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P= 3x ( 4x - 11 ) + 5x2 ( x - 1 ) - 4x ( 3x + 9 ) + x ( 5x - 5x2 )
\(P=12x^2-33x+5x^3-5x^2-12x^2-36x+5x^2-5x^3\)
\(P=-69x\)
b)\(TH1:x=2\). PT có dạng:
\(-69x=-69.2=-138\)
\(TH2:x=-2\). PT có dạng
\(-69x=-69.\left(-2\right)=138\)
c)Tại P=207 ta đc:
\(-69x=207\Rightarrow x=-3\)
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/ \(P=3x\left(4x-11\right)+5x^2\left(x-1\right)-4x\left(3x+9\right)+x\left(5x-5x^2\right)\)
\(P=3x\left(4x-11\right)+5x^2\left(x-1\right)-4x.3\left(x+3\right)+x.5x\left(1-x\right)\)
\(P=3x\left(4x-11\right)-5x^2\left(1-x\right)-12x\left(x+3\right)+5x^2\left(1-x\right)\)
\(P=3x\left[4x-11-4\left(x+3\right)\right]\)
\(P=3x\left(4x-11-4x-12\right)\)
\(P=3x.132\)
\(P=396x\)
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 \(P=207\).
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)
\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)
\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)
\(=-10x^3+19x^2+74x+1\)
\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)
\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)
\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)
\(=-5x^4-11x^3+24x^2+12x+7\)
\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)
\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)
\(=-2x^2-27x+57\)
\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)
\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)
\(=-x^3+4x^2+22x+5\)
\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)
\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)
\(=-9x^3-55x^2+4x+35\)
\(g,\left(x-1\right)^2-\left(x+2\right)^2\)
\(=x^2-2x+1-x^2-4x-4\)
\(=-6x-3\)
a: \(P=\dfrac{x}{x+3}-\dfrac{x^2-5x-6}{\left(x-3\right)\left(x+3\right)}+\dfrac{3}{x-3}\)
\(=\dfrac{x^2-3x-x^2+5x+6+3x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{5}{x-3}\)
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