Question

bài 1:

a)(3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)

b)3(2x-1)(3x-1)-(2x-3)(9x-1)=0

bài 2:

Cho biểu thức M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x²

tính M theo a,b,c biết rằng x=\(\frac{a+b+c}{2}\)

bài 3:

C=x⁵-36x⁴+37x³-69x²-34x+15 với x=35

Answer 1

Bài 1:
a)

\((3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)\)

\(\Leftrightarrow (6x^2+27x+4x+18)-(6x^2+x+12x+2)=7\)

\(\Leftrightarrow 18x+16=7\Leftrightarrow 18x=-9\Leftrightarrow x=\frac{-1}{2}\)

b)

\(3(2x-1)(3x-1)-(2x-3)(9x-1)=0\)

\(\Leftrightarrow 3(6x^2-2x-3x+1)-(18x^2-2x-27x+3)=0\)

\(\Leftrightarrow 3(6x^2-5x+1)-(18x^2-29x+3)=0\)

\(\Leftrightarrow 14x=0\Leftrightarrow x=0\)

Answer 2

Bài 2:

\(M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x^2\)

\(=(x^2-ax-bx+ab)+(x^2-bx-cx+bc)+(x^2-cx-ax+ac)+x^2\)

\(=4x^2-2(a+b+c)x+(ab+bc+ac)\)

\(=(2x)^2-(a+b+c).2x+(ab+bc+ac)\)

\(=(a+b+c)^2-(a+b+c).(a+b+c)+(ab+bc+ac)\)

\(=(a+b+c)^2-(a+b+c)^2+(ab+bc+ac)=ab+bc+ac\)

Answer 3

Bài 3:

Với $x=35$ thì $x-35=0$. Ta có:

\(C=x^5-36x^4+37x^3-69x^2-34x+15\)

\(=x^4(x-35)-x^3(x-35)+2x^2(x-35)+x(x-35)+x+15\)

\(=x^4.0-x^3.0+2x^2.0+x.0+35+15=50\)