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Bài 1a):
Ta có:
\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\left(a+b\right).\dfrac{a+b}{ab}=\dfrac{a^2+2ab+b^2}{ab}=\dfrac{a^2+b^2}{ab}+2\)
Lại có: (a - b)2 = a2 - 2ab + b2 \(\ge\) 0
\(\Rightarrow\) a2 + b2 \(\ge\) 2ab
\(\Rightarrow\) \(\dfrac{a^2+b^2}{ab}\ge2\)
\(\Rightarrow\) \(\dfrac{a^2+b^2}{ab}+2\ge4\)
Vậy \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
Bài 2a):
Ta có: \(\left(\sqrt{a}-\sqrt{b}\right)^2=a-2\sqrt{ab}+b\ge0\)
\(\Rightarrow a+b\ge2\sqrt{ab}\)
Vậy ta có đpcm
Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)
=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)
=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)
=> \(x=\dfrac{-1}{11}\)
Đây toán 8 mà? :v
a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)
\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)
\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)
\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)
\(\Leftrightarrow\left(11+1\right)x=0\)
\(\Leftrightarrow11x+1=0;x=0\)
\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)
Vậy....
a) \(\dfrac{a}{b}+\dfrac{-a}{b+1}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}+\dfrac{-a.b}{b\left(b+1\right)}=\dfrac{ab+a-ab}{b\left(b+1\right)}=\dfrac{a}{b\left(b+1\right)}\)
b) \(\dfrac{a}{b+1}+\dfrac{-a}{b}=\dfrac{ab}{b\left(b+1\right)}+\dfrac{-a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab-ab-a}{b\left(b+1\right)}=\dfrac{-a}{b\left(b+1\right)}\)
TH1:a+b+c=0
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
\(\Rightarrow H=\dfrac{b+a}{b}.\dfrac{c+b}{c}.\dfrac{a+c}{a}=\dfrac{\left(-c\right)\left(-b\right)\left(-a\right)}{b.c.a}=-1\)
TH2:\(a+b+c\ne0\)
Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
\(\Rightarrow H=\dfrac{b+a}{b}.\dfrac{c+b}{c}.\dfrac{a+c}{a}=\dfrac{\left(2c\right)\left(2b\right)\left(2a\right)}{b.c.a}=8\)
Vậy H=-1 hoặc H=8
c)
Ta có \(a< b< c< d< m< n\)
\(\Rightarrow\left\{{}\begin{matrix}a< b\\c< d\\m< n\end{matrix}\right.\)
\(\Rightarrow a+c+m\le b+d+n\)
\(\dfrac{a+c+m}{a+b+c+d+m+n}< \dfrac{1}{2}\)
\(\Leftrightarrow2a+2c+2m< a+b+c+d+m+n\)
\(\Leftrightarrow a+c+m< b+d+n\) ( thỏa mãn đề bài )
\(\Rightarrow\) đpcm
áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=1\)
\(\Rightarrow\dfrac{a+b-c}{c}=1\Leftrightarrow a+b-c=c\Leftrightarrow a+b=2c\)
\(\Rightarrow\dfrac{b+c-a}{a}=1\Leftrightarrow b+c-a=a\Leftrightarrow b+c=2a\)
ta có
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\dfrac{a+b}{a}\times\dfrac{c+a}{c}\times\dfrac{b+c}{b}=\dfrac{2c}{a}\times\dfrac{2b}{c}\times\dfrac{2a}{b}=8\)
\(\Rightarrow M=8\)
bt lm thì lm đi Hung nguyen , mình cx chưa bt làm thế nào, khó vãi
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
=> (a+b).\(\left(\dfrac{1}{b}+\dfrac{1}{b}\right)\ge\left(a+b\right).\dfrac{4}{a+b}=4\left(dpcm\right)\)
b)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+b+c}\)
=>\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right).\dfrac{9}{a+b+c}=9\left(dpcm\right)\)