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Câu 1: Đặt tính chia, kết quả là x - 3 dư 6
Câu 2:
a) \(P=\left(x+1\right)^3+\left(x+1\right)\left(6-x^2\right)-12\)
\(\Leftrightarrow P=\left(x+1\right)\left(x^2+2x+1\right)+\left(x+1\right)\left(6-x^2\right)-12\)
\(\Leftrightarrow P=\left(x+1\right)\left(x^2+2x+1+6-x^2\right)-12\)
\(\Leftrightarrow P=\left(x+1\right)\left(7+2x\right)-12\)
b) \(x=-\dfrac{1}{2}\) thì giá trị P là:
\(\Leftrightarrow P=\left(-\dfrac{1}{2}+1\right)\left(7-2.\dfrac{1}{2}\right)-12\)
\(\Leftrightarrow P=3-12\)
\(\Leftrightarrow P=-9\)
c) \(P=0\)
\(\Leftrightarrow\left(x+1\right)\left(7+2x\right)-12=0\)
\(\Leftrightarrow\left(x+1\right)\left(7+2x\right)=12\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
(Sai sót mong bạn thông cảm)
Câu 1:
\(\dfrac{x^2-6x+15}{x-3}=\dfrac{x^2-6x+9+6}{x-3}=\left(x-3\right)+\dfrac{6}{x-3}\)
=>Số dư là 6
Câu 2:
a: \(P=x^3+3x^2+3x+1+6x-x^3+6-x^2-12\)
\(=2x^2+9x-5\)
b: Khi x=-1/2 thì \(P=2\cdot\dfrac{1}{4}-\dfrac{9}{2}-5=\dfrac{1}{2}-\dfrac{9}{2}-5=-9\)
c: Để P=0 thì 2x^2+9x-5=0
hay \(x\in\left\{\dfrac{1}{2};-5\right\}\)
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(P=\left(\dfrac{3x^2+3x-3}{x^2+x-2}+\dfrac{1}{x-1}+\dfrac{1}{x+2}-2\right):\dfrac{1}{x^2-1}\left(dk:x\ne-2,x\ne\pm1\right)\)
\(=\left(\dfrac{3x^2+3x-3}{\left(x-1\right)\left(x+2\right)}+\dfrac{1}{x-1}+\dfrac{1}{x+2}-2\right).\left(x^2-1\right)\)
\(=\left(\dfrac{3x^2+3x-3+x+2+x-1-2\left(x^2+x-2\right)}{\left(x-1\right)\left(x+2\right)}\right).\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{3x^2+5x-2-2x^2-2x+4}{x+2}.\left(x+1\right)\\ =\dfrac{x^2+3x+2}{x+2}.\left(x+1\right)\)
\(=\dfrac{x^2+x+2x+2}{x+2}.\left(x+1\right)\\ =\dfrac{x\left(x+1\right)+2\left(x+1\right)}{x+2}.\left(x+1\right)\\ =\dfrac{\left(x+1\right)^2\left(x+2\right)}{x+2}\\ =x^2+2x+1\)
Ta có :
\(x^2-x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\ \Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)
Với \(x=3\) thì \(P=x^2+2x+1=\left(x+1\right)^2=\left(3+1\right)^2=16\)
Vậy ...
\(a,P=\dfrac{2x^2+2x+2+2x-1+x^2+6x+2}{\left(x-1\right)\left(x^2+x+1\right)}\\ P=\dfrac{3x^2+10x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(P=\dfrac{-x^4+2x^3-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)
\(=\dfrac{\left(1-x^2\right)\left(1+x^2\right)+2x\left(x^2-1\right)}{4x^2-1}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(=\dfrac{\left(1-x^2\right)\left(1+x^2-2x\right)}{4x^2-1}+\dfrac{2}{2x+1}\)
\(=\dfrac{\left(1-x^2\right)\left(x^2-2x+1\right)+4x-2}{4x^2-1}\)
1/\(\left(x^2-6x+15\right):\left(x-3\right)\)
Đặt cột dọc ta được x-3 dư 6
2/a/\(p=\left(x+1\right)^3+\left(x+1\right)\left(6-x^2\right)-12\)
\(=x^3+3x^2+3x+1+6x-x^3+6-x^2-12\)
\(=2x^2+9x-11\)
b/thay x = -1/2 ta đc \(2.-\left(\frac{1}{2}\right)^2+9.-\frac{1}{2}-11\)
\(=\frac{1}{2}+\left(-\frac{9}{2}\right)-11\)
\(=\left(-15\right)\)