mình tưởng lopw 9 mới học căn

em chịu bác lớp 7 học rui

lớp 7 đã học đâu?

\(x\sqrt{x}-3x+4\sqrt{x}-2=x\sqrt{x}-x-2x+2\sqrt{x}+2\sqrt{x}-2\)

\(=x\left(\sqrt{x}-1\right)-2\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)\)

\(=\left(\sqrt{x}-1\right)\left(x-2\sqrt{x}+2\right)\)

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

Đặt \(x^2+3x+1=a,\)ta được:

\(a\left(a+1\right)-6\)

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

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

Thay \(a=x^2+3x+1,\)ta được:

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

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

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

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

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

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

Đặt \(t=x^2+3x\) thì biểu thức có dạng \(t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2=\left(x^2+3x+1\right)^2\)

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

Đặt \(k=x^2-x+2\) thì biểu thức có dạng

k²+4(k-3)=k²+4k-12=k²-2k+6k-12=k(k-2)+6(k-2)=(k-2)(k+6)=(x²-x)(x²-x+8)=(x-1)x(x²-x+8)

c)làm tương tự câu a

Câu b đề sai nha bạn.

a) \(x^3+9x^2+27x+27=\left(x+3\right)^3\)

b) \(3\sqrt{3x^3}+18x^2+12\sqrt{3x}+8=\left(\sqrt{3x}+2\right)^3\)

c) \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)

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

\(x-4\sqrt{x-2}+2=\left(x-2\right)-4\sqrt{x-2}+4=\left(\sqrt{x-2}-2\right)^2\)

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

\(x-4\sqrt{x-2}+2=\left(\sqrt{x-2}+4\right)^2\)