Hình như bạn viết sai đề bài thì phải.

\(b,x^2+4x+3=x^2+3x+x+3.\)

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

\(c,16x-5x^2-3=x-5x^2+15x-3\)

\(=x\left(1-5x\right)+3\left(5x-1\right)\)

\(=\left(x+3\right)\left(1-5x\right)\)

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

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

\(e,x^3-2x^2+x-xy^2=x\left(x^2-2x+1-y^2\right).\)

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

a, sửa đề : x² + 7x - 8 

\(x^2+7x-8=x^2-x+8x-8\)

\(=x\left(x-1\right)+8\left(x-1\right)=\left(x-1\right)\left(x+8\right)\)

x khác 1

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

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

\(=\frac{x}{x^2+x+1}\)

Xét hiệu 1/3-N=\(\frac{1}{3}-\frac{x}{x^2+x+1}=\frac{x^2+x+1-3x}{3\left(x^2+x+1\right)}=\frac{x^2-2x+1}{3\left(x^2+x+1\right)}=\frac{\left(x-1\right)^2}{3\left(x^2+x+1\right)}>0\)với mọi x khác 1

=> 1/3 >N

Em xét hai tam giác đồng dạng DEC và DCI

\(S_{ABC}=\frac{CE.CD}{2},S_{ABC}=\frac{DE.CI}{2}\Rightarrow CE.CD=DE.CI\)

a) M xác định khi \(x+1\ne0\)

\(x^2+1\ne0\)

\(x^2+2x+1=\left(x+1\right)^2\ne0\)

\(\Leftrightarrow x\ne\pm1\)

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

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

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

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

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

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

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

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

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

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

\(=\frac{\left(x^4-1\right)}{\left(x+1\right)\left(x^4-1\right)}+\frac{-2\left(x-x^3\right)}{\left(x^4-1\right)}\)\(=\frac{1}{x+1}+\frac{-2\left(x-x^3\right)}{\left(x^4-1\right)}\)??? Chắc hết rút được rồi :v

Câu b) hơi dài quá rồi.Làm lại

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

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

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

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

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

\(=\frac{1}{x+1}+\frac{2x}{\left(x^2+1\right)\left(x+1\right)}=\frac{x+1}{x^2+1}\) (Quy đồng và rút gọn)

chị đã ngủ chưa

@NguyễnLamGiang

Bn nghĩ có thể vừa ngủ vừa đăng câu hỏi ư ???

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Câu hỏi của Hoàng Liên - Toán lớp 9 - Học toán với OnlineMath Em tham khảo tại link này nhé !