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\(a,=16x^8+8x^6\\ b,=4x^4-6x^5-4x^3\\ c,=15x^6+9x^3y-10x^3y-6y^2\\ =15x^6-x^3y-6y^2\\ d,=2a^4-a^3b+6a^2b-3ab^2-3ab^2+b^3\\ =2a^4-a^3b+6a^2b-6ab^2+b^3\)
1.
\(x^2-5x+6=0\\ \Rightarrow x^2-2x-3x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
2.
\(\left(x+4\right)^2-\left(3x-1\right)^2=0\\ \Rightarrow\left(x+4-3x+1\right)\left(x+4+3x-1\right)=0\\ \Rightarrow\left(-2x+5\right)\left(4x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2x+5=0\\4x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
3.
\(x^2-2x+24=0\\ \Rightarrow\left(x^2-2x+1\right)+23=0\\ \Rightarrow\left(x-1\right)^2+23=0\)
Vì (x-1)2≥0
23>0
\(\Rightarrow\left(x-1\right)^2+23>0\)
Vậy x vô nghiệm
4.
\(9x^2-4=0\\ \Rightarrow\left(3x-4\right)\left(3x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-4=0\\3x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
5.
\(x^2+2x-8=0\\ \Rightarrow\left(x^2+2x+1\right)-9=0\\ \Rightarrow\left(x+1\right)^2-3^2=0\\ \Rightarrow\left(x-2\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
a) \(2x\left(x^2-7x-3\right)=2x.x^2-2x.7x-2x.3=2x^3-14x^2-6x\)
b) \(\left(-2x^3+y^2-7xy\right)4xy^2=\left(-2x^3\right)4xy^2+y^24xy^2-7xy.4xy^2=-8x^4y^2+4xy^4-28x^2y^3\)
c) \(\left(-5x^3\right)\left(2x^2+3x-5\right)=-5x^32x^2-5x^33x-5x^3.-5=-10x^5-15x^4+25x^3\)
d) \(\left(2x^2-xy+y^2\right)\left(-3x^3\right)=-3x^32x^2-3x^3.-xy-3x^3y^2=-6x^5+3x^4y-3x^3y^2\)
e) \(\left(x^2-2x+3\right)\left(x-4\right)=x\left(x^2-2x+3\right)-4\left(x^2-2x+3\right)=x^3-2x^2+3x-4x^2+8x-12=x^3-6x^2+11x-12\)
f) \(\left(2x^3-3x-1\right)\left(5x+2\right)=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)=10x^4-15x^2-5x+4x^3-6x-2=10x^4+4x^3-15x^2-11x-2\)
\(A=4-6x-x^2=-\left(x^2+6x-4\right)=-\left(x^2+6x+9-13\right)\)
\(=-\left[\left(x+3\right)^2-13\right]=-\left(x+3\right)^2+13\le13\)
Vậy \(A_{max}=13\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
\(B=3x^2-6x+1=\left(\sqrt{3}x\right)^2-2.\sqrt{3}x.\sqrt{3}+3-2\)
\(=\left(\sqrt{3}x-\sqrt{3}\right)^2-2\ge-2\)
Vậy \(B_{min}=-2\Leftrightarrow\sqrt{3}x-\sqrt{3}=0\Leftrightarrow x=1\)
\(C=5x^2-2x-3=\left(\sqrt{5}x\right)^2-2.\sqrt{5}x.\frac{1}{\sqrt{5}}+\frac{1}{5}-\frac{16}{5}\)
\(=\left(\sqrt{5}x-\frac{1}{\sqrt{5}}\right)^2-\frac{16}{5}\ge-\frac{16}{5}\)
Vậy \(C_{min}=-\frac{16}{5}\Leftrightarrow\sqrt{5}x-\frac{1}{\sqrt{5}}=0\Leftrightarrow\sqrt{5}x=\frac{1}{\sqrt{5}}\Leftrightarrow x=\frac{1}{5}\)
a: x^3-7x-6
=x^3-x-6x-6
=x(x-1)(x+1)-6(x+1)
=(x+1)(x^2-x-6)
=(x-3)(x+2)(x+1)
b: =2x^3+x^2-2x^2-x+6x+3
=x^2(2x+1)-x(2x+1)+3(2x+1)
=(2x+1)(x^2-x+3)
c: =2x^3-3x^2-2x^2+3x+2x-3
=x^2(2x-3)-x(2x-3)+(2x-3)
=(2x-3)(x^2-x+1)
d: =2x^3+x^2+2x^2+x+2x+1
=(2x+1)(x^2+x+1)
e: =3x^3+x^2-3x^2-x+6x+2
=(3x+1)(x^2-x+2)
f: =27x^3-9x^2-18x^2+6x+12x-4
=(3x-1)(9x^2-6x+4)
a) \(x^3-7x-6\)
\(=x^3-x-6x-6\)
\(=\left(x^3-x\right)-\left(6x+6\right)\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)\)
b) \(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(x^2-x+3\right)\left(2x+1\right)\)
c) \(2x^3-5x^2+5x+1\)
\(=2x^3-3x^2-2x^2+3x+2x-3\)
\(=\left(2x^3-3x^2\right)-\left(2x^2-3x\right)+\left(2x-3\right)\)
\(=x^2\left(2x-3\right)-x\left(2x-3\right)+\left(2x-3\right)\)
\(=\left(x^2-x+1\right)\left(2x-3\right)\)
d) \(2x^3+3x^2+3x+1\)
\(=2x^3+x^2+2x^2+x+2x+1\)
\(=\left(2x^3+x^2\right)+\left(2x^2+x\right)+\left(2x+1\right)\)
\(=x^2\left(2x+1\right)+x\left(2x+1\right)+\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(3x^3-2x^2+5x+2\)
\(=3x^3+x^2-3x^2-x+6x+2\)
\(=\left(3x^3+x^2\right)-\left(3x^2+x\right)+\left(6x+2\right)\)
\(=x^2\left(3x+1\right)-x\left(3x+1\right)+2\left(3x+1\right)\)
\(=\left(3x-1\right)\left(x^2-x+2\right)\)
f) \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
a) 2x. (x2 – 7x -3)
= 2x3- 14x2- 6x
b) ( -2x3 + y2 -7xy). 4xy2
= -8x4y2+ 4xy4- 28x2y3
c)(-5x3).(2x2+3x-5)
= -10x5-15x4+25x3
d) (2x2 - xy+ y2).(-3x3)
=-6x5+ 3x4y -3x3y2
e)(x2 -2x+3). (x-4)
=x3-2x2+3x -4x2+8x-12
=x3-6x2+11x-12
f) ( 2x3 -3x -1). (5x+2)
=10x4-15x2-5x +4x3-6x-2
=10x4+4x3-15x2-11x-2
\(a,\left(3x+1\right)\left(3x-1\right)-\left(18x^3+5x^2-2x\right):2x\\ =\left(9x^2-1\right)-\left(9x^2+\dfrac{5}{2}x-1\right)\\ =9x^2-1-9x^2-\dfrac{5}{2}x+1=\dfrac{5}{2}x\)
\(b,3x\left(x-2021\right)-x+2021=0\\ \Rightarrow b,3x\left(x-2021\right)-\left(x-2021\right)=0\\ \Rightarrow\left(x-2021\right)\left(3x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{3}\end{matrix}\right.\)
1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)
\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)
\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)
\(\Leftrightarrow-41x=-115\)
hay \(x=\dfrac{115}{41}\)
2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)
\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)
\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)
\(\Leftrightarrow x^3=64\)
hay x=4
3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)
Bài 1:
a) \(3x\left(5x^2-2x+1\right)\)
\(=15x^3-6x^2+3x\)
b) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^2\left(x^2-1\right)+2x\left(x^2-1\right)\)
\(=x^4-x^2+2x^3-2x\)
\(=x^4+2x^3-x^2-2x\)
Bài 2:
a) \(3x^2=2x\)
\(\Leftrightarrow3x^2-2x=0\)
\(\Leftrightarrow x\left(3x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)
b)\(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-2x-1=12x-5\)
\(\Leftrightarrow14x=4\Leftrightarrow x=\frac{2}{7}\)
