Các câu hỏi tương tự
Cho 3 số thực a,b,c thỏa mãn điều kiện: \(a^2+b^2+c^2-7a-8b-9c+25=0\)
Tính giá trị biểu thức: \(D=\left(a-2\right)^{2014}+\left(b-3\right)^{2015}+\left(c-4\right)^{2016}\)
Cho a,b,c thỏa mãn a2 +b2+c2-7a-8b-9c+25=0
Tính giá trị biểu thức : A= (a-2)2014 + (b-3)2015 + (c-4)2016
Cho ba số a, b, c thỏa mãn
\(\frac{a}{2014}=\frac{b}{2015}=\frac{c}{2016}\)
tính giá trị của biểu thức:
\(M=4\left(a-b\right)\left(b-c\right)-\left(c-a\right)^2\)
Cho \(\left(a+\sqrt{a^2+2016}\right)\left(b+\sqrt{b^2+2016}\right)=1\)
Tính \(\frac{a^3+b^3}{a^{2016}+b^{2015}+2014}\)
Cho \(\left(a+\sqrt{a^2+2016}\right)\left(b+\sqrt{b^2+2016}\right)=1\)
Tính \(\frac{a^3+b^3}{a^{2016}+b^{2015}+2014}\)
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)\(CM:B=\sqrt{1+2014^2+\frac{2014^2}{2015^2}}+\frac{2014}{2015}nguyen\)
30 tháng 9 2021 lúc 22:00
1. Tính
a. \(\left(3\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-3\sqrt{2}\right)\)
b. \(\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}\)
c. \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}\)
d. \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}\)
Pfrac{a}{sqrt{left(b+1right)left(b^2-b+1right)}}+frac{b}{sqrt{left(c+1right)left(c^2-c+1right)}}+frac{c}{sqrt{left(a+1right)left(a^2-a+1right)}}gefrac{2a}{b^2+2}+frac{2b}{c^2+2}+frac{2c}{a^2+2}left(a+b+cright)-left(frac{ab^2}{b^2+2}+frac{bc^2}{c^2+2}+frac{ca^2}{a^2+2}right)6-left(frac{2ab^2}{b^2+4+b^2}+frac{2bc^2}{c^2+4+c^2}+frac{2ca^2}{a^2+4+a^2}right)ge6-left(frac{2ab}{b+4}+frac{2bc}{c+4}+frac{2ca}{a+4}right)6-left(2a+2b+2c-frac{8a}{b+4}-frac{8b}{c+4}-frac{8c}{a+4}right)frac{8a}{b+4}+frac{8b}{...
Đọc tiếp
\(P=\frac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}+\frac{b}{\sqrt{\left(c+1\right)\left(c^2-c+1\right)}}+\frac{c}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\)
\(\ge\frac{2a}{b^2+2}+\frac{2b}{c^2+2}+\frac{2c}{a^2+2}=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+2}+\frac{bc^2}{c^2+2}+\frac{ca^2}{a^2+2}\right)\)
\(=6-\left(\frac{2ab^2}{b^2+4+b^2}+\frac{2bc^2}{c^2+4+c^2}+\frac{2ca^2}{a^2+4+a^2}\right)\ge6-\left(\frac{2ab}{b+4}+\frac{2bc}{c+4}+\frac{2ca}{a+4}\right)\)
\(=6-\left(2a+2b+2c-\frac{8a}{b+4}-\frac{8b}{c+4}-\frac{8c}{a+4}\right)\)
\(=\frac{8a}{b+4}+\frac{8b}{c+4}+\frac{8c}{a+4}-6=\frac{8a^2}{ab+4a}+\frac{8b^2}{bc+4b}+\frac{8c^2}{ca+4c}-6\)
\(\ge\frac{8\left(a+b+c\right)^2}{\left(ab+bc+ca\right)+4\left(a+b+c\right)}-6\ge\frac{288}{\frac{\left(a+b+c\right)^2}{3}+24}-6=2\)
a)tính giá trị biểu thức: \(A=\frac{2.1+1}{\left(1^2+1\right)^2}+\frac{2.2+1}{\left(2^2+2\right)^2}+\frac{2.3+1}{\left(3^2+3\right)^2}+...+\frac{2.2015+1}{\left(2015^2+2015\right)^2}+\frac{2.2016+1}{\left(2016^2+2016\right)^2}\)
b) cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\), tính giá trị biểu thức: \(M=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}\)