6x^3-295x-7
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Phân tích đa thức thành nhân tử:
\(a,x^3-6x^2+11x-6\)
\(=x^3-x^2-5x^2+5x+6x-6\)
\(=x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-5x+6\right)\)
\(=\left(x-1\right)\left(x^2-3x-2x+6\right)\)
\(=\left(x-1\right)\left[x\left(x-3\right)-2\left(x-2\right)\right]\)
\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
bạn không biết làm câu b) à ? câu a) dễ hơn mà nên ai mà chả bít làm, trừ bạn "Sana" ở trên thôi !!!!!
Đặt \(\sqrt{2x^3+7}=a\)
=>6ax=3a^2+1+2x-4a
=>a=2x+1 hoặc a=1/3
=>2x^3+7=(2x+1)^2 hoặc 2x^3+7=1/3
=>\(x\in\left\{1;\dfrac{1-\sqrt{13}}{2};\sqrt[3]{-\dfrac{31}{9}}\right\}\)
a) \(6x^3-6x=0\Leftrightarrow6x\left(x^2-1\right)=0\Leftrightarrow6x\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)b) \(2x\left(3x+7\right)-6x^2=28\Leftrightarrow6x^2+14x-6x^2=28\Leftrightarrow14x=28\Leftrightarrow x=2\)
c) \(2\left(4x+4\right)-5\left(x-3\right)=0\Leftrightarrow8x+8-5x+15=0\Leftrightarrow3x=-23\Leftrightarrow x=-\dfrac{23}{3}\)
\(\frac{2x-3}{\left(7-6x\right)^2}+\frac{x-2}{\left(7-6x\right)^2}=\frac{6x-3}{\left(3x-5\right)^2}-\frac{12x-10}{\left(3x-5\right)^2}\)
\(\Leftrightarrow\frac{2x-3+x-2}{\left(7-6x\right)^2}=\frac{6x-3-12x+10}{\left(3x-5\right)^2}\)
\(\Leftrightarrow\frac{3x-5}{\left(7-6x\right)^2}=\frac{7-6x}{\left(3x-5\right)^2}\)
\(\Leftrightarrow\left(7-6x\right)^3=\left(3x-5\right)^3\)
\(\Leftrightarrow7-6x=3x-5\)
\(\Leftrightarrow7+5=3x+6x\)
\(\Leftrightarrow12=9x\)
\(\Leftrightarrow x=\frac{4}{3}\)
Vậy \(x=\frac{4}{3}\)
a. x2 - 6x = -9
<=> x2 - 6x + 9 = 0
<=> (x - 3)2 = 0
<=> x - 3 = 0
<=> x = 3
b. 2(x + 3) - x2 + 3x = 0
<=> 2(x + 3) - x(x + 3) = 0
<=> (2 - x)(x + 3) = 0
<=> \(\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
a/ (x-1)2-(4x+3)(2-x)=x2-2x+1-(8x-4x2+6-3x)
=x2-2x+1-8x+4x2-6+3x=5x2-7x-6
b/ (15x3y2 - 6x2y3) : 3x2y2 = 5x - 2y
c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)