a) \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2+3x+1\right)=x-4\)

\(\Leftrightarrow\left(3x\right)^3-2^3-\left(3x^3\right)+1=x-4\)

\(\Leftrightarrow x=13\)

9(2x+1)=4(x-5)²

<=> 18x+9=4(x²-10x+25)

<=> 4x²-58x+91=0

\(\Leftrightarrow x=\frac{29\pm3\sqrt{53}}{4}\)

x³-4x²-12x+27=0

<=> (x+3)(x²-7x+9)=0

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}\)

\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{7}{6x+30}\)đkxđ \(x\ne\pm5\)

\(\Leftrightarrow45+9x-90-14x+70=0\)

\(\Leftrightarrow25-5x=0\)

\(\Leftrightarrow-5\left(x-5\right)=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\left(\inđkxđ\right)\)

\(\Leftrightarrow x\in\varnothing\)

0 đấy làm rồi

\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\)đkxđ \(x\ne-1;-3\)

\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{6}{x+3}+\frac{3}{x+1}=1\)

\(\Leftrightarrow\frac{4x-6x-6+3x+9}{\left(x+1\right)\left(x+3\right)}=1\)

\(\Leftrightarrow\frac{x+3}{\left(x+3\right)\left(x+1\right)}=1\)

\(\Leftrightarrow\frac{1}{x+1}=1\)

\(\Leftrightarrow x+1=1\)

\(\Leftrightarrow x=0\left(tm\right)\)

\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\)

\(\frac{4x}{\left(x+1\right)\left(x+3\right)}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\)

\(\frac{4x}{\left(x+1\right)\left(x+3\right)}-1=6\left(\frac{1}{x+3}-\frac{1}{2x-2}\right)\)

\(4x-\left(x+1\right)\left(x+3\right)=6\left(\frac{1}{x+3}-\frac{1}{2\left(x+1\right)}\right)\left(x+1\right)\left(x+2\right)\)

\(-x^2-3=\frac{6x^2}{x+3}+\frac{24x}{x+3}+\frac{18}{x+3}-\frac{3x^2}{x+1}-\frac{12x}{x+1}-\frac{9}{x+1}\)

\(-x^4-4x^3-6x^2-12x=3x^3+9x^2-3x\)

\(-x^4-4x^3-6x^2-12x=3x^3+9x^2-3x\)

\(-x^4-4x^3-6x^2-12x-3x^3-9x^2+3x=0\)

\(x^4+7x^3+15x^2+9x=0\)

\(x\left(x^3+6x+9\right)\left(x+1\right)=0\)

\(x\left(x+3\right)^2\left(x+1\right)=0\)

\(x=0;-3;-1\)

\(\frac{3x}{x-1}-\frac{2x}{x-3}+\frac{4x}{\left(x-1\right)\left(x-3\right)}=0\)đkxd \(x\ne1;3\)

\(\Leftrightarrow3x^2-9x-2x^2-2x+4x=0\)

\(\Leftrightarrow x^2-7x=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-7=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=7\end{cases}\left(tm\right)}\)

\(\frac{3x}{x-1}-\frac{2x}{x-3}+\frac{4x}{\left(x-1\right)\left(x-3\right)}=0\)\(ĐKXĐ:x\ne1;3\)

\(3x\left(x-3\right)\left(x+3\right)-2x\left(x-1\right)\left(x+3\right)+4x\left(x-3\right)=0\)

\(x^3-33x=0\)

\(x\left(x^2-33\right)=0\)

\(x=0;\pm\sqrt{33}\)

a) \(\frac{1}{x+3}+\frac{x}{x^2-6x+9}\left(x\ne\pm3\right)\)

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

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

anhdun_•Ŧ๏áйツɦọς• giải a r nha , tớ giải b+c cho 

\(b,\frac{2x}{x^2-9}-\frac{x-1}{x+3}\)

\(\frac{2x}{x^2-3^2}-\frac{x-1}{x+3}\)

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

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

\(\frac{2x-x^2+3x+x-3}{\left(x+3\right)\left(x-3\right)}\)

\(\frac{\left(2x+3x+x\right)-x^2-3}{\left(x+3\right)\left(x-3\right)}\)

\(\frac{6x-x^2-3}{\left(x+3\right)\left(x-3\right)}\)

\(\frac{2}{2x+1}-\frac{3}{1-2x}=\frac{3x+8}{4x^2-1}\)đkxđ \(x\ne\pm\frac{1}{2}\)

\(\Leftrightarrow4x-2+6x+3-3x-8=0\)

\(\Leftrightarrow7x-7=0\)

\(\Leftrightarrow x=1\left(tm\right)\)

tự kẻ hình

a, xét tam giác ABC và tam giác HBA có : góc B chung

góc BAC = góc BHA = 90

=> tam giác ABC đồng dạng với tam giác HBA (g-g)

=>  AB/BH = AC/AH 

=> AB.AH = BH.AC 

b, xét tam giác BAH vuông tại H => HB^2 + HA^2 = AB^2 (Pytago)

BH = 3; AB = 5(gt)

=> 3^2 + AH^2 = 5^2

=> AH^2 = 16

=> AH = 4 do AH > 0

xét tam giác ABH có : BI là pg của góc ABH (gt)

=> AI/AB = IH/BH (tính chất)

=> AI+IH/AB+BH = AI/AB = IH/BH

=> AH/AB + BH = AI/AB = IH/BH 

có: AH = 4; AB = 5; BH = 3

=> 4/3+5 = AI/5 = IH/3

=> AI/5 = IH/3 = 1/2

=> AI = 5/2 và IH = 3/2

c,  góc CAH = 90 - góc HAB 

góc HBA = 90 - góc HAB 

=> góc CAH = góc HBA 

xét tam giác AHC và tam giác BHA có: góc AHC = góc BHA = 90

=> tam giác AHC đồng dạng với tam giác BHA (g-g)

=>  AC/AB = AH/HB

=> AC/AH = AB/HB 

BI là pg của tam giác AHB => AI/AH = AB/AB

CK là pg của tam giác AHC => CK/KH = AC/AH

=> AI/AH = CK/KH

=> KI // AC