\(x^4-x^3-x^2+1\)

\(\text{ Phân tích thành nhân tử}\)

\(\left(x-1\right)\left(x^3-x-1\right)\)

\(-x-y^2+x^2-y\)

\(\text{ Phân tích thành nhân tử}\)

\(\left(-\left(y-x+1\right)\right)\left(y+x\right)\)

\(x^2-y^2-x-y\)

\(\text{ Phân tích thành nhân tử}\)

\(\left(-\left(y-x+1\right)\right)\left(y+x\right)\)

\(x^2-y^2+4-4x\)

\(\text{ Phân tích thành nhân tử}\)

\(\left(-\left(y-x+2\right)\right)\left(y-x+2\right)\)

a ) x^4 - x^3 - x^2 +1

=từ từ

b ) - x - y^2 + x^2 - y

=(x+y)(x-y) - (x+y)

= (x+y) (x-y+1)

c ) x^2 - y^2 - x - y

= Giống câu b

d ) x^2 - y^2 + 4 - 4x

= (x^2 - 2x + 4) - y^2

= (x-2)^2 - y^2 

= (x+y-2) (x-y-2)

a ) x^4 - x^3 - x^2 +1

= (1+x) (1-x) - x^3(1-x)

= (-x^3 +x+1) (1-x)

giải giúp mình câu 2 với ạ...câu 1 mk đã làm rồi ạ

\(\left(x^2+x\right)^2-2x^2-2x-15\)

\(=\left(x^2+x\right)^2-\left(2x^2+2x+15\right)\)

\(=\left(x^2+x\right)^2-\left[\left(2x^2+2x\right)+15\right]\)

\(=\left(x^2+x\right)^2-\left[2.\left(x^2+x\right)+15\right]\)

\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\) \(\left(1\right)\)

đặt \(x^2+x=t\)

\(\left(1\right)\)\(=\)  \(t^2-2t-15\)

            \(=\left(t-1\right)^2-16\)

            \(=\left(t-1-4\right)\left(t-1+4\right)\)

           \(=\left(t-5\right)\left(t+3\right)\)

thay \(t=x^2+x\) ta có

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

các câu còn lại tương tự nha

học tốt 

\(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)

\(=\left(x^8-x^6+x^5-x^3+x^2\right)\)

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

\(+\left(x^6-x^4+x^3-x+1\right)\)

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

\(x^5+x+1\)

\(=x^5-x^2+x^2+x+1\)

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

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

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

Bài làm:

a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

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

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

Đặt \(x^2+5x+5=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)

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

b) Tương tự như a phân tích và đặt ra được: \(t^2-1-24=t^2-25=\left(t-5\right)\left(t+5\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)=x\left(x+5\right)\left(x^2+5x+10\right)\)

c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

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

Đặt \(x^2+8x+11=t\)\(\Rightarrow\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1\)

\(=\left(t-1\right)\left(t+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

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

d) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+11=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

Làm mẫu cho 1 vd:

a, (x+1)(x+2)(x+3)(x+4)+1

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

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

Đặt \(y=x^2+5x+5\)

Khi đó ::

(1) = \(\left(y-1\right)\left(y+1\right)+1\)

\(=y^2-1+1=y^2\)

Thay vào ta được: \(\left(x^2+5x+5\right)^2\)

Cái này chưa học bt làm mấy câu

b. x^2 + 2x - 3

= x^2 + 3x - x - 3

= x ( x - 1 ) + 3 ( x - 1 )

= ( x + 3 ) ( x - 1 )

\(4x^2-3x-4\)

\(=\left(2x\right)^2-2.2x.\frac{3}{4}+\frac{9}{16}-\frac{73}{16}\)

\(=\left(2x-\frac{3}{4}\right)^2-\frac{73}{16}\)

\(=\left(2x-\frac{3}{4}\right)^2-\left(\frac{\sqrt{73}}{4}\right)^2\)

\(=\left(2x-\frac{3}{4}-\frac{\sqrt{73}}{4}\right)\left(2x-\frac{3}{4}+\frac{\sqrt{73}}{4}\right)\)

\(=\left(2x-\frac{3+\sqrt{73}}{4}\right)\left(2x+\frac{-3+\sqrt{73}}{4}\right)\)

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

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

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

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

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\) \(\left(1\right)\)

đặt \(x^2+5x+5=t\)

\(\left(1\right)\)\(=\) \(\left(t-1\right)\left(t+1\right)-24\)

            \(=t^2-1-24\)

            \(=t^2-25\)

            \(=\left(t-5\right)\left(t+5\right)\)

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

               \(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

                \(=x\left(x+5\right)\left(x^2+5x+10\right)\)

học tốt

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

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

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

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

d ) 

=(x²-3x)(x²-3x+2)-24

đặt x²-3x+1=a ta đc 

(a-1)(a+1)-24

=a²-1-24=a²-25

=(a-5)(a+5)

=(x²-3x+1+5)(x²-3x+1-5)

=(x²-3x+6)(x²-3x-4)

=(x²-3x+6)(x²-4x+x-4)

=(x²-3x+1)[x(x-4)+(x-4)]

=(x-4)(x+1)(x²-3x+1)

mấy câu kia làm tương tự nhé