a, Xét △ ABC vuông tại A có: 

BC² = AC² + AB² (định lý Pytago)

=> BC² = 6² + 8² = 100

=> BC = 10 cm

Vì AD là phân giác \(\widehat{BAC}\) (gt)

\(\Rightarrow\frac{CD}{AC}=\frac{BD}{AB}=\frac{CD+BD}{AC+AB}=\frac{BC}{6+8}=\frac{10}{14}=\frac{5}{7}\)(áp dụng t/c dãy tỉ số bằng nhau)

Do đó: \(\frac{CD}{AC}=\frac{5}{7}\) \(\Rightarrow\frac{CD}{6}=\frac{5}{7}\) \(\Rightarrow CD=\frac{6.5}{7}=\frac{30}{7}\)(cm)

\(\frac{BD}{AB}=\frac{5}{7}\)\(\Rightarrow\frac{BD}{8}=\frac{5}{7}\)\(\Rightarrow BD=\frac{8.5}{7}=\frac{40}{7}\)(cm)

b, Xét △AHB vuông tại H và △AEH vuông tại E

Có: \(\widehat{HAB}\)là góc chung

=> △AHB ᔕ △AEH (g.g)

\(\Rightarrow\frac{AH}{AE}=\frac{AB}{AH}\)

=> AH . AH = AE . AB

=> AH² = AE . AB

c, Xét △AHC vuông tại H và △AFH vuông tại F

Có: \(\widehat{HAC}\)là góc chung

=> △AHC ᔕ △AFH (g.g)

\(\Rightarrow\frac{AH}{AF}=\frac{AC}{AH}\)

=> AH² = AF . AC

mà AH² = AE . AB (cmt)

=> AE . AB = AF . AC

 a : xét tg ABD và tg ACE có :

góc A chung

góc BAD = góc CEA (=90 độ)

ngoặc 2 dòng trên suy ra tg ABD đồng dạng vs tg ACE (g.g)

2) Đặt \(x+4=t\)

\(PT\Leftrightarrow\left(t-1\right)^4+\left(t+1\right)^4=16\)

\(\Leftrightarrow2t^4+12t^2-14=0\)

\(\Leftrightarrow t^4+6t^2-7=0\)

\(\Leftrightarrow\left(t^2-1\right)\left(t^2+6\right)=0\)

\(\Rightarrow t^2-1=0\Leftrightarrow t^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}t=1\\t=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)

( x + 6 )⁴ + ( x + 4 )⁴ = 82

Đặt t = x + 5

pt <=> ( t + 1 )⁴ + ( t - 1 )⁴ = 82

<=> t⁴ + 4t³ + 6t² + 4t + 1 + t⁴ - 4t³ + 6t² - 4t + 1 - 82 = 0

<=> 2t⁴ + 12t² - 80 = 0

<=> t⁴ + 6t² - 40 = 0

Đặt a = t²

<=> a² + 6a - 40 = 0

<=> a² - 4a + 10a - 40 = 0

<=> a( a - 4 ) + 10( a - 4 ) = 0

<=> ( a - 4 )( a + 10 ) = 0

<=> ( t² - 4 )( t² + 10 ) = 0

<=> ( t - 2 )( t + 2 )( t² + 10 ) = 0

<=> ( x + 5 - 2 )( x + 5 + 2 )[ ( x + 5 )² + 10 ] = 0

<=> ( x + 3 )( x + 7 )[ ( x + 5 )² + 10 ] = 0

Vì ( x + 5 )² + 10 > 0 

=> x + 3 = 0 hoặc x + 7 = 0 

<=> x = -3 hoặc x = -7

Vậy ...

Vì \(x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1>0\left(\forall x\right)\)

\(\Rightarrow8x-4=0\Leftrightarrow x=\frac{1}{2}\)

Vậy x = 1/2

(8x - 4)(x² + 2x + 2) = 0

<=> \(\orbr{\begin{cases}8x-4=0\\x^2+2x+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x\in\varnothing\end{cases}}\Leftrightarrow x=\frac{1}{2}\)

Vậy x = 1/2 là nghiệm phương trình

2)  \(\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2=0\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x+2\right)\right]+4x^2=0\)

\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2=0\)

Nếu x = 0 thì PT vô nghiệm

Nếu x khác 0, chia cả 2 vế cho x² ta được:

\(PT\Leftrightarrow\left(x-\frac{8}{x}-7\right)\left(x-\frac{8}{x}-2\right)+4=0\)

Đặt \(x-\frac{8}{x}=b\) khi đó: 

\(\left(b-7\right)\left(b-2\right)+4=0\)

\(\Leftrightarrow b^2-9b+14+4=0\)

\(\Leftrightarrow b^2-9b+18=0\)

\(\Leftrightarrow\left(b-3\right)\left(b-6\right)=0\Leftrightarrow\orbr{\begin{cases}b-3=0\\b-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}b=3\\b=6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{8}{x}=3\\x-\frac{8}{x}=6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-8=3x\\x^2-8=6x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-3x-8=0\\x^2-6x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3\pm\sqrt{41}}{2}\\x=3\pm\sqrt{17}\end{cases}}\)

Vậy ...

3) \(\left(x-2\right)\left(x-4\right)\left(x-5\right)\left(x-10\right)-54x^2=0\)

\(\Leftrightarrow\left[\left(x-2\right)\left(x-10\right)\right]\left[\left(x-4\right)\left(x-5\right)\right]-54x^2=0\)

\(\Leftrightarrow\left(x^2-12x+20\right)\left(x^2-9x+20\right)-54x^2=0\)

Nếu x = 0 thì PT vô nghiệm

Nếu x khác 0 thì chia cả 2 vế cho x² ta được:
\(PT\Leftrightarrow\left(x+\frac{20}{x}-12\right)\left(x+\frac{20}{x}-9\right)-54=0\)

Đặt \(x+\frac{20}{x}=c\) nên khi đó:

\(\left(c-12\right)\left(c-9\right)-54=0\)

\(\Leftrightarrow c^2-21c+108-54=0\)

\(\Leftrightarrow c^2-21c+54=0\)

\(\Leftrightarrow\left(c-3\right)\left(c-18\right)=0\Leftrightarrow\orbr{\begin{cases}c-3=0\\c-18=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}c=3\\c=18\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{20}{x}=3\\x+\frac{20}{x}=18\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+20=3x\\x^2+20=18x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-3x+20=0\\x^2-18x+20=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-\frac{3}{2}\right)^2=-\frac{71}{4}\left(ktm\right)\\x=9\pm\sqrt{61}\end{cases}}\)

Vậy ...

nhìn căng nhể :))

a) ( x - 1 )( x - 3 )( x + 5 )( x + 7 ) - 297 = 0

<=> [ ( x - 1 )( x + 5 ) ][ ( x - 3 )( x + 7 ) ] - 297 = 0

<=> ( x² + 4x - 5 )( x² + 4x - 21 ) - 297 = 0

Đặt t = x² + 4x - 5

pt <=> t( t - 16 ) - 297 = 0

<=> t² - 16t - 297 = 0

<=> t² - 27t + 11t - 297 = 0

<=> t( t - 27 ) + 11( t - 27 ) = 0

<=> ( t - 27 )( t + 11 ) = 0

<=> ( x² + 4x - 5 - 27 )( x² + 4x - 5 + 11 ) = 0

<=> ( x² + 4x - 32 )( x² + 4x + 6 ) = 0

<=> ( x² - 4x + 8x - 32 )( x² + 4x + 6 ) = 0

<=> [ x( x - 4 ) + 8( x - 4 ) ]( x² + 4x + 6 ) = 0

<=> ( x - 4 )( x + 8 )( x² + 4x + 6 ) = 0

Đến đây dễ rồi :)

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

\(\Leftrightarrow2x^2-3x-9+x^2-2x=3\left(x^2-4x+4\right)\)

\(\Leftrightarrow3x^2-5x-9=3x^2-12x+12\)

\(\Leftrightarrow7x=21\Rightarrow x=3\)

b) \(\left(4x+7\right)\left(x-3\right)-x^2=3x\left(x+2\right)\)

\(\Leftrightarrow4x^2-5x-21-x^2=3x^2+6x\)

\(\Leftrightarrow11x=-21\Rightarrow x=-\frac{21}{11}\)

a,X=3

CBHT

a) \(5\left(\frac{1}{x}-1\right)=\left(\frac{1}{x}-1\right)\left(x^2+1\right)\) (x khác 0)

\(\Leftrightarrow\left(x^2+1\right)\left(\frac{1}{x}-1\right)-5\left(\frac{1}{x}-1\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}-1\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}-1=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=1\\x^2=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=\pm2\end{cases}}\)

b) đk: \(x\ne-2\)

Ta có: \(\left(x^2+x+1\right)\cdot\frac{3x+1}{x+2}=\left(x^2+x+1\right)\cdot\frac{x}{2\left(x+2\right)}\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(\frac{3x+1}{x+2}-\frac{x}{2\left(x+2\right)}\right)=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\cdot\frac{5x+2}{2\left(x+2\right)}=0\)

Vì \(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow\frac{5x+2}{2\left(x+2\right)}=0\Rightarrow5x+2=0\Rightarrow x=-\frac{2}{5}\)

567-589=

a) Với \(x\ne1\)ta có:

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

\(=\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x}{x^2+x+1}-\frac{1}{x-1}\right].\frac{2}{x-1}\)

\(=\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right].\frac{2}{x-1}\)

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

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

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

b) \(A=\frac{2}{3}\)\(\Leftrightarrow\frac{2}{x^2+x+1}=\frac{2}{3}\)

\(\Leftrightarrow x^2+x+1=3\)\(\Leftrightarrow x^2+x-2=0\)

\(\Leftrightarrow x^2-x+2x-2=0\)\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(ktmĐKXĐ\right)\\x=-2\left(tmĐKXĐ\right)\end{cases}}\)

Vậy \(A=\frac{2}{3}\)\(\Leftrightarrow x=-2\)

b) Ta có: \(x^2+x+1=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

\(\Rightarrow x^2+x+1\ge\frac{3}{4}\forall x\)\(\Rightarrow\frac{1}{x^2+x+1}\le\frac{4}{3}\forall x\)

\(\Rightarrow\frac{2}{x^2+x+1}\le\frac{8}{3}\forall x\)\(\Rightarrow A\le\frac{8}{3}\)

Dấu " = " xảy ra \(\Leftrightarrow x+\frac{1}{2}=0\)\(\Leftrightarrow x=-\frac{1}{2}\)( thỏa mãn ĐKXĐ )

Vậy \(maxA=\frac{8}{3}\Leftrightarrow x=-\frac{1}{2}\)