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

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

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

\(=\left(x^2+x-6\right)\left(2x^2+2x-13\right)\)

\(=\left(x-2\right)\left(x+3\right)\left(2x^2+2x-13\right)\)

\(C=2\left(x^2+x-5\right)^2-5\left(x^2+x\right)+28\)

Đặt t=\(x^2+x\)

\(\Rightarrow C=2\left(t-5\right)^2-5t+28=2t^2-20t+50-5t+28=2t^2-25t+78=2\left(t-\dfrac{13}{2}\right)\left(t-6\right)\)

Thay t: \(C=2\left(t-\dfrac{13}{2}\right)\left(t-6\right)=2\left(x^2+x-\dfrac{13}{2}\right)\left(x^2+x-6\right)=2\left(x-2\right)\left(x+3\right)\left(x^2+x-\dfrac{13}{2}\right)\)

Ta có: \(C=2\left(x^2+x-5\right)^2-5\left(x^2+x\right)+28\)

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

\(=\left(x-2\right)\left(x+3\right)\left(2x^2+2x-13\right)\)

\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)

Sao đề là phân tích mà lại "= 0" vậy bạn?

=(x^2+8x)^2+23(x^2+8x)+135

Cái này ko phân tích được nha bạn

\(\left(x^2+8x+8\right)\left(x^2+8x+15\right)+15\\ \Leftrightarrow\left(x^4+8x^3+15x^2+8x^3+64x^2+120x+8x^2+64x+120\right)+15\\ \Leftrightarrow x^4+16x^3+87x^2+184x+135\)

Gọi `A=(x^2+8x+8)(x^2+8x+15)+15`

Đặt `t=x^2+8x+11,5`

`=>A=(t-3,5)(t+3,5)+15=t^2-3,5^2+15=t^2-2,75=(t-sqrt(2,75))(t+sqrt(2,75))=(x^2+8x+11,5-(sqrt11)/2)(x^2+8x+11,5+(sqrt11)/2)=(x^2+8x+(23-\sqrt11)/2)(x^2+8x+(23+\sqrt11)/2)`

x³+27+(x+3)(x+9)

= (x+3)(x²-3x+9)+(x+3)(x+9)

= (x+3)(x²-3x+9+x+9)

=(x+3)(x²-2x+18)

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

x³+27+(x+3)(x+9)

= (x+3)(x²-3x+9)+(x+3)(x+9)

= (x+3)(x²-3x+9+x+9)

=(x+3)(x²-2x+18)

`(x+3)^4+(x+5)^4-2`

`={[(x+3)^2]^2-1^2}+{[(x+5)^2]^2 -1^2}`

`=[(x+3)^2-1^2][(x+3)^2+1]+[(x+5)^2-1^2][(x+5)^2+1]`

`=(x+3-1)(x+3+1)[(x+3)^2+1]+(x+5-1)(x+5+1)[(x+5)^2+1]`

`=(x+2)(x+4)[(x+3)^2+1]+(x+4)(x+6)[(x+5)^2+1]`

`=(x+4){(x+2)[(x+3)^2+1]+(x+6)[(x+5)^2+1]}`

`=(x+4)(2x^3+24x^2+108x+176)`

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

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

\(=\left[\left(x^2+6x+9-1\right)\left(x^2+6x+9+1\right)\right]+\left[\left(x^2+10x+25-1\right)\left(x^2+10x+25+1\right)\right]\)

\(=\left(x^2+6x+8\right)\left(x^2+6x+10\right)+\left(x^2+10x+24\right)\left(x^2+10x+26\right)\)

\(=\left(x+2\right)\left(x+4\right)\left(x^2+6x+10\right)+\left(x+4\right)\left(x+6\right)\left(x^2+10x+26\right)\)

\(=\left(x+4\right)\left[\left(x+2\right)\left(x^2+6x+10\right)+\left(x+6\right)\left(x^2+10x+26\right)\right]\)

\(=\left(x+4\right)\left(x^3+6x^2+10x+2x^2+12x+20+x^3+10x^2+26x+6x^2+60x+156\right)\)

\(=\left(x+4\right)\left(2x^3+24x^2+108x+176\right)\)

\(=2\left(x+4\right)\left(x^3+12x^2+54x+88\right)\)

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

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

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

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

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

\(=4.2x\)

\(=8x\)

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

(x + 2)² - (x - 2)²

= (x + 2 - x + 2).(x + 2 + x - 2)

= 4.2x

= 8x