{x^2+1}\right)=1\)
CMR : \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)
Cho 3 số dương x,y,z thỏa mãn x + y + z = xyz. Cmr:
\(A=\frac{\sqrt{\left(1+y^2\right)\left(1+z^2\right)}-\sqrt{1+y^2}-\sqrt{1+z^2}}{yz}+\frac{\sqrt{\left(1+z^2\right)\left(1+x^2\right)}-\sqrt{1+x^2}-\sqrt{1+z^2}}{xz}+\frac{\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-\sqrt{1+x^2}-\sqrt{1+y^2}}{xy}=0\)
@Akai Haruma, Nguyen, Nguyễn Thị Ngọc Thơsvtkvtm
Bạn tham khảo tại đây:
Câu hỏi của Vũ Sơn Tùng - Toán lớp 9 | Học trực tuyến
Cho x,y biết \(\left(x+\sqrt{1+y^2}\right)\left(y+\sqrt{1+x^2}\right)=1\)
CMR \(x\left(\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)=1\)
Cho \(\left(x+\sqrt{y^2+1}\right)\left(y+\sqrt{x^2+1}\right)=1\)
\(CMR:\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)
Bài 2. Cho A=\(\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}\) :\([\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\dfrac{1}{xy+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)]\)
Bạn cần làm gì với biểu thức này?
\(\frac{\sqrt{x}\left(\sqrt{x}-2\right)+\sqrt{y}\left(\sqrt{y}+2\right)-2\sqrt{xy}+1}{\sqrt{x}\left(\sqrt{x}-2\sqrt{y}\right)+\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\)
\(=\frac{x-2\sqrt{x}+y+2\sqrt{y}-2\sqrt{xy}+1}{x-2\sqrt{xy}+y-1}\)\(=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2-2\left(\sqrt{x}-\sqrt{y}\right)+1}{\left(\sqrt{x}-\sqrt{y}\right)^2-1}\)
\(=\frac{\left(\sqrt{x}-\sqrt{y}-1\right)^2}{\left(\sqrt{x}-\sqrt{y}+1\right)\left(\sqrt{x}-\sqrt{y}-1\right)}=\frac{\sqrt{x}-\sqrt{y}-1}{\sqrt{x}-\sqrt{y}+1}\)
Cho x,y là các số thực dương thỏa mãn đồng thời các điều kiên:
1) \(\left(x+2\right)\left(y+2\right)=3\left(x^2+y^2+\sqrt{xy}\right)\)
2) \(\left(\sqrt{x}+\sqrt{y}\right)^3=4\left(x^3+y^3\right)\)
CMR: \(\sqrt{x}+\sqrt{y}=2\)
Giải hệ phương trình
\(\left\{{}\begin{matrix}\sqrt{\left(x-1\right)^2+\left(y-2\right)^2}=\sqrt{\left(x+1\right)^2+\left(y-1\right)^2}\\\sqrt{\left(x-1\right)^2+\left(y-2\right)^2}=\sqrt{\left(x-5\right)^2+\left(y+1\right)^2}\end{matrix}\right.\)
\(ĐK:x,y\in R\)
Từ 2 PT \(\Leftrightarrow\sqrt{\left(x+1\right)^2+\left(y-1\right)^2}=\sqrt{\left(x-5\right)^2+\left(y+1\right)^2}\)
\(\Leftrightarrow x^2+2x+y^2-2y+2=x^2-10x+y^2+2y+26\\ \Leftrightarrow12x-4y-24=0\\ \Leftrightarrow3x-y-6=0\\ \Leftrightarrow y=3x-6\)
Thay vào \(PT\left(1\right)\Leftrightarrow\sqrt{\left(x-1\right)^2+\left(3x-8\right)^2}=\sqrt{\left(x+1\right)^2+\left(3x-7\right)^2}\)
\(\Leftrightarrow10x^2-50x+65=10x^2-40x+50\\ \Leftrightarrow10x=15\Leftrightarrow x=\dfrac{3}{2}\Leftrightarrow y=-\dfrac{3}{2}\)
Vậy hệ có nghiệm \(\left(x;y\right)=\left(\dfrac{3}{2};-\dfrac{3}{2}\right)\)
Cho các số thực x,y thỏa mãn : \(\left(x+\sqrt{1+y^2}\right)\left(y+\sqrt{1+x^2}\right)=1\)
cmr: \(\left(x+\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)=1\)
Cho 3 số x y z thỏa mãn x+y+z=xyz.Cm:\(\dfrac{\sqrt{\left(1+y^2\right)\left(1+z^2\right)}-\sqrt{1+y^2}-\sqrt{1+z^2}}{yz}+\dfrac{\sqrt{\left(1+z^2\right)\left(1+x^2\right)}-\sqrt{1+z^2}-\sqrt{1+x^2}}{zx}+\dfrac{\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-\sqrt{1+x^2}-\sqrt{1+z^2}}{yz}=0\)
Lời giải:
Từ \(x+y+z=xyz\Rightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)
Đặt \((\frac{1}{a}, \frac{1}{b}, \frac{1}{c})=(x,y,z)\), trong đó $a,b,c>0$ thì ta có:
\(ab+bc+ac=1\) và cần phải CMR:
\(P=\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}+\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}+\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}\)
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Ta có:
\(\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}=\sqrt{(b^2+1)(c^2+1)}-b\sqrt{c^2+1}-c\sqrt{b^2+1}\)
\(=\sqrt{(b^2+ab+bc+ac)(c^2+ac+bc+ab)}-b\sqrt{c^2+ac+bc+ab}-c\sqrt{b^2+ab+bc+ac}\)
\(=\sqrt{(b+a)(b+c)(c+a)(c+b)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}\)
\(=(b+c)\sqrt{(a+b)(a+c)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}(1)\)
Tương tự:
\(\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}=(a+c)\sqrt{(b+a)(b+c)}-a\sqrt{(c+a)(c+b)}-c\sqrt{(a+b)(a+c)}(2)\)
\(\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}=(a+b)\sqrt{(c+a)(c+b)}-b\sqrt{(a+b)(a+c)}-a\sqrt{(b+c)(b+a)}(3)\)
Từ \((1);(2);(3)\Rightarrow P=(b+c-c-b)\sqrt{(a+b)(a+c)}+(a+c-c-a)\sqrt{(b+a)(b+c)}+(a+b-b-a)\sqrt{(c+a)(c+b)}\)
\(=0\)
Ta có đpcm.