\(\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1}{\left(x-3\right)\left(2x-1\right)}=\frac{2x+5}{\left(x-3\right)\left(2x-1\right)}\)

\(\frac{\left(x-3\right)\left(x+4\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\frac{\left(x+1\right)\left(x-2\right)}{\left(x-3\right)\left(2x-1\right)\left(x-2\right)}=\frac{\left(2x+5\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}\)

\(\Rightarrow x^2+x-12+x^2-x-2=2x^2+x-10\Leftrightarrow x=-4\)

\(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)

\(\Rightarrow\frac{x+4}{2x^2-5x+2}=\frac{2x-5}{2x^2-7x+3}-\frac{x+1}{2x^2-7x+3}\)

\(\Rightarrow\frac{x+4}{2x^2-5x+2}=\frac{x+4}{2x^2-7x+3}\)

TH1:\(x+4\ne0\)

\(\Rightarrow2x^2-5x+2=2x^2-7x+3\)

\(\Rightarrow-5x+2=-7x+3\)

\(\Rightarrow2x=1\)

\(\Rightarrow x=\frac{1}{2}\)

TH2:\(x+4=0\)

\(\Rightarrow x=-4\)

Ta có:\(\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+y^2}\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+y^2\right)< \left(x+y\right)\left(x^2-y^2\right)\)

\(\Leftrightarrow x^3+xy^2-yx^2-y^3< x^3+x^2y-y^2x-y^3\)

\(\Leftrightarrow xy^2-yx^2< x^2y-y^2x\)

\(\Rightarrow2xy^2< 2yx^2\)

\(\Rightarrow y< x\)(luôn đúng)

Vậy \(\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+y^2}\)

\(\frac{x^2-y^2}{x^2+y^2}>\frac{x-y}{x+y}\)

\(\Leftrightarrow\frac{x^2-y^2}{x^2+y^2}-\frac{x-y}{x+y}>0\)

\(\Leftrightarrow\frac{\left(x^2-y^2\right)\left(x+y\right)-\left(x-y\right)\left(x^2+y^2\right)}{\left(x^2+y^2\right)\left(x+y\right)}>0\)

\(\Leftrightarrow\frac{\left(x-y\right)\left(x+y\right)^2-\left(x-y\right)\left(x^2+y^2\right)}{\left(x^2+y^2\right)\left(x+y\right)}>0\)

\(\Leftrightarrow\frac{\left(x-y\right)\left(x^2+y^2+2xy-x^2-y^2\right)}{\left(x^2+y^2\right)\left(x+y\right)}>0\)

\(\Leftrightarrow\frac{\left(x-y\right)2xy}{\left(x^2+y^2\right)\left(x+y\right)}>0\)( luôn đúng vì x>y>0)

\(\Rightarrow\frac{x^2-y^2}{x^2+y^2}>\frac{x-y}{x+y}\)

đpcm

a) \(x^4+2019x^2+2018x+2019\)

\(=\left(x^4-x\right)+\left(2019x^2+2019x+2019\right)\)

\(=x\left(x^3-1\right)+2019\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2019\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+2019\right]\)

\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)

b) \(E=2x^2-8x+1=2x^2-8x+8-7\)

\(=2\left(x^2-4x+4\right)-7=2\left(x-2\right)^2-7\)

Vì \(2\left(x-2\right)^2\ge0\forall x\Rightarrow E\ge-7\)

Dấu "=" xảy ra <=> \(2\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy MinE = -7 <=> x = 2

b) \(E=2x^2-8x+1\)

\(E=2\left(x^2-4x+\frac{1}{2}\right)\)

\(E=2\left(x^2-2\cdot x\cdot2+2^2+\frac{7}{2}\right)\)

\(E=2\left[\left(x-2\right)^2+\frac{7}{2}\right]\)

\(E=2\left(x-2\right)^2+7\ge7\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy....

mỗi bn 3 k cho người giải dc

1, bn c/m tam giác AEH đồng dạng vs tam giác ABO (g-g)

=>\(\frac{AE}{AB}=\frac{AH}{AO}\Rightarrow AE.AO=AB.AH\)

=> AO.AE +AB.BH=AB.AH+AB.BH

                              =AB.(AH+BH)=AB.AB=AB²(ĐPCM)

2,

 c/m tam giác CAH đòng dạng vs Tam giác BAH (G-G)( có 2 góc=nhau=90^o, góc HCA=BCH)

=> \(\frac{AH}{HC}=\frac{CH}{BH}\Rightarrow AH.BH=CH^2\)

T² bn c/m tam giác CAH đòng dạng vs tam giác BAC(g-g)

=> \(\frac{AB}{AC}=\frac{AC}{AH}\Rightarrow AB.AH=AC^2\)

Mà AB.AH=AO.AE (theo phần 1)

=> AO.AE=AC²

Xét tam giác CAH vuông tại H theo định lí Py-ta-go ta có

\(AH^2+CH^2=AC^2\Rightarrow AH^2=AC^2-CH^2\)

=> AH²=AO.AE - HA.HB (đpcm)

tk cho mk nhé

*****Chúc bạn học gỏi*****

Ta có pt

\(\frac{43}{x}+\frac{40}{60}+\frac{120}{1,2x}=\frac{163}{x}\)

\(\frac{43}{x}+\frac{100}{x}-\frac{163}{x}=\frac{-2}{3}\Rightarrow\frac{-20}{x}=\frac{-2}{3}\Rightarrow x=30\)

Vậy vận tốc lúc đầu là 30km/h