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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 \(P=3\)
\(e,\)
\(\left(\dfrac{1}{3}a^3b+\dfrac{1}{3}a^2b^2-\dfrac{1}{4}ab^3\right):5ab\)
\(=\dfrac{1}{15}a^2+\dfrac{1}{15}ab-\dfrac{1}{20}b^2\)
\(f,\)
\(\left(-\dfrac{2}{3}x^5y^2+\dfrac{3}{4}x^4y^3-\dfrac{4}{5}x^3y^4\right):6x^2y^2\)
\(=-\dfrac{1}{9}x^3+\dfrac{1}{8}x^2y-\dfrac{2}{15}xy^2\)
\(g,\)
\(\left(\dfrac{3}{4}a^6b^3+\dfrac{6}{5}a^3b^4-\dfrac{5}{10}ab^5\right):\left(\dfrac{3}{5}ab^3\right)\)
\(=\dfrac{5}{4}a^5+2a^2b-\dfrac{5}{6}b^2\)
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
\(ab=x;bc=y;ac=z\)
\(\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left[\left(x+y\right)+z\right]\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=0\\x=y=z\end{matrix}\right.\)
Tự full nhé?
a)
\(a^2+b^2+c^2+d^2+m^2-a(b+c+d+m)\)
\(=\frac{4a^2+4b^2+4c^2+4d^2+4m^2-4a(b+c+d+m)}{4}\)
\(=\frac{(a^2+4b^2-4ab)+(a^2+4c^2-4ac)+(a^2+4d^2-4ad)+(a^2+4m^2-4am)}{4}\)
\(=\frac{(a-2b)^2+(a-2c)^2+(a-2d)^2+(a-2m)^2}{4}\geq 0\) (đpcm)
Dấu "=" xảy ra khi \(a=2b=2c=2d=2m\)
b)
Xét hiệu
\(\frac{1}{x}+\frac{1}{y}-\frac{4}{x+y}=\frac{x+y}{xy}-\frac{4}{x+y}=\frac{(x+y)^2-4xy}{xy(x+y)}\)
\(=\frac{x^2+y^2-2xy}{xy(x+y)}=\frac{(x-y)^2}{xy(x+y)}\geq 0, \forall x,y>0\)
\(\Rightarrow \frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}\) (đpcm)
Dấu "=" xảy ra khi $x=y$
c)
Xét hiệu:
\((a^2+c^2)(b^2+d^2)-(ab+cd)^2\)
\(=(a^2b^2+a^2d^2+c^2b^2+c^2d^2)-(a^2b^2+2abcd+c^2d^2)\)
\(=a^2d^2-2abcd+b^2c^2=(ad-bc)^2\geq 0\)
\(\Rightarrow (a^2+c^2)(b^2+d^2)\geq (ab+cd)^2\) (đpcm)
Dấu "=" xảy ra khi \(ad=bc\)
d)
Xét hiệu:
\(a^2+b^2-(a+b-\frac{1}{2})=a^2+b^2-a-b+\frac{1}{2}\)
\(=(a^2-a+\frac{1}{4})+(b^2-b+\frac{1}{4})\)
\(=(a-\frac{1}{2})^2+(b-\frac{1}{2})^2\geq 0\)
\(\Rightarrow a^2+b^2\geq a+b-\frac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
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
Đặt \(ab=x;bc=y;ca=z\) thì có \(x^3+y^3+z^3=3xyz\) dễ nhé
Bài 1:
a^2-5ab-6b^2=0
=>a^2-6ab+ab-6b^2=0
=>a*(a-6b)+b(a-6b)=0
=>(a-6b)(a+b)=0
=>a=-b hoặc a=6b
TH1: a=-b
\(A=\dfrac{-2b-b}{-3b-b}+\dfrac{5b+b}{-3b+b}=\dfrac{-3}{-4}+\dfrac{6}{-2}=\dfrac{3}{4}-3=-\dfrac{9}{4}\)
TH2: a=6b
\(A=\dfrac{12b-b}{18b-b}+\dfrac{5b-6b}{18b+b}=\dfrac{11}{17}+\dfrac{-1}{19}=\dfrac{192}{323}\)
Lời giải:
\((a^2+b^2)^3=(a^3+b^3)^2\)
\(\Leftrightarrow a^6+3a^4b^2+3a^2b^4+b^6=a^6+2a^3b^3+b^6\)
\(\Rightarrow 3a^2b^2(a^2+b^2)=2a^3b^3\)
\(\Rightarrow 3(a^2+b^2)=2ab\) (chia cả 2 vế cho \(a^2b^2\neq 0\) )
\(\Rightarrow 2(a^2+b^2)=-(a^2+b^2)+2ab=-(a-b)^2(*)\)
Với mọi \(a,b\neq 0\) thì \(2(a^2+b^2)>0; -(a-b)^2\leq 0\), do đó $(*)$ vô lý
Vậy không tồn tại $a,b$ thỏa mãn đkđb, kéo theo không tìm được giá trị của $A$