\(\sqrt{\left(x+y\right)^2}+\sqrt{\left(y-2005\right)^2}< 0\)0
\(\sqrt{\left(x+y\right)^2}+\sqrt{\left(y-2005\right)^2}< 0\)
\(\sqrt{\left(x-y\right)^2}+\sqrt{\left(y-2005\right)^2}< hoac=0\)
Vì \(\sqrt{\left(x-y\right)^2}=\left|x-y\right|\ge0\forall x;y\)
\(\sqrt{\left(y-2015\right)^2}=\left|y-2016\right|\ge0\forall y\)
\(\Rightarrow\sqrt{\left(x-y\right)^2}+\sqrt{\left(y-2015\right)^2}=\left|x-y\right|+\left|y-2015\right|\ge0\forall x;y\)
Để \(\sqrt{\left(x-y\right)^2}+\sqrt{\left(y-2005\right)^2}\le0\Leftrightarrow\hept{\begin{cases}\left|x-y\right|=0\\\left|y-2005\right|=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-y=0\\x-2005=0\end{cases}\Rightarrow x=y=2005}\)
Vậy \(x=y=2005\)
Cho: \(\left(x+\sqrt{x^2+\sqrt{2005}}\right)\left(y+\sqrt{y^2+\sqrt{2005}}\right)=\sqrt{2005}\)
1) Chứng minh: \(y+\sqrt{y^2+\sqrt{2005}}=-\left(x-\sqrt{x^2+\sqrt{2005}}\right)\)
2) Tính S = x + y
Làm đầy đủ và chi tiết nhé mọi người
1) \(\left(x+\sqrt{x^2+\sqrt{2005}}\right)\left(\sqrt{x^2+\sqrt{2005}}-x\right)=\sqrt{2005}\)
Kết hợp với giả thiết ta được:
\(\sqrt{x^2+\sqrt{2005}}-x=y+\sqrt{y^2+\sqrt{2005}}\)
suy ra: đpcm
2) \(\left(x+\sqrt{x^2+\sqrt{2005}}\right)\left(y+\sqrt{y^2+\sqrt{2005}}\right)=\sqrt{2005}\)
Ta có: \(\hept{\begin{cases}\left(x+\sqrt{x^2+\sqrt{2005}}\right)\left(\sqrt{x^2+\sqrt{2005}}-x\right)=\sqrt{2005}\\\left(y+\sqrt{y^2+\sqrt{2005}}\right)\left(\sqrt{y^2+\sqrt{2005}}-y\right)=\sqrt{2005}\end{cases}}\)
Kết hợp với giả thiết ta có:
\(\hept{\begin{cases}\sqrt{x^2+\sqrt{2005}}-x=y+\sqrt{y^2+\sqrt{2005}}\\\sqrt{y^2+\sqrt{2005}}-y=x+\sqrt{x^2+\sqrt{2005}}\end{cases}}\)
suy ra: \(x+y=-\left(x+y\right)\)
\(\Rightarrow\)\(S=x+y=0\)
cho a,b,c ,x,y,z là các số dương thỏa \(x+y+x=a;x^2+y^2+z^2=b;a^2=b+4010\)
tính \(M=\sqrt[x]{\frac{\left(2005+y^2\right)\left(2005+z^2\right)}{2005+x^2}}+\sqrt[y]{\frac{\left(2005+x^2\right)\left(2005+z^2\right)}{2005+y^2}}\)
\(+\sqrt[z]{\frac{\left(2005+x^2\right)\left(2005+y^2\right)}{2005+z^2}}\)
giải giup mik vs
Cho\(\left(x+\sqrt{x^2+2005}\right)\left(y+\sqrt{y^2\sqrt{2005}}\right)=\sqrt{2005}\)
Tính x+y
chp a,b,c,x,y,z là các số nguyên dương thỏa \(x+y+z=a\) ;\(x^2+y^2+z^2=b\);\(a^2=b+4010\)
tính \(M=\sqrt[x]{\frac{\left(2005+y^2\right)\left(2005+z^2\right)}{\left(2005+x^2\right)}}+\sqrt[y]{\frac{\left(2005+x^2\right)\left(2005+z^2\right)}{2005+y^2}}\)\(+\sqrt[z]{\frac{\left(2005+x^2\right)\left(2005+y^2\right)}{2005+z^2}}\)
\(a^2=b+4010\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+4010\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz=x^2+y^2+z^2+4010\)
\(\Rightarrow2xy+2yz+2xz=4010\Rightarrow xy+yz+xz=2005\)
\(x\sqrt{\frac{\left(2015+y^2\right)\left(2005+z^2\right)}{\left(2005+x^2\right)}}=x\sqrt{\frac{\left(xz+yz+xy+y^2\right)\left(xy+xz+yz+z^2\right)}{\left(xy+yz+x^2+xz\right)}}\)
\(=x\sqrt{\frac{\left(z\left(x+y\right)+y\left(x+y\right)\right)\left(x\left(y+z\right)+z\left(y+z\right)\right)}{\left(y\left(x+z\right)+x\left(x+z\right)\right)}}=x\sqrt{\frac{\left(y+z\right)^2\left(x+y\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}=x\left(y+z\right)=xy+xz\)
tương tự : \(y\sqrt{\frac{\left(2015+x^2\right)\left(2015+z^2\right)}{2015+y^2}}=xy+yz;z\sqrt{\frac{\left(2005+x^2\right)\left(2005+y^2\right)}{2015+z^2}}=xz+yz\)
\(\Rightarrow M=xy+xz+xy+yz+xz+yz=2\left(xy+yz+xz\right)=2\cdot2005=4010\)
Biết \(0< x\le y\)và \(\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)+2\left(x+2y\right)}\right)+\left(\frac{y}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}\right)=\frac{5}{3}\)
Tính \(\frac{x}{y}\)
a:\(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}\left(b>0;a\ne4\right)\)
b:\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne0\right)\)
c:\(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}\left(a>0;b\ne2\right)}\)
d:\(\dfrac{x}{\left(y-3\right)^2}.\sqrt{\dfrac{\left(y-3\right)^2}{x^2}\left(x>0;y\ne3\right)}\)
e:2x +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\)
b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\sqrt{x}=a,\sqrt{y}=b\)
Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)
\(\Rightarrow B=x+\sqrt{xy}+y\)
Vậy...
c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\)
d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\)
a:b(a−4)2.√(a−4)4b2(b>0;a≠4)b(a−4)2.(a−4)4b2(b>0;a≠4)
= \(\dfrac{b}{\left(a-4\right)}.\dfrac{\sqrt{\left[\left(a-4\right)^2\right]^2}}{\sqrt{b^2}}\)
=\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\)
= 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn)
b:x√x−y√y√x−√y(x≥0;y≥0;x≠0)xx−yyx−y(x≥0;y≥0;x≠0)
=\(\dfrac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức )
= (x+\(\sqrt{xy}\)+y)
c:a(b−2)2.√(b−2)4a2(a>0;b≠2)a(b−2)2.(b−2)4a2(a>0;b≠2)
Tương tự câu a
d:x(y−3)2.√(y−3)2x2(x>0;y≠3)x(y−3)2.(y−3)2x2(x>0;y≠3)
tương tự câu a
e:2x +√1−6x+9x23x−1
= \(2x+\dfrac{\sqrt{\left(3x\right)^2-6x+1}}{3x-1}\)
= 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức)
=2x+\(\dfrac{3x-1}{3x-1}\)
=2x+1
Tìm các số x,y,z thỏa mãn đẳng thức:\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)| = 0