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PT
0
22 tháng 1 2019
a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\\x=5\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)
b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)
\(\Leftrightarrow3x\left(x-4\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;4\right\}\)
c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)
\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)
\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)
d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)
\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)
e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(6x+1-x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)
g. \(\left(2x-1\right)^2=49\)
\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)
16 tháng 6 2023
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
22 tháng 9 2021
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
F
5 tháng 10 2020
a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)
\(\Leftrightarrow2x=-40\)
\(\Rightarrow x=-20\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=-12\)
\(\Rightarrow x=-3\)
c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)
\(\Leftrightarrow-14x=14\)
\(\Rightarrow x=-1\)
F
5 tháng 10 2020
d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(\Leftrightarrow17x=-34\)
\(\Rightarrow x=-2\)
e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x=24\)
\(\Rightarrow x=1\)
(3x+4)2-(3x-1).(3x+1)=49
<=> 9x2+24x+16-(9x2-1)=49
<=>9x2+24x+16-9x2+1=49
<=>24x+17=49
<=>24x =32
<=>x =4/3
Vậy ...
(x+2).(x^2-2x+4)-x.(x+3).(x-3)
=x3+8-x(x2-9)
=x3+8-x3+9x
=9x+8
(3x+4)2-(3x-1).(3x+1)=49
<=> 9x2+24x+16-(9x2-1)=49
<=>9x2+24x+16-9x2+1=49
<=>24x+17=49
<=>24x =32
<=>x =4/3
Vậy ...
(x+2).(x^2-2x+4)-x.(x+3).(x-3)
=x3+8-x(x2-9)
=x3+8-x3+9x
=9x+8