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1: (3x+2)(x+2)(2x-1)
=(3x^2+6x+2x+4)(2x-1)
=(3x^2+8x+4)(2x-1)
=6x^3-3x^2+16x^2-8x+8x-4
=6x^3+13x^2-4
2: (5x+1)(x-1)+3x(2x+2)
=5x^2-5x+x-1+6x^2+6x
=11x^2+10x-1
3: 4x(2x+1)(x-1)+(x+5)(x-3)
=4x(2x^2-2x+x-1)+x^2+2x-15
=8x^3-4x^2-4x+x^2+2x-15
=8x^3-3x^2-2x-15
4: (2x-1)(x+2)(x-2)+(3x-1)(x-1)
=(2x-1)(x^2-4)+3x^2-4x+1
=2x^3-8x-x^2+4+3x^2-4x+1
=2x^3+2x^2-12x+5
\(a,-3x^2+7x-9+\left(x-1\right)\left(x+2\right)\\ =-3x^2+7x-9+x^2-x+2x-2\\ =\left(-3x^2+x^2\right)+\left(7x-x+2x\right)-\left(9+2\right)\\ =-2x^2+8x-11\\ b,x\left(x-5\right)-2x\left(x+1\right)\\ =x^2-5x-2x^2-2x\\ =\left(x^2-2x^2\right)-\left(5x+2x\right)\\ =-3x^2-7x\\ c,4x\left(x^2-x+1\right)-\left(x-1\right)\left(x^2-x\right)\\ =4x^3-4x^2+4x-x\left(x^2-x\right)+x^2-x\\ =4x^3-4x^2+4x-x^3+x^2+x^2-x\\ =\left(4x^3-x^3\right)+\left(-4x^2+x^2+x^2\right)+\left(4x-x\right)\\ =3x^3-2x^2+3x\\ =x\left(3x^2-2x+3\right)\)
\(d,-5x\left(x-5\right)+\left(x-3\right)\left(x^2-7\right)\\ =-5x^2+25x+x\left(x^2-7\right)-3\left(x^2-7\right)\\ =-5x^2+25x+x^3-7x-3x^2+21\\ =\left(-5x^2-3x^2\right)+\left(25x-7x\right)+x^3+21\\ =-8x^2+x^3+18x+21\)
\(A=2x^3+6x^2-3x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}^3+6\cdot\dfrac{1}{3}^2-3\cdot\dfrac{1}{3}+\dfrac{1}{2}\)
=13/54
b) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\Rightarrow\frac{3}{5}x=\left(-\frac{1}{7}\right)+\frac{1}{2}\)
\(\Rightarrow\frac{3}{5}x=\frac{5}{14}\)
\(\Rightarrow x=\frac{5}{14}:\frac{3}{5}\)
\(\Rightarrow x=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}.\)
c) \(5-\left|3x-1\right|=3\)
\(\Rightarrow\left|3x-1\right|=5-3\)
\(\Rightarrow\left|3x-1\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3:3\\x=\left(-1\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{1;-\frac{1}{3}\right\}.\)
d) \(\left(1-2x\right)^2=9\)
\(\Rightarrow\left(1-2x\right)^2=\left(\pm3\right)^2\)
\(\Rightarrow1-2x=\pm3.\)
\(\Rightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-2\right):2\\x=4:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{-1;2\right\}.\)
Chúc bạn học tốt!
a) 1/4(x-3)+2=1/5
1/4.(x-3) = 1/5-2
1/4.(x-3) = -9/5
x-3 = (-9/5):1/4
x-3 = -36/5
x = -36/5+3
x= -21/5
\(=\dfrac{6x^4-2x^3+5x^2-2}{3x^2-x+1}\)
\(=\dfrac{6x^4-2x^3+2x^2+3x^2-x+1+x-3}{3x^2-x+1}\)
\(=2x^2+1+\dfrac{x-3}{3x^2-x+1}\)
a: \(=\dfrac{5}{9}\cdot\left(-3\right)\cdot x^2y\cdot x^3y^6=-\dfrac{5}{3}x^4y^7\)
Bậc là 11
b: \(=\dfrac{-1}{3}\cdot\dfrac{3}{2}\cdot x^2y\cdot xy^3=-\dfrac{1}{2}x^3y^4\)
Bậc là 7
`(3x-1)(x-3)-2(x-3)=9`
`-> 3x(x-3)-1(x-3)-2x+6=9`
`-> 3x^2-9x-x+3-2x+6=9`
`-> 3x^2-12x+9=9`
`-> 3x^2-12x=0`
`-> x(3x-12)=0`
`->`\(\left[{}\begin{matrix}x=0\\3x-12=0\end{matrix}\right.\)
`->`\(\left[{}\begin{matrix}x=0\\3x=12\end{matrix}\right.\)
`->`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy, `x={0 ; 4}`.
\(\Leftrightarrow\left(3x+5\right)^2=3^2=\left(-3\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}3x+5=3\\3x+5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-2\\3x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-\dfrac{8}{3}\end{matrix}\right.\)
(3x + 5)2 = 9
(3x + 5)2 = 32
3x + 5 = 3
3x = 3 - 5
3x = -2
x = -2 : 3
X = \(\dfrac{-2}{3}\)
3
a)2|3x−1|+1=52|3x−1|=4|3x−1|=2⇒[3x−1=23x−1=−2+)3x−1=23x=3x=1.+)3x−1=−23x=−1x=−13.�)2|3�−1|+1=52|3�−1|=4|3�−1|=2⇒[3�−1=23�−1=−2+)3�−1=23�=3�=1.+)3�−1=−23�=−1�=−13.
vậy x = 1 hoặc x=−13.
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