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a: =>x+2022=0
hay x=-2022
b: \(\Leftrightarrow x-198=0\)
hay x=198
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a: \(\Leftrightarrow\left(x+8-3x+7\right)\left(x+8+3x-7\right)=0\)
=>(-2x+15)(4x+1)=0
=>x=15/2 hoặc x=-1/4
b: \(\Leftrightarrow\left(x-5\right)\left(x+6-x+5\right)=0\)
=>x-5=0
hay x=5
c: \(\Leftrightarrow x^2+15x+56-x^2+6x-9=0\)
=>21x=-47
hay x=-47/21
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(3) \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
\(\Leftrightarrow\) \(\dfrac{a^2+b^2}{ab}\ge2\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\left(luôn đúng\right)\)
(4)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a +b+c}\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\ge9\)
\(\Leftrightarrow1+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{a}{b}+1+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{b}{c}+1\ge9\)
\(\Leftrightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge3+2+2+2\ge9\) (đpcm)
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a: Xét ΔABM và ΔADM có
AB=AD
\(\widehat{BAM}=\widehat{DAM}\)
AM chung
Do đó: ΔABM=ΔADM
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a: ABCD là hình thang có MN//AB
nên AM/MD=BN/NC
=>AM/4=BN/1=6/5
=>AM=4,8cm
b: ABCD là hình thag có MN//AB//CD
nên BN/NC=AM/MD
=>4/2=AM/3
=>AM=6cm
=>AD=9cm
c; BN/NC=AM/MD=1
=>BN=5cm
Bài 2:
Gọi tử là x
Mẫu là x+6
Theo đề, ta có:
\(\dfrac{x+3}{x+5}=\dfrac{4}{5}\)
=>5x+15=4x+20
=>x=5