Cho a+b+c=0
Tính GTBT:\(B=\frac{\text{a}b}{\text{a}^2+b^2-c^2}+\frac{bc}{b^2+c^2-\text{a}^2}+\frac{c\text{a}}{c^2+\text{a}^2-b^2}\)
Cho a+b+c=0
Tính GTBT:B=\(\frac{\text{ab}}{\text{a}^2+b^2-c^2}+\frac{bc}{b^2+c^2-\text{a}^2}+\frac{c\text{a}}{c^2+\text{a}^2-b^2}\)
Ta có a+b+c=0
<=> a+b=-c <=>a2+b2-c2=-2ab
b+c=-a <=> b2+c2-a2=-2bc
c+a=-b <=> c2+a2-b2=-2ca
Thay vào biểu thức ta có
\(B=\frac{ab}{-2ab}-\frac{bc}{2bc}-\frac{ca}{2ca}=\frac{-3}{2}\)
Cho 3 sô dương a,b,c . Chứng mình rằng
\(\sqrt[3]{\frac{\left(a\text{+}b\right)\left(b\text{+}c\right)\left(c\text{+}a\right)}{abc}}\ge\frac{4}{3}\left(\frac{a^2}{a^2\text{+}bc}\frac{b^2}{b^2\text{+}ab}\frac{c^2}{c^2\text{+}ac}\right)\)
Mấy bạn giúp mình câu này với ;-;
Trong tam giác ABC.Chứng minh rằng:
\(\frac{b^2-c^2}{c\text{os}B+c\text{os}C}\)+\(\frac{c^2-a^2}{c\text{os}C+c\text{os}A}\)+\(\frac{a^2-b^2}{c\text{os}A+c\text{os}B}\)=0
bài này khó quá chắc mình không giải được rồi
Cho các số thực không âm a,b,ca,b,c thoả mãn a+b+c=1a+b+c=1. Chứng minh rằng :
\(\sqrt{a+\frac{\left(b-c\right)^2}{4}}+\sqrt{b+\frac{\left(c-a\right)^2}{4}}+\sqrt{c+\frac{\left(a-b\right)^2}{4}}\le\sqrt{3}+\left(1-\frac{\sqrt{3}}{2}\right)\left(\text{|
}a-b\text{|
}\right)+\text{|
}b-c\text{|
}+\text{|
}c-a\text{|
}.\)
Cho C=\(\text{}\text{}\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\left(a>0,b>0,c>0\right)\)và D=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}\)
Chứng minh C>D
\(C=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(D< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(\Rightarrow D< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow D< 1-\frac{1}{2017}< 1\)
Vậy C > D
cho a,b,c,d la các số thực dương co tong bang 1. Cmr
\(\frac{\text{a}^2}{\text{a}+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+\text{a}}\ge\frac{1}{2}\)
Cách 1. Áp dụng BĐT AM-GM :
\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\ge\frac{\left(a+b+c+d\right)^2}{2\left(a+b+c+d\right)}\)
\(\Rightarrow\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\ge\frac{a+b+c+d}{2}=\frac{1}{2}\)
Cách 2. Áp dụng BĐT Cauchy : \(\frac{a^2}{a+b}+\frac{a+b}{4}\ge2\sqrt{\frac{a^2}{a+b}.\frac{a+b}{4}}=a\)
Tương tự : \(\frac{b^2}{b+c}+\frac{b+c}{4}\ge b\) , \(\frac{c^2}{c+d}+\frac{c+d}{4}\ge c\), \(\frac{d^2}{d+a}+\frac{d+a}{4}\ge d\)
Cộng theo vế : \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}+\frac{1}{4}.2.\left(a+b+c+d\right)\ge a+b+c+d\)
\(\Leftrightarrow\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\ge\frac{a+b+c+d}{2}=\frac{1}{2}\)
\(\text{Cho }a,b,c>0\text{ thỏa mãn }a+b+c=3\)
\(\text{CMR: }\frac{1+b}{1+4a^2}+\frac{1+c}{1+4b^2}+\frac{1+a}{1+4c^2}\ge\frac{6}{5}\)
1.Cho ab/b = bc/c=ca/a. Tính A= (a-b)(b-c)(c-a) + 2016
2. Cho (ab + bc)/ ( a+b) = ( bc + ca )/(b+c)= ( ca + ab) / (c+a)
Tính M=\(\left(\frac{b}{a}+1\right)\left(\frac{c}{b}+1\right)\left(\frac{a}{c}+1\right)+2016\)
3. Cho a+b+c+d khác 0 và \(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)
Tìm giá trị của A=\(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
\(1)\)\(\frac{\overline{ab}}{b}=\frac{\overline{bc}}{c}=\frac{\overline{ca}}{a}\)
\(\Leftrightarrow\)\(\frac{10a+b}{b}=\frac{10b+c}{c}=\frac{10c+a}{a}\)
\(\Leftrightarrow\)\(\frac{10a}{b}=\frac{10b}{c}=\frac{10c}{a}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{10a}{b}=\frac{10b}{c}=\frac{10c}{a}=\frac{10a+10b+10c}{a+b+c}=\frac{10\left(a+b+c\right)}{a+b+c}=10\)
Do đó :
\(\frac{10a}{b}=10\)\(\Leftrightarrow\)\(a=b\)
\(\frac{10b}{c}=10\)\(\Leftrightarrow\)\(b=c\)
\(\frac{10c}{a}=10\)\(\Leftrightarrow\)\(c=a\)
\(\Rightarrow\)\(a=b=c\)
\(\Rightarrow\)\(A=\left(a-b\right)\left(b-c\right)\left(c-a\right)+2016=2016\)
\(2)\)\(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{2\left(\overline{ab}+\overline{bc}+\overline{ca}\right)}{2\left(a+b+c\right)}=\frac{\overline{ab}+\overline{bc}+\overline{ca}}{a+b+c}\)
\(=\frac{10a+b+10b+c+10c+a}{a+b+c}=\frac{11a+11b+11c}{a+b+c}=\frac{11\left(a+b+c\right)}{a+b+c}=11\)
Do đó :
\(\frac{\overline{ab}+\overline{bc}}{a+b}=11\)\(\Leftrightarrow\)\(10a+11b+c=11a+11b\)\(\Leftrightarrow\)\(c=a\)
\(\frac{\overline{bc}+\overline{ca}}{b+c}=11\)\(\Leftrightarrow\)\(10b+11c+a=11b+11c\)\(\Leftrightarrow\)\(a=b\)
\(\frac{\overline{ca}+\overline{ab}}{c+a}=11\)\(\Leftrightarrow\)\(10c+11a+b=11c+11a\)\(\Leftrightarrow\)\(b=c\)
\(\Rightarrow\)\(a=b=c\)
\(\Rightarrow\)\(M=\left(\frac{b}{a}+1\right)\left(\frac{c}{b}+1\right)\left(\frac{a}{c}+1\right)+2016=2.2.2+2016=2024\)
Chúc bạn học tốt ~
Ta có: \(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)
\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)
hay \(\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)
Do các tử số trên bằng nhau nên các mẫu số cũng bằng nhau hay \(b+c+d=a+c+d=a+b+d=a+b+c\)
Suy ra a = b =c =d
\(\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
\(3)\)\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)
\(\Leftrightarrow\)\(\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)
\(\Leftrightarrow\)\(\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)
Vì các tử bằng nhau nên mẫu cũng bằng nhau :
+) Với \(b+c+d=a+c+d\)\(\Leftrightarrow\)\(a=b\)
+) Với \(a+b+d=a+b+c\)\(\Leftrightarrow\)\(c=d\)
+) Với \(a+c+d=a+b+d\)\(\Leftrightarrow\)\(b=c\)
\(\Rightarrow\)\(a=b=c=d\)
\(\Rightarrow\)\(A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
Chúc bạn học tốt ~
\(\text{Cho }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
\(\text{Tìm GTLN của }P=\frac{1}{\sqrt{a^2-ab+b^2}}+\frac{1}{\sqrt{b^2-bc+c^2}}+\frac{1}{\sqrt{c^2-ca+a^2}}\)
\(P=\frac{1}{\sqrt{\frac{1}{2}\left(a-b\right)^2+\frac{1}{2}\left(a^2+b^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(b-c\right)^2+\frac{1}{2}\left(b^2+c^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(c-a\right)^2+\frac{1}{2}\left(c^2+a^2\right)}}\)
\(\Rightarrow P\le\frac{1}{\sqrt{\frac{1}{2}\left(a^2+b^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(b^2+c^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(c^2+a^2\right)}}\)
\(\Rightarrow P\le\frac{1}{\sqrt{\frac{1}{4}\left(a+b\right)^2}}+\frac{1}{\sqrt{\frac{1}{4}\left(b+c\right)^2}}+\frac{1}{\sqrt{\frac{1}{4}\left(c+a\right)^2}}\)
\(\Rightarrow P\le\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\)
\(\Rightarrow P\le\frac{2}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
\(\Rightarrow P_{max}=3\) khi \(a=b=c=1\)
Không có điều kiện a;b;c dương thì ko biết giải kiểu gì đâu bạn