Thay 11= 10+1 ta có

x¹⁰-(10+1)x⁹+(10+1)x⁸-(10+1)x⁷+(10+1)x⁶-(10+1)x⁵+(10+1)x⁴-(10+1)x³+(10+1)x²-(10+1)x+2

= x¹⁰-(x+1)x⁹+(x+1)x⁸-(x+1)x⁷+(x+1)x⁶-(x+1)x⁵+(x+1)x⁴-(x+1)x³+(x+1)x²-(x+1)x+2

= x¹⁰-x¹⁰-x⁹+x⁹+x⁸-x⁸-x⁷+x⁷+x⁶-x⁶-x⁵+x⁵+x⁴-x⁴-x³+x³+x²-x²-x+2

= -x+2 

Thay x=10 vào bt

= -10+2

= -8

thay 11=x+1 ta có:

f(x)= \(x^{10}\)-11\(x^9\)+11\(x^8\)-11\(x^7\)+....+11\(x^2\)-11x+100

     =\(x^{10}\)-(x+1)\(x^9\)+(x+1)\(x^8\)-(x+1)\(x^7\)+...+(x+1)\(x^2\)-(x+1)x+100

     =\(x^{10}\)-\(x^{10}\)-\(x^9\)+\(x^9\)+\(x^8\)-\(x^8\)-\(x^7\)+......+\(x^3\)+\(x^2\)-\(x^2\)-x+100

     =-x+100

=> f(10)=-10+100=90      

Thay 11 = x + 1 ta có:

f(x) = \(x^{10}-11x^9+11x^8-11x^7+...+11x^2-11x+100\)

      \(=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...+\left(x+1\right)x^2-\left(x+1\right)x^2-\left(x+1\right)x+100\)

      = \(x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^3+x^2-x^2-x+100\)

      = -x+100

=>f(10)= - 10 + 100 = 90

o ti k hieu

1.

$-7(5-x)-2(x-10)=15$

$-35+7x-2x+20=15$

$5x-15=15$

$5x=30$

$x=30:5=6$

2.

$3(x-4)-(8-x)=12$

$3x-12-8+x=12$

$4x-20=12$

$4x=12+20=32$

$x=32:4=8$

3.

$7(x-9)-5(6-x)=(-6)+11x$

$7x-63-30+5x=11x-6$

$12x-93=11x-6$

$12x-11x=93-6$

$x=87$

       7(x-3)-5(3-x)=11x-5

<=>7(x-3)+5(x-3)-11x+5=0

<=>12(x-3)-11x+5=0

<=>12x-36-11x+5=0

<=>x-31=0

<=>x=31

a, => 7x-63-30+5x=11x-6

=> 12x-93=11x-6

=> 12x=11x-6+93 = 11x+87

=> 12x-11x=87

=> x=87

b, => 7x-21-15+5x=11x-5

=> 12x-36=11x-5

=> 12x=11x-5+36 = 11x+31

=> 12x-11x=31

=> x=31

Tk mk nha