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

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

\(=y^2+8y+15\)

\(=y^2+3y+5y+15\)

\(=y\left(y+3\right)+5\left(y+3\right)\)

\(=\left(y+5\right)\left(y+3\right)\)

(*)\(x^7+x^2+1\)

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

\(=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)

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

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

(*)\(\left(a+b\right)^3-c^3=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)

(*)\(x^4+x^3-x^2+x-2=x^4-x^3+2x^3-2x^2+x^2-x+2x-2\)

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

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

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

............................................

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

b: =>x(8-7)=-33

=>x=-33

c: =>-12x+60+21-7x=5

=>-19x=-76

hay x=4

d: =>-2x-2-x+5+2x=0

=>3-x=0

hay x=3

x+1)(x+3)(x+5)(x+7) + 15 = [ (x+1)(x+7) ].[ (x+3)(x+5) ] + 15 

= (x² + 7x + x + 7).(x² + 5x + 3x + 15) + 15 

= (x² + 8x + 7).(x² + 8x + 15) + 15 

= (x² + 8x + 11 - 4)(x² + 8x + 11 + 4) + 15. Đặt x² + 8x + 11 = y (1) ta được. 

(t - 4)(t + 4) + 15 = t² - 16 + 15 = t² - 1 = (t+1)(t-1) (2). 

Thay (1) vào (2) ta được: đa thức trên được phân tích thành: 

(x² + 8x + 11 + 1)(x² + 8x + 11 - 1) = x² + 8x + 12)(x² + 8x + 10). 

Lưu ý: phương pháp này có tên là "Đặt ẩn phụ". 

\(\left(x^2+8x+7\right)\cdot\left(x^2+8x+15\right)+15=\left(x^2+8x+11\right)^2-16+15\)

    \(\left(x^2+8x+11\right)^2-1=\left(x^2+8x+10\right)\cdot\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\cdot\left(x+2\right)\cdot\left(x+6\right)\)

( x² + 8x + 7 )( x² + 8x + 15 ) + 15 (1)

Đặt t = x² + 8x + 7 

(1) <=> t( t + 8 ) + 15

       = t² + 8t + 15

       = t² + 3t + 5t + 15

       = t( t + 3 ) + 5( t + 3 )

       = ( t + 3 )( t + 5 )

       = ( x² + 8x + 7 + 3 )( x² + 8x + 7 + 5 )

       = ( x² + 8x + 10 )( x² + 8x + 12 )

       = ( x² + 8x + 10 )( x² + 2x + 6x + 12 )

       = ( x² + 8x + 10 )[ x( x + 2 ) + 6( x + 2 ) ]

       = ( x² + 8x + 10 )( x + 2 )( x + 6 )

Đặt \(x^2+8x+11=y\)

Từ đó: \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

\(=\left(x^2+8x+11-4\right)\left(x^2+8x+11+4\right)+15\)

\(=\left(y-4\right)\left(y+4\right)+15\)

\(=y^2-16+15\)

\(=y^2-1\)

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

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

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

x=7

nên x+1=8

\(A=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-5\)

\(=x-5=7-5=2\)

a, \(x\) + 99: 3 = 55

    \(x\) + 33 = 55

     \(x\)        = 55 - 33

      \(x\)       = 22

b, (\(x\) - 25):15 = 20

       \(x\) - 25  = 20 x 15

       \(x\) - 25  = 300

        \(x\)         = 300 + 25

          \(x\)        = 325

c, (3\(x\) - 15).7 = 42

      3\(x\) - 15 = 42:7

       3\(x\) - 15 = 6

        3\(x\)         = 6 + 15

         3\(x\)         = 21

            \(x\)         = 21: 3

             \(x\)         = 7

d, (8\(x\) - 16).(\(x\) -5) = 0

       \(\left[{}\begin{matrix}8x-16=0\\x-5=0\end{matrix}\right.\)

         \(\left[{}\begin{matrix}8x=16\\x=5\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=16:8\\x=5\end{matrix}\right.\)

           \(\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Vậy \(x\) \(\in\) {2; 5}

vào câu hỏi tương tự

tích nha