cho \(\left(a+\sqrt{a^2+2015}\right).\left(b+\sqrt{b^2+2015}\right)=2015.\)
Chứng minh \(a^{2015}+b^{2015}=0\)
Cho 3 số dương x , y , z thỏa mãn điều kiện :
\(xy+yz+zx=2015\) và :
\(P=x\sqrt{\frac{\left(2015+y^2\right)\left(2015+z^2\right)}{2015+x^2}+y\sqrt{\frac{\left(2015+x^2\right)\left(2015+z^2\right)}{2015+y^2}}+z\sqrt{\frac{\left(2015+x^2\right)\left(2015+y^2\right)}{2015+z^2}}}\)
Chứng minh rằng P không phải là số chính phương .
Ta có\(x\sqrt{\frac{\left(2015+y^2\right)\left(2015+z^2\right)}{2015+x^2}}=x\sqrt{\frac{\left(xy+yz+zx+y^2\right)\left(xy+yz+zx+z^2\right)}{xy+yz+zx+x^2}}\)
\(=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}=xy+xz\)
Tương tự:\(y\sqrt{\frac{\left(2015+x^2\right)\left(2015+z^2\right)}{2015+y^2}}=yx+yz\)
\(z\sqrt{\frac{\left(2015+x^2\right)\left(2015+y^2\right)}{2015+z^2}}=zx+zy\)
Ta có :\(P=xy+xz+yx+yz+zx+zy=2\left(xy+yz+zx\right)=4030\)
=>P không phải là số chính phương
CM:a)\(2\left(\sqrt{a}-\sqrt{b}\right)< \frac{1}{\sqrt{b}}< 2\left(\sqrt{a}-\sqrt{b}\right)biet:a=b+1=c+2\left(c>0\right).\)
b)\(CM:B=\sqrt{1+2014^2+\frac{2014^2}{2015^2}}+\frac{2014}{2015}nguyen\)
b, Ta có \(2015^2=\left(2014+1\right)^2=2014^2+2.2014+1\)
=> \(2014^2+1=2015^2-2.2014\)
=> \(B=\sqrt{1+2014^2+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\)
= \(\sqrt{2015^2-2.2014+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\)
= \(\sqrt{\left(2015-\frac{2014}{2015}\right)^2}+\frac{2014}{2015}\) = \(2015-\frac{2014}{2015}+\frac{2014}{2015}=2015\)
=> đpcm
Cho a,b>0 thoa mãn ab>2015+2016b. CMR: \(a+b>\left(\sqrt{2015}+\sqrt{2016}\right)^2\)
sửa lại tí nha
Cho a,b>0 thoa mãn ab>2015a+2016b. CMR: \(a+b>\left(\sqrt{2015}+\sqrt{2016}\right)^2\)
Cho a,b,c>0 thỏa mãn \(a+b+c+2\sqrt{abc}=1\)Chứng minh biểu thức
A=\(\sqrt{a\left(1-b\right)\left(1-c\right)}+\sqrt{b\left(1-c\right)\left(1-a\right)}+\sqrt{c\left(1-a\right)\left(1-b\right)}-\sqrt{abc}+2015\)là hằng số
Có: \(a+b+c+2\sqrt{abc}=1\Rightarrow\hept{\begin{cases}a+2\sqrt{abc}=1-b-c\\b+2\sqrt{abc}=1-a-c\\c+2\sqrt{abc}=1-a-b\end{cases}}\)
\(A=\sqrt{a\left(1-b\right)\left(1-c\right)}+\sqrt{b\left(1-c\right)\left(1-a\right)}+\sqrt{c\left(1-a\right)\left(1-b\right)}-\sqrt{abc}+2015\)
\(A=\sqrt{a\left(1-b-c+bc\right)}+\sqrt{b\left(1-a-c+ac\right)}+\sqrt{c\left(1-a-b+ab\right)}-\sqrt{abc}+2015\)
\(A=\sqrt{a\left(a+2\sqrt{abc}+bc\right)}+\sqrt{b\left(b+2\sqrt{abc}+ac\right)}+\sqrt{c\left(c+2\sqrt{abc}+ab\right)}-\sqrt{abc}+2015\)
\(A=\sqrt{\left(a^2+2a\sqrt{abc}+abc\right)}+\sqrt{\left(b^2+2b\sqrt{abc}+abc\right)}+\sqrt{\left(c^2+2c\sqrt{abc}+abc\right)}-\sqrt{abc}+2015\)
\(A=\sqrt{\left(a+\sqrt{abc}\right)^2}+\sqrt{\left(b+\sqrt{abc}\right)^2}+\sqrt{\left(c+\sqrt{abc}\right)^2}-\sqrt{abc}+2015\)
\(A=a+\sqrt{abc}+b+\sqrt{abc}+c+\sqrt{abc}-\sqrt{abc}+2015\)
\(A=a+b+c+2\sqrt{abc}+2015\)
\(A=1+2015=2016\)
Vậy:....
Cho các số không âm thỏa mãn \(a\ge b\)và \(\sqrt{a-b+c}=\sqrt{a-b}+\sqrt{c}\). CMR:\(\left(\sqrt{a^2c^2+2015}+bc\right)\left(\sqrt{b^2c^2+2015}-ac\right)=2015\)
Giúp tớ nhé~
Cho a, b, c>0. Chứng minh rằng
a2016+b2016+c2016>=\(\frac{\left(b+c\right).a^{2015}}{2}\)+\(\frac{\left(c+a\right).b^{2015}}{2}\)+\(\frac{\left(a+b\right).c^{2015}}{2}\)
Giải hệ phương trình:
\(\left(x+\sqrt{x^2+\sqrt{2015}}\right)\left(y+\sqrt{y+\left(y^2 +\sqrt{2015}\right)}\right)=\sqrt{2015}\)
\(\left(x+\sqrt{x^2+\sqrt{2015}}\right)\left(y+\sqrt{y^2+\sqrt{2015}}\right)=\sqrt{2015}\)
Chứng minh rằng
a, \(2\left(\sqrt{a}-\sqrt{b}\right)< \frac{1}{\sqrt{b}}< 2\left(\sqrt{b}-\sqrt{c}\right)\)\
Biết a,b,c là 3 số thự thỏa mãn điều kiện: a=b+1=c+2 và c>0
b, Biểu thức B=\(\sqrt{1+2014^2+\frac{2014^2}{2015^2}}+\frac{2014}{2015}\)có giá trị là 1 số nguyên
a,a=b+1
suy ra a-b=1 suy ra(\(\sqrt{a}+\sqrt{b}\))(\(\sqrt{a}-\sqrt{b}\))=1
suy ra \(\sqrt{a}-\sqrt{b}\)=\(\frac{1}{\sqrt{a}+\sqrt{b}}\)(1)
vì a=b+1 suy ra a>b suy ra \(\sqrt{a}>\sqrt{b}\)suy ra \(\sqrt{a}+\sqrt{b}>2\sqrt{b}\)
suy ra \(\frac{1}{\sqrt{a}+\sqrt{b}}< \frac{1}{2\sqrt{b}}\)(2)
từ (1) ,(2) suy ra\(\sqrt{a}-\sqrt{b}< \frac{1}{2\sqrt{b}}\)suy ra \(2\left(\sqrt{a}-\sqrt{b}\right)< \frac{1}{\sqrt{b}}\)(*)
ta lại có b+1=c+2 suy ra b-c =1 suy ra\(\left(\sqrt{b}-\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)=1\)
suy ra \(\sqrt{b}-\sqrt{c}=\frac{1}{\sqrt{b}+\sqrt{c}}\)(3)
vì b>c suy ra \(\sqrt{b}>\sqrt{c}\) suy ra \(\sqrt{b}+\sqrt{c}>2\sqrt{c}\)
suy ra \(\frac{1}{\sqrt{b}+\sqrt{c}}< \frac{1}{2\sqrt{c}}\)(4)
Từ (3),(4) suy ra \(\sqrt{b}-\sqrt{c}< \frac{1}{2\sqrt{c}}\) suy ra\(2\left(\sqrt{b}+\sqrt{c}\right)< \frac{1}{\sqrt{c}}\)(**)
từ (*),(**) suy ra đccm
Chứng minh :
\(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=\left|1+\frac{1}{a}-\frac{1}{a+1}\right|\)
Áp dụng tính: \(\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)