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1/a +1/b +1/c = 0=> 1/a +1/b= 1/-c
=> 1/(a+ b)^3 = 1/(-c)^3
=> 1/ a^3+ 3a^2b+ 3ab^2+ b^3 = 1/-c^3
=> 1/a+ 1/b^3+ 1/c^3= 3/ -a^2b- ab^2
= -3/ ab(-c)= 3/abc
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
=>\(\dfrac{bc+ac+ab}{abc}=\dfrac{1}{a+b+c}\)
=> (bc+ac+ab)(a+b+c)=abc
=> abc+b2c+bc2+a2c+abc+ac2+a2b+ab2+abc=abc
=>abc+b2c+bc2+a2c+abc+ac2+a2c+ab2+abc-abc=0
=>(a2c+2abc+b2c)+(a2b+ab2)+(ac2+bc2)=0
=>c(a+b)2+ab(a+b)+c2(a+b)=0
=>(a+b)[c(a+b)+ab+c2]=0
=>(a+b)(ac+bc+ab+c2)=0
=>(a+b)[a(c+b)+c(b+c)]=0
=>(a+b)(c+b)(a+c)=0
=> a+b=0, c+b=0, a+c=0
nếu a+b=0=>a=-b
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{1}{-b^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{1}{c^3}\)(1)
và \(\dfrac{1}{a^3+b^3+c^3}=\dfrac{1}{-b^3+b^3+c^3}=\dfrac{1}{c^3}\) (2)
từ (1) và (2) suy ra đpcm
Bài 3:
a) Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{xy}+\frac{2}{x^2+y^2}=2\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\) \(\geq 2.\frac{(1+1)^2}{2xy+x^2+y^2}=\frac{8}{(x+y)^2}=8\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
b) Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{xy}+\frac{1}{x^2+y^2}=\frac{1}{2xy}+\left (\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\geq \frac{1}{2xy}+\frac{(1+1)^2}{2xy+x^2+y^2}\)
\(=\frac{1}{2xy}+\frac{4}{(x+y)^2}\)
Theo BĐT AM-GM:
\(xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}\Rightarrow \frac{1}{2xy}\geq 2\)
Do đó \(\frac{1}{xy}+\frac{1}{x^2+y^2}\geq 2+4=6\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
Bài 1: Thiếu đề.
Bài 2: Sai đề, thử với \(x=\frac{1}{6}\)
Bài 4 a) Sai đề với \(x<0\)
b) Áp dụng BĐT AM-GM:
\(x^4-x+\frac{1}{2}=\left (x^4+\frac{1}{4}\right)-x+\frac{1}{4}\geq x^2-x+\frac{1}{4}=(x-\frac{1}{2})^2\geq 0\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} x^4=\frac{1}{4}\\ x=\frac{1}{2}\end{matrix}\right.\) (vô lý)
Do đó dấu bằng không xảy ra , nên \(x^4-x+\frac{1}{2}>0\)
Bài 6: Áp dụng BĐT AM-GM cho $6$ số:
\(a^2+b^2+c^2+d^2+ab+cd\geq 6\sqrt[6]{a^3b^3c^3d^3}=6\)
Do đó ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=d=1\)
5) a) Đặt b+c-a=x;a+c-b=y;a+b-c=z thì 2a=y+z;2b=x+z;2c=x+y
Ta có:
\(\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\)
Vậy ta suy ra đpcm
b) Ta có: a+b>c;b+c>a;a+c>b
Xét: \(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)
.Tương tự:
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c};\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+c}\)
Vậy ta có đpcm
6) Ta có:
\(a^2+b^2+c^2+d^2+ab+cd\ge2ab+2cd+ab+cd=3\left(ab+cd\right)\)
\(ab+cd=ab+\dfrac{1}{ab}\ge2\)
Suy ra đpcm
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=3^2\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=9\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=9\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{c}{abc}+\dfrac{a}{abc}+\dfrac{b}{abc}\right)=9\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\dfrac{a+b+c}{abc}=9\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2.3abc}{abc}=9\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+6=9\)
\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\)
Không biết đề có sai không nhỉ? 
5. phân tích ra : \(1+\dfrac{a}{b}+\dfrac{b}{a}+1\)
áp dụng bđ cosy
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
=> đpcm
6. \(x^2-x+1=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
hay với mọi x thuộc R đều là nghiệm của bpt
7.áp dụng bđt cosy
\(a^4+b^4+c^4+d^4\ge2\sqrt{a^2.b^2.c^2.d^2}=4abcd\left(đpcm\right)\)
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
<=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
<=> \(2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=0\)
<=> \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=0\)
<=>\(\dfrac{a+b+c}{abc}=0\)
=> a+b+c =0
=> a3+b3+c3=3abc ⋮3 (bn tự cm)
=> a3+b3+c3 ⋮3 (đpcm)
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
ta có:\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
=>\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\)
=>\(\dfrac{a+b}{ab}+\dfrac{a+b}{ac+bc+c^2}=0\)
=>\(\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{ac+bc+c^2}\right)=0\)
=>\(\left(a+b\right)\left[\dfrac{ac+bc+c^2+ab}{ab\left(ac+bc+c^2\right)}\right]=0\)
=>\(\left(a+b\right).\dfrac{c\left(a+c\right)+b\left(a+c\right)}{ab\left(ac+bc+c^2\right)}=0\)
=>\(\left(a+b\right).\dfrac{\left(c+b\right)\left(a+c\right)}{ab\left(ac+bc+c^2\right)}=0\)
=>(a+b)(a+c)(c+b)=0
=>a+b=0<=>a=-b
a+c=0<=>a=-c
c+b=0<=>c=-b
thay a=-b ta đc \(\dfrac{1}{\left(-b\right)^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{1}{c^3}\left(1\right)\)
\(\dfrac{1}{a^3+b^3+c^3}=\dfrac{1}{-b^3+b^3+c^3}=\dfrac{1}{c^3}\left(2\right)\)
=>(1)=(2),các trường hợp còn lài cm tương tự
vậy.............
Lời giải:
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow \frac{a+b}{ab}+\frac{a+b}{c(a+b+c)}=0\)
\(\Leftrightarrow (a+b)\left(\frac{1}{ab}+\frac{1}{c(a+b+c)}\right)=0\)
\(\Leftrightarrow (a+b).\frac{c(a+b+c)+ab}{abc(a+b+c)}=0\Leftrightarrow \frac{(a+b)(b+c)(c+a)}{abc(a+b+c)}=0\)
\(\Leftrightarrow (a+b)(b+c)(c+a)=0\)
Khi đó:
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3+b^3+c^3}\)\(=\frac{a^3+b^3}{a^3b^3}+\frac{a^3+b^3}{c^3(a^3+b^3+c^3)}\)
\(=(a^3+b^3)\left(\frac{1}{a^3b^3}+\frac{1}{c^3(a^3+b^3+c^3)}\right)=(a^3+b^3).\frac{a^3b^3+c^3(a^3+b^3+c^3)}{a^3b^3c^3(a^3+b^3+c^3)}\)
\(=\frac{(a^3+b^3)(b^3+c^3)(c^3+a^3)}{a^3b^3c^3(a^3+b^3+c^3)}=\frac{(a+b)(b+c)(c+a)(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)}{a^3b^3c^3(a^3+b^3+c^3)}\)
\(=0\) do \((a+b)(b+c)(c+a)=0\)
Suy ra \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3+b^3+c^3}\)