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Cho a, b là các số thực thỏa mãn : a + b = 1. Chứng minh: a2 +b2 > hoặc = \(\frac{1}{2}\)
\(gt\Rightarrow\left(a+b\right)^2=1\Leftrightarrow a^2+2ab+b^2=1\) (1)
Do theo BĐT AM-GM (Cô si) \(a^2+b^2\ge2\left|ab\right|\ge2ab\)
Thay vào (1) suy ra \(1=a^2+2ab+b^2\ge4ab\)
Suy ra \(ab\le\frac{1}{4}\).Từ đây ta có: \(a^2+b^2=\left(a+b\right)^2-2ab=1-2ab\ge\frac{1}{2}^{\left(đpcm\right)}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a^2=b^2\\a+b=1\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\a+b=1\end{cases}}\Leftrightarrow a=b=\frac{1}{2}\)
Phép chứng minh hoàn tất!
nhìn kinh vậy thôi dẽ mà @quế anh
2)
\(M=a^2+b^2+c^2-ab-ac-bc\) \(a\ne b\ne c\Rightarrow M\ne0\)
\(T=a^3+b^3+c^3-3abc=\left(a+b+c\right).M\)
\(A=\dfrac{T}{M}=\dfrac{\left(a+b+c\right).M}{M}=\left(a+b+c\right)=2016\)
1)
\(P=\left(4a^2+b^2+9+4ab-12a-6b\right)+3\left(b^2-2b+1\right)\)
\(P=\left(2a+b-3\right)^2+3\left(b-1\right)^2\ge0\)
DS: Pmin=0 ; tại b=1, a=1
1a)\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
b)\(\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{9}\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
2a)\(a^2+\dfrac{b^2}{4}\ge ab\)
\(\Leftrightarrow a^2-ab+\dfrac{b^2}{4}\ge0\)
\(\Leftrightarrow a^2-2\cdot\dfrac{1}{2}b\cdot a+\left(\dfrac{1}{2}b\right)^2\ge0\)
\(\Leftrightarrow\left(a-\dfrac{1}{2}b\right)^2\ge0\)(luôn đúng)
b)Đã cm
c)\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)
Dấu bằng xảy ra khi a=b=1
a ) \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+2.0=0\)
\(\Leftrightarrow a^2+b^2+c^2=0\)
Do \(a^2\ge0;b^2\ge0;c^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=0\) ( * )
Thay * vào biểu thức M , ta được :
\(M=\left(0-1\right)^{1999}+0^{2000}+\left(0+1\right)^{2001}\)
\(=-1^{1999}+0+1^{2001}\)
\(=-1+0+1\)
\(=0\)
Vậy \(M=0\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc+ac+ab-1}{abc}=0\)
\(\Leftrightarrow bc+ac+ab-1=0\)
\(\Leftrightarrow bc+ac+ab=1\)
Mà \(a^2+b^2+c^2=1\)
\(\Rightarrow bc+ac+ab=a^2+b^2+c^2\)
\(\Rightarrow2bc+2ac+2ab=2a^2+2b^2+2c^2\)
\(\Rightarrow2a^2+2b^2+2c^2-2bc-2ac-2ab=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Mà \(P=\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}\)
\(\Rightarrow P=\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\)
\(\Rightarrow P=1+1+1=3\)
Vậy \(P=3\)
+ \(2a^2+a=3b^2+b\)
\(\Rightarrow3a^2-3b^2+a-b=a^2\)
\(\Rightarrow3\left(a-b\right)\left(a+b\right)+\left(a-b\right)=a^2\)
\(\Rightarrow\left(a-b\right)\left(3a+3b+1\right)=a^2\) (*)
+ Gọi \(d=\left(a-b;3a+3b+1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a-b⋮d\\3a+3b+1⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-3b⋮d\\3a+3b+1⋮d\end{matrix}\right.\)
\(\Rightarrow3a+3b+1+3a-3b⋮d\)
\(\Rightarrow6a+1⋮d\) (1)
+ \(\left\{{}\begin{matrix}a-b⋮d\\3a+3b+1⋮d\end{matrix}\right.\)
\(\Rightarrow\left(a-b\right)\left(3a+3b+1\right)⋮d^2\)
\(\Rightarrow a^2⋮d^2\Rightarrow a⋮d\Rightarrow6a⋮d\) (2)
+ Từ (1) và (2) \(\Rightarrow1⋮d\Rightarrow d=1\)
=> a - b và 3a + 3b + 1 là 2 số nguyên tố cùng nhau (**)
+ Từ (*) và (**) => đpcm
P/s : nếu tích 2 số nguyên tố cùng nhau là số cp thì mỗi số đều là số chính phương
1,https://diendantoanhoc.net/topic/157361-t%C3%ACm-c%C3%A1c-s%E1%BB%91-nguy%C3%AAn-x-y-tho%E1%BA%A3-m%C3%A3n-x3y32016/
1) \(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left[\left(b^3-c^3\right)+\left(a^3-b^3\right)\right]+c\left(a^3-b^3\right)\)
\(\left(do\left[\left(b^3-c^3\right)+\left(a^3-b^3\right)\right]=-\left(c^3-a^3\right)\right)\)
\(=\left(a-b\right)\left(b^3-c^3\right)+\left(c-b\right)\left(a^3-b^3\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(b^2+bc+c^2\right)-\left(a^2+ab+b^2\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(c^2-a^2\right)+\left(bc-ab\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
2) \(\dfrac{a-b}{b+c}+\dfrac{b-a}{c+a}+\dfrac{c-b}{a+b}=1\)
\(\Rightarrow\dfrac{a-c}{b+c}+1+\dfrac{b-a}{c+a}+1+\dfrac{c-b}{a+b}+1=4\)
\(\Rightarrow\dfrac{a-c+b+c}{b+c}+\dfrac{b-a+c+a}{c+a}+\dfrac{c-b+a+b}{a+b}=4\)
\(\Rightarrow\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}=4\)
đăng từng câu 1 thôi, nhiều nhất là 3 câu/ 1 lần hỏi vì đâu có giới hạn số lần hỏi
\(ab=x;bc=y;ac=z\)
\(\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left[\left(x+y\right)+z\right]\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=0\\x=y=z\end{matrix}\right.\)
Tự full nhé?
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{1+1+1}=\dfrac{\left(\dfrac{3}{2}\right)^2}{3}=\dfrac{9}{\dfrac{4}{3}}=\dfrac{9}{12}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi: \(a=b=c=\dfrac{1}{2}\)
có cách khác ko bn ?