a) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)-4=\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\)

Đặt \(t=x^2+6x+5\)

\(PT=t\left(t+3\right)-4=t^2+3t-4=\left(t-1\right)\left(t+4\right)\)

Thay t: \(PT=\left(x^2+6x+5-1\right)\left(x^2+6x+5+4\right)=\left(x^2+6x+4\right)\left(x^2+6x+9\right)=\left(x^2+6x+4\right)\left(x+3\right)^2\)

b)  Đặt \(t=\left(2x+1\right)^2\)

\(PT=t^2-3t+2=\left(t^2-3t+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(t+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(t+1\right)\left(t+2\right)\)

Thay t:

\(PT=\left[\left(2x+1\right)^2+1\right]\left[\left(2x+1\right)^2+2\right]=\left[4x^2+4x+2\right]\left[4x^2+4x+3\right]=2\left[2x^2+2x+1\right]\left[4x^2+4x+3\right]\)

a: \(=x^4-5x^3+4x^3-20x^2+7x^2-35x+4x-20\)

\(=\left(x-5\right)\left(x^3+4x^2+7x+4\right)\)

\(=\left(x-5\right)\left(x^3+x^2+3x^2+3x+4x+4\right)\)

\(=\left(x-5\right)\left(x+1\right)\left(x^2+3x+4\right)\)

b: Đề sai rồi bạn

Ta có : (x+2)(x+4)(x+6)(x+8) + 16

=[(x+2).(x+8)].[(x+4)(x+6)]+16

=(x²+10x+16).(x²+10x+24)+16 (1)

Đặt x^2+10x+16=a thì (1) trở thành:

a.(a+8)+16=a²+8a+16=(a+4)²=(x^2+10x+20)²

Bài 1 :

Mình nghĩ phải sửa đề ntn :

\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\7x+23=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{-23}{7}\end{cases}}}\)

Vậy....

b) \(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt \(q=x^2+x+1\)ta có :

\(A=q\left(q+1\right)-12\)

\(A=q^2+q-12\)

\(A=q^2+4q-3q-12\)

\(A=q\left(q+4\right)-3\left(q+4\right)\)

\(A=\left(q+4\right)\left(q-3\right)\)

Thay \(q=x^2+x+1\)ta có :

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

\(A=\left(x^2+x+5\right)\left(x^2+x-2\right)\)

\(A=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)\)

\(A=\left(x^2+x+5\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)

\(A=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)

Cảm ơn ạ><

Đặt \(x^2+x+1=t\) 

Ta có: \(\left(x^2+x+1\right)^2+3x\left(x^2+x+1\right)+2x^2\)

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

\(=t^2+xt+2xt+2x\)

\(=t\left(t+x\right)+2x\left(t+x\right)\)

\(=\left(t+x\right)\left(t+2x\right)\)

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

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

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

Chúc bạn học tốt.

Đặt x^2+2x=t =>3t^2-2t-1=3t^2-3t+t-1=3t(t-1)+(t-1)=(t-1)(3t+1)

=>(x^2+2x-1)(3x^2+6x+1)

Câu hỏi của Nguyễn Tấn Phát - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo câu a nhé!