a) Ta thấy: \(AB.AC=BC.AH\)

\(\Leftrightarrow AB^2.AC^2=BC^2.AH^2\)

\(\Leftrightarrow AH^2=\frac{AB^2.AC^2}{BC^2}\)

\(\Leftrightarrow AH^2=\frac{AB^2.AC^2}{AB^2+AC^2}\)

\(\Leftrightarrow\frac{1}{AH^2}=\frac{AB^2+AC^2}{AB^2.AC^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\)

Ta có: \(\frac{AB}{AC}=\frac{5}{7}\Rightarrow AB:AC=\frac{5}{7}\Rightarrow AB=\frac{5}{7}AC\)

Áp dụng công thức trên: \(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{15^2}=\frac{1}{\frac{25}{49}AC^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{225}=\frac{49}{25}.\frac{1}{AC^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{225}=\frac{1}{AC^2}\left(\frac{49}{25}+1\right)\)

\(\Rightarrow\frac{1}{225}=\frac{1}{AC^2}.\frac{74}{25}\Rightarrow\frac{1}{AC^2}=\frac{1}{225}.\frac{25}{74}=\frac{1}{666}\Rightarrow AC^2=666\Rightarrow AC=\sqrt{666}=3\sqrt{74}cm\)

Do đó: \(AB=\frac{5}{7}.3\sqrt{74}=\frac{15\sqrt{74}}{7}cm\)

Xét tam giác ABH có: \(AH^2+BH^2=AB^2\Leftrightarrow15^2+BH^2=\left(\frac{15\sqrt{74}}{7}\right)^2\Leftrightarrow BH^2=\frac{16650}{49}-225=\frac{5625}{49}\)

\(\Rightarrow BH=\frac{\sqrt{5625}}{\sqrt{49}}=\frac{75}{7}cm\)

Xét tam giác ACH có: \(AH^2+HC^2=AC^2\Leftrightarrow15^2+HC^2=666\Leftrightarrow HC^2=666-225=441\)

\(\Rightarrow HC=\sqrt{441}=21cm\)

Vậy: \(BH=\frac{75}{7}cm\) và \(HC=21cm\)

b) Chu vi tam giác ABC là: \(AB+AC+BC=\frac{15\sqrt{74}}{7}+3\sqrt{74}+21+\frac{75}{7}\approx76cm\)

Vì tam giác ABC vuông tại A => góc B + góc C = 90^o

Vì tam giác HAC vuông tại H => góc HAC + góc C = 90^o

=> góc HAC = góc B

Xét tam giác HAC và tam giác HBA có:

     góc HAC = góc B (cmt)

     góc AHC = góc AHB (=90^o)

=> tam giác HAC đồng dạng với tam giác HBA (TH3)

=> \(\frac{AC}{AB}=\frac{AH}{BH}=\frac{HC}{AH}=\frac{7}{5}\)

=> \(HC=15.\frac{7}{5}=21\left(cm\right);HB=15.\frac{5}{7}=\frac{75}{7}\left(cm\right)\)

Sau đó tính AB; AC; BC. Ngại là lắm, làm nốt nhá ._.

Ta có:

 \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+16\right)+8\left(x^2+10x+16\right)+16\)

\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)

\(=\left(x^2+10x+16+4\right)^2=\left(x^2+10+20\right)^2\)

k nha!!

\(\text{( x + 2 ) ( x + 4 ) ( x + 6 ) ( x + 8 ) + 16}\)

\(\text{Phân tích thành nhân tử :}\)

\(\left(x^2+10x+20\right)^2\)

ĐKXĐ: \(\hept{\begin{cases}x\ne10\\x\ne-70\end{cases}}\)

\(\frac{x-20}{x-10}=\frac{x+40}{x+70}\)

\(\Rightarrow\left(x-20\right)\left(x+70\right)=\left(x+40\right)\left(x-10\right)\)

\(\Rightarrow x^2+70x-20x-1400=x^2-10x+40x-400\)

\(\Rightarrow20x=1000\Rightarrow x=50\)

                                                 Vậy x = 50

a, \(x^3+6x^2+11x+6\)

\(=x^3+3x^2+3x^2+9x+2x+6\)

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

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

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

\(=\left(x+3\right)\text{[}x\left(x+1\right)+2\left(x+1\right)\text{]}\)

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

b, \(2x^3+3x^2+3x+2\)

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

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

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

c, \(x^3-4x^2-8x+8\)

\(=x^3+2x^2-6x^2-12x+4x+8\)

\(=x^2\left(x+2\right)-6x\left(x+2\right)+4\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-6x+4\right)\)

\(x^3+6x^2-13x-42\)

\(x^3+6x^2-13x-42\)

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

b, \(2x^3-x^2+3x+6\)

\(=2x^3+2x^2-3x^2-3x+6x+6\)

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

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

\(A=\frac{-1}{2x+3}\)
Để A có giá trị nguyên thì -1 phải chia hết cho 2x+3
                            hay 2x+3\(\in\)Ư(-1)={1;-1}
                             =>x={-1;-2}

1) \(\left(x^2+8x+7\right).\left(x+3\right).\left(x+5\right)+15\)

\(=\left(x^2+8x+7\right).\left(x^2+5x+3x+15\right)+15\)

\(=\left(x^2+8x+7\right).\left(x^2+8x+15\right)+15\)

Ta đặt: \(x^2+8x+7=n\)

\(=n.\left(n+8\right)+15\)

\(=n^2+8n+15\)

\(=n^2+3n+5n+15\)

\(=\left(n^2+3n\right)+\left(5n+15\right)\)

\(=n.\left(n+3\right)+5.\left(n+3\right)\)

\(=\left(n+3\right).\left(n+5\right)\)

\(=\left(x^2+8x+7+3\right).\left(x^2+8x+7+5\right)\)

\(=\left(x^2+8x+10\right).\left(x^2+8x+12\right)\)

\(=\left(x^2+8x+10\right).\left(x^2+2x+6x+12\right)\)

\(=\left(x^2+8x+10\right).[x.\left(x+2\right)+6.\left(x+2\right)]\)

\(=\left(x^2+8x+10\right).\left(x+2\right).\left(x+6\right)\)

2) \(x^2-2xy+3x-3y-10+y^2\)

\(=\left(x-y\right)^2+3.\left(x-y\right)-10\)

Ta đặt: \(x-y=n\)

\(=n^2+3n-10\)

\(=n^2-2n+5n-10\)

\(=\left(n^2-2n\right)+\left(5n-10\right)\)

\(=n.\left(n-2\right)+5.\left(n-2\right)\)

\(=\left(n-2\right).\left(n+5\right)\)

\(=\left(x-y-2\right).\left(x-y+5\right)\)

à thôi mn không cần phải làm nữa