a)B=(3x.4x-11)+5x².(x-1)-4x.(3x+9)+x.5x-5x²

=12x²-11+5x³-5x²-12x²-36x+5x²-5x²

=5x³+(12x²-12x²+5x²-5x²-5x²)-36x-11

=5x³-5x²-36x-11

b)|x|=2=>x=2 hoặc x=-2

*)x=2 =>B=5.2³-5.2²-36.2-11=40-20-72-11=-63

*)x=-2 =>B=5.(-2)³-5.(-2)²-36.(-2)-11=-40-20+72-11=1

c)B=207

=>5x³-5x²-36x-11=207

<=>5x³-5x²-36x-218=0(bó tay)

a)

\(P=12x-33+5x^3-5x^2-12x^2-36x+5x^2-5x^3\)

\(P=-24x-33-12x^2\)

b) |x| = 2 => x= -2 hoặc x = 2

ta có 

\(P_{\left(2\right)}=-24.2-33-12.2^2=-129\)

\(P_{\left(-2\right)}=-24.\left(-2\right)-33-12.\left(-2\right)^2=-33\)

c) để P = 207 thì -48x-33-12x² = 207

                                \(< =>-24x-33-12x^2-207=0\)

                                \(< =>-12x^2-24x-240=0\)

                               \(< =>-12\left(x^2+2x+20\right)=0\)

                                  \(< =>x^2+2x+20=0\)

                                 \(< =>x^2+2x+1+19=0\)

                                 \(< =>\left(x+1\right)^2+19=0\)

vì (x+1)² luôn lớn hơn hoặc bằng 0 với mọi x nên \(\left(x+1\right)^2+19>0\)

=> phương trình vô nghiệm 

vậy không có giá trị nào của x đê P = 207

toán lớp mấy v bn ? 

nhanh giùm mình được không

Bài 1: 

a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)

\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)

a)Ta có:  P = 3x(4x - 11) + 5x²(x - 1) - 4x(3x + 9) + x(5x - 5x²)

P = 12x² - 33x + 5x³ - 5x² - 12x² - 36x + 5x² - 5x³

P = -69x

b) Ta có: x = 2

=> P = -69.2 = -138

c) Ta có: P = 207

=> -69x = 207

=> x = 207 : (-69)

=>  x = -3

\(a,P=3x\left(4x-11\right)+5x^2\left(x-1\right)-4x\left(3x+9\right)+x\left(5x-5x^2\right)\)

\(=12x^2-33x+5x^3-5x^2-12x^2-36x+5x^2-5x^3\)

\(=\left(12x^2-5x^2-12x^2+5x^2\right)-\left(33x+36x\right)+\left(5x^3-5x^3\right)\)

\(=-33x-36x=-69x\)

\(b,\)Khi \(x=2\Leftrightarrow P=-69.2=-138\)

\(c,\)Để \(P=207\Leftrightarrow-69x=207\Leftrightarrow x=-3\)

a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

a/ \(P=3x\left(4x-11\right)+5x^2\left(x-1\right)-4x\left(3x+9\right)+x\left(5x-5x^2\right)\)

\(P=3x\left(4x-11\right)+5x^2\left(x-1\right)-4x.3\left(x+3\right)+x.5x\left(1-x\right)\)

\(P=3x\left(4x-11\right)-5x^2\left(1-x\right)-12x\left(x+3\right)+5x^2\left(1-x\right)\)

\(P=3x\left[4x-11-4\left(x+3\right)\right]\)

\(P=3x\left(4x-11-4x-12\right)\)

\(P=3x.132\)

\(P=396x\)

b/ Ta có \(\left|x\right|=2\)

<=> \(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

Thay x = 2 vào biểu thức P, ta có: P = 792

Tương tự với x = -2, ta cũng có: P = -792

Vậy \(P=\pm792\)khi \(\left|x\right|=2\)

c/ Để \(P=207\)

<=> \(396x=207\)

<=> \(x=\frac{207}{396}\)

Vậy \(x=\frac{207}{396}\)thì \(P=207\).

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

\(=4\cdot\left(-\dfrac{3}{2}\right)^2-6\cdot\dfrac{-3}{2}\)

\(=4\cdot\dfrac{9}{4}+6\cdot\dfrac{3}{2}\)

=9+9

=18