câu 1 tìm x,biết:
a) 2/3x(x2 - 4)=0
b) 2x2 - x -6=0
câu 2 tìm a để đa thức 2x3 - 3x2 + x + a chia hết cho x + 2
câu 3 tìm giá trị nhỏ nhất của biểu thức A= 4x2 - 4x +5
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Câu 2:
a: \(\Leftrightarrow3x^2+2x-1=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
b: \(\Leftrightarrow x^3-4x-x^3-8=4\)
hay x=-3
a)\(f\left(x\right)=2x^2-x-3+5=\left(x+1\right)\left(2x-3\right)+5\)
Để \(f\left(x\right)⋮g\left(x\right)\Leftrightarrow\left(x+1\right)\left(2x-3\right)+5⋮\left(x+1\right)\)
\(\Leftrightarrow5⋮\left(x+1\right)\)
mà \(x+1\in Z\Rightarrow x+1\in U\left(5\right)=\left\{-1;1;5;-5\right\}\)
\(\Leftrightarrow x\in\left\{-2;0;4;-6\right\}\)
Vậy...
b) \(f\left(x\right)=3x^2-4x+6=\left(3x^2-4x+1\right)+5=\left(3x-1\right)\left(x-1\right)+5\)
Để \(f\left(x\right)⋮g\left(x\right)\Leftrightarrow\left(3x-1\right)\left(x-1\right)+5⋮\left(3x-1\right)\)
\(\Leftrightarrow5⋮\left(3x-1\right)\) mà \(3x-1\in Z\Rightarrow3x-1\in U\left(5\right)=\left\{-1;1;5;-5\right\}\)
\(\Leftrightarrow x\in\left\{0;\dfrac{2}{3};2;-\dfrac{4}{3}\right\}\) mà x nguyên\(\Rightarrow x\in\left\{0;2\right\}\)
Vậy...
c)\(f\left(x\right)=\left(-2x^3-7x^2-5x+2\right)+3\)\(=\left(-2x^3-4x^2-3x^2-6x+x+2\right)+3\)\(=\left[-2x^2\left(x+2\right)-3x\left(x+2\right)+\left(x+2\right)\right]+3\)
\(=\left(x+2\right)\left(-2x^2-3x+1\right)+3\)
Làm tương tự như trên \(\Rightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Leftrightarrow x\in\left\{-5;-3;-1;1\right\}\)
Vậy...
d)\(f\left(x\right)=x^3-3x^2-4x+3=x\left(x^2-3x-4\right)+3=x\left(x+1\right)\left(x-4\right)+3\)
Làm tương tự như trên \(\Rightarrow x+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow x\in\left\{-4;-2;0;2\right\}\)
Vậy...
\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(a,f\left(x\right)⋮g\left(x\right)\\ \Leftrightarrow\dfrac{-x^4+2x^2-3x+5}{x-1}\in Z\\ \Leftrightarrow\dfrac{-x^4+x^3-x^3+x^2+x^2-x-2x+2+3}{x-1}\in Z\\ \Leftrightarrow\dfrac{-x^3\left(x-1\right)-x^2\left(x-1\right)+x\left(x-1\right)-2\left(x-1\right)+3}{x-1}\in Z\\ \Leftrightarrow-x^3-x^2+x-2+\dfrac{3}{x-1}\in Z\\ \Leftrightarrow3⋮x-1\\ \Leftrightarrow x-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\\ Mà.x< 0\\ \Leftrightarrow x=-2\\ b,B=\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4+4y^2-2024\\ B=\left(x-y\right)^2+4\left(x-y\right)+4+4y^2-2024\\ B=\left(x-y-2\right)^2+4y^2-2024\ge-2024\\ B_{min}=-2024\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)