Ta có:

\(A=\frac{6x+8}{x^2+1}=\frac{\left(x^2+6x+9\right)-\left(x^2+1\right)}{x^2+1}=\frac{\left(x+3\right)^2}{\left(x+1\right)^2}-1\)

ta có:  \(\frac{\left(x+3\right)^2}{\left(x+1\right)^2}=\left(\frac{x+3}{x+1}\right)^2\ge0\forall x\ne-1\)

\(\Rightarrow A=\frac{\left(x+3\right)^2}{\left(x+1\right)^2}-1\ge-1\forall x\ne-1\)

dấu bằng xảy ra \(\Leftrightarrow\left(x+3\right)^2=0\Leftrightarrow x=-3\)

vậy \(MinA=-1\Leftrightarrow x=-3\)

Mình khuyên bạn nên làm cách thứ 2 (cách thứ nhất là của lớp 9)

Cách 1: Nhận thấy \(x^2+1\ne0\)nên ta có \(A=\frac{6x+8}{x^2+1}\)\(\Leftrightarrow A\left(x^2+1\right)=6x+8\)\(\Leftrightarrow Ax^2-6x+A-8=0\)(*)

Ta có \(\Delta'=\left(-3\right)^2-A\left(A-8\right)=9-A^2+8A\)

Để pt (*) có nghiệm thì \(\Delta'\ge0\)hay \(9-A^2+8A\ge0\)\(\Leftrightarrow A^2-8A-9\le0\)\(\Leftrightarrow A^2+A-9A-9\le0\)\(\Leftrightarrow A\left(A+1\right)-9\left(A+1\right)\le0\)\(\Leftrightarrow\left(A+1\right)\left(A-9\right)\le0\)

Ta xét 2 trường hợp:

TH1: \(\hept{\begin{cases}A+1\ge0\\A-9\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}A\ge-1\\A\le9\end{cases}}\Leftrightarrow-1\le A\le9\)(nhận)

TH2: \(\hept{\begin{cases}A+1\le0\\A-9\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}A\le-1\\A\ge9\end{cases}}\)(loại)

Vậy GTNN của A là \(-1\)khi \(\frac{6x+8}{x^2+1}=-1\Leftrightarrow x^2+1=-6x-8\Leftrightarrow x^2+6x+9=0\Leftrightarrow\left(x+3\right)^2=0\Leftrightarrow x=-3\)

GTLN của A là \(9\)khi \(\frac{6x+8}{x^2+1}=9\Leftrightarrow9\left(x^2+1\right)=6x+8\Leftrightarrow9x^2-6x+1=0\Leftrightarrow\left(3x-1\right)^2=0\Leftrightarrow x=\frac{1}{3}\)

Cách 2: Ta có \(A=\frac{6x+8}{x^2+1}=\frac{-\left(x^2+1\right)+\left(x^2+6x+9\right)}{x^2+1}=\frac{\left(x+3\right)^2}{x^2+1}-1\)

Vì \(\left(x+3\right)^2\ge0;x^2+1>0\)\(\Leftrightarrow\frac{\left(x+3\right)^2}{x^2+1}\ge0\)\(\Leftrightarrow A\ge-1\)

Dấu "=" xảy ra khi \(x+3=0\Leftrightarrow x=-3\)

Mặt khác \(A=\frac{6x+8}{x^2+1}=\frac{9\left(x^2+1\right)-\left(9x^2-6x+1\right)}{x^2+1}=9-\frac{\left(3x-1\right)^2}{x^2+1}\)

Vì \(\left(3x-1\right)^2\ge0;x^2+1>0\)\(\Leftrightarrow\frac{\left(3x-1\right)^2}{x^2+1}\ge0\)\(\Leftrightarrow-\frac{\left(3x-1\right)^2}{x^2+1}\le0\)\(\Leftrightarrow A\le9\)

Dấu "=" xảy ra khi \(3x-1=0\Leftrightarrow x=\frac{1}{3}\)

Vậy GTNN của A là \(-1\)khi \(x=-3\)

GTLN của A là \(9\)khi \(x=\frac{1}{3}\)

Có biết nhưng ko nói :)))))))))))

\(P=\dfrac{3x^2+1-3x^2+6x-3}{3x^2+1}=1-\dfrac{3\left(x-1\right)^2}{3x^2+1}\le1\)

\(P_{max}=1\) khi \(x=1\)

a) Xét \(\Delta ABC\)và \(\Delta HBA\), ta có:

\(\widehat{B}\)chung, \(\widehat{BAC}=\widehat{BHA}\left(=90^o\right)\)

\(\Rightarrow\Delta ABC~\Delta HBA\left(g.g\right)\)(đpcm)

b) \(\Delta ABC\)vuông tại A \(\Rightarrow BC^2=AB^2+AC^2\)\(\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{30^2+40^2}=\sqrt{900+1600}=\sqrt{2500}=50\left(cm\right)\)

Ta có \(S_{ABC}=\frac{1}{2}AB.AC=\frac{1}{2}AH.BC\)\(\Rightarrow AB.AC=AH.BC\)\(\Rightarrow AH=\frac{AB.AC}{BC}=\frac{30.40}{50}=24\left(cm\right)\)

Vậy \(AH=24cm\)

Bạn vào thống kê hỏi đáp của mình xem nhé.

A B C K M H

a, Xét \(\Delta ABH\)và \(\Delta BCA\)có :

\(\widehat{B}\)chung

\(\widehat{BAC}=\widehat{BHA}\)

\(\Rightarrow\Delta ABH~\Delta BCA\)

b, Xét \(\Delta AKH\)và \(\Delta AHC\)có :

\(\widehat{KAH}\)chung

\(\widehat{AKH}=\widehat{AHC}=90^0\)

= > \(\Delta AKH~\Delta AHC\)

= > \(\frac{HK}{HC}=\frac{AH}{AC}\)( 1 )

Xét \(\Delta ABC\)có \(\widehat{A}=90^0\)

Áp dụng đinh lí Pytago trong tam giác ABC  vuông tại A có :

\(AB^2+AC^2=BC^2\)

\(\Leftrightarrow6^2+8^2=BC^2\)

\(\Leftrightarrow100=BC^2\Rightarrow BC=\sqrt{100}=10\)

Xét \(\Delta ABC\)có : \(\widehat{A}-90^0\), \(AH\perp BC,H\in BC\)

\(\Rightarrow AH.BC=AB.AC\)

\(\Rightarrow AH.10=6.8\)

\(\Rightarrow AH=4,8\)

( 1 ) \(\Rightarrow\frac{HK}{HC}=\frac{AH}{AC}\Rightarrow\frac{HK}{6,4}=\frac{4,8}{8}\)

\(\Rightarrow HK=3,84\)

c, Bạn làm nốt nhé

Chắc đề bài là:

\(P=\dfrac{1}{\left(a+1\right)^2+b^2+1}+\dfrac{1}{\left(b+1\right)^2+c^2+1}+\dfrac{1}{\left(c+1\right)^2+a^2+1}\)

Ta có:

\(P=\dfrac{1}{a^2+b^2+2a+2}+\dfrac{1}{b^2+c^2+2b+2}+\dfrac{1}{c^2+a^2+2c+2}\)

\(P\le\dfrac{1}{2ab+2a+2}+\dfrac{1}{2bc+2b+2}+\dfrac{1}{2ca+2c+2}\)

\(P\le\dfrac{1}{2}\left(\dfrac{1}{ab+a+1}+\dfrac{1}{bc+b+1}+\dfrac{1}{ca+c+1}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{1}{ab+a+1}+\dfrac{a}{abc+ab+a}+\dfrac{ab}{ab.ca+abc+ab}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{1}{ab+a+1}+\dfrac{a}{1+ab+a}+\dfrac{ab}{a+1+ab}\right)\) (do \(abc=1\))

\(P\le\dfrac{1}{2}\left(\dfrac{ab+a+1}{ab+a+1}\right)=\dfrac{1}{2}\)

\(P_{max}=\dfrac{1}{2}\) khi \(a=b=c=1\)

\(x^2+3x+3+x^2-x-1-2x^2+2x+1=1\)

\(\Leftrightarrow4x+2=0\Leftrightarrow x=-\dfrac{1}{2}\)