a)(x-1)(x^2+5x-2)-x^3+1=0
b)5(x^2+3x)-9(3x+3)=x^2-36
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11 tháng 3 2020
a) ( 4x - 1 ) (x - 3) - ( x - 3 ) ( 5x + 2 ) = 0
<=> (x - 3)(4x - 1 - 5x - 2) = 0
<=> (x - 3)(-x - 3) = 0
<=> x = 3 hoặc x = -3
b) ( x + 3 ) ( x - 5 ) + ( x + 3 ) ( 3x - 4) = 0
<=> (x + 3)(x - 5 + 3x - 4) = 0
<=> (x + 3)(4x - 9) = 0
<=> x = -3 hoặc x = 9/4
c) ( x + 6 ) ( 3x - 1 )+ x2 - 36 = 0
<=> 3x^2 + 17x - 6 + x^2 - 36 = 0
<=> 4x^2 + 17x - 42 = 0
<=> 4x^2 + 24x - 7x - 42 = 0
<=> 4x(x + 6) - 7(x + 6) = 0
<=> (4x - 7)(x + 6) = 0
<=> x = -6 hoặc x = 7/4
d) ( x + 4 ) ( 5x + 9 ) - x2 + 16 = 0
<=> 5x^2 + 29x + 36 - x^2 + 16 = 0
<=> 4x^2 + 29x + 52 = 0
<=> 4x^2 + 16x + 13x + 42 = 0
<=> 4x(x + 4) + 13(x + 4) = 0
<=> (4x + 13)(x + 4) = 0
<=> x = -13/4 và x = -4
7 tháng 5 2020
a)
\(\left(5x+3\right)\cdot\left(x^2+4\right)\cdot\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x+3=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{5}\\x=4\end{matrix}\right.\)
b)
\(\left(4x-1\right)\cdot\left(x-3\right)-\left(x-2\right)\cdot\left(5x+2\right)=0\\ \Leftrightarrow4x^2-12x-x+3-5x^2-2x+10x+4=0\\ \Leftrightarrow-x^2-5x+7=0\\ \Rightarrow x=\left[{}\begin{matrix}-\frac{5+\sqrt{53}}{2}\\-\frac{5-\sqrt{53}}{2}\end{matrix}\right.\)
c)
\(\left(x+3\right)\cdot\left(x-5\right)+\left(x+3\right)\cdot\left(3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(x-5+3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(4x-9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)
d)
\(\left(x+6\right)\cdot\left(3x-1\right)+x^2-36=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x^2-36\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x+6\right)\cdot\left(x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1+x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(4x-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)
e)
\(0.75x\cdot\left(x+5\right)=\left(x+5\right)\cdot\left(3-1.25x\right)\\ \Leftrightarrow0.75x\cdot\left(x+5\right)-\left(x+5\right)\cdot\left(3-1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(0.75x-3+1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(2x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.\)
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
24 tháng 7 2019
Bài 1:
a) -6x + 3(7 + 2x)
= -6x + 21 + 6x
= (-6x + 6x) + 21
= 21
b) 15y - 5(6x + 3y)
= 15y - 30 - 15y
= (15y - 15y) - 30
= -30
c) x(2x + 1) - x2(x + 2) + (x3 - x + 3)
= 2x2 + x - x3 - 2x2 + x3 - x + 3
= (2x2 - 2x2) + (x - x) + (-x3 + x3) + 3
= 3
d) x(5x - 4)3x2(x - 1) ??? :V
Bài 2:
a) 3x + 2(5 - x) = 0
<=> 3x + 10 - 2x = 0
<=> x + 10 = 0
<=> x = -10
=> x = -10
b) 3x2 - 3x(-2 + x) = 36
<=> 3x2 + 2x - 3x2 = 36
<=> 6x = 36
<=> x = 6
=> x = 5
c) 5x(12x + 7) - 3x(20x - 5) = -100
<=> 60x2 + 35x - 60x2 + 15x = -100
<=> 50x = -100
<=> x = -2
=> x = -2
6 tháng 2 2020
Bài 1 :
a, \(\left(4x-1\right)\left(x-3\right)-\left(x-3\right)\left(5x+2\right)=0\)
=> \(\left(x-3\right)\left(4x-1-5x-2\right)=0\)
=> \(\left(x-3\right)\left(-x-3\right)=0\)
=> \(\left[{}\begin{matrix}x-3=0\\-x-3=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=\pm3\) .
b, \(\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\)
=> \(\left(x+3\right)\left(x-5+3x-4\right)=0\)
=> \(\left(x+3\right)\left(4x-9\right)=0\)
=> \(\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=-3,x=\frac{9}{4}\) .
c, \(\left(x+6\right)\left(3x-1\right)+x^2-36=0\)
=> \(\left(x+6\right)\left(3x-1\right)+\left(x-6\right)\left(x+6\right)=0\)
=> \(\left(x+6\right)\left(3x-1+x-6\right)=0\)
=> \(\left(x+6\right)\left(4x-7\right)=0\)
=> \(\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=-6,x=\frac{7}{4}\) .
6 tháng 2 2020
a) ( 4x - 1 ) ( x - 3 ) - ( x - 3 ) ( 5x + 2 ) = 0
⇔ ( x - 3 ) ( 4x - 1 - 5x - 2 ) = 0
⇔ ( x - 3 ) ( -x - 3 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Ý b) tương tự ý a) thôi.
c) ( x + 6 ) ( 3x - 1 ) + x2 - 36 = 0
⇔ ( x + 6 ) ( 3x - 1 ) + ( x + 6 ) ( x - 6 ) = 0
⇔ (x+6)(3x-1+x-6)=0
⇔ (x+6)(4x-7)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)
PN
13 tháng 6 2019
a) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)-3=-3\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3-3=-3\)
\(\Leftrightarrow14x=0\)
\(\Leftrightarrow x=0\)
Vậy pt có nghiệm duy nhất x = 0.
b) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)
\(\Leftrightarrow6x^2+19x-7-6x^2-x+5=7\)
\(\Leftrightarrow18x-2=7\)
\(\Leftrightarrow18x=9\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy pt có nghiệm duy nhất \(x=\frac{1}{2}\)
c) \(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\)
\(\Leftrightarrow36x^2-24x+4+25x^2-20x+4-60x^2+33x-8=0\)
\(\Leftrightarrow x^2-11x=0\)
\(\Leftrightarrow x\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=11\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{0;11\right\}\)
d) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow x^2-6x+9-x^2-4x+32=1\)
\(\Leftrightarrow41-10x=1\)
\(\Leftrightarrow-10x=40\)
\(\Leftrightarrow x=-4\)
Vậy pt có nghiệm duy nhất x = -4.
e) \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)
\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-4x+1-7x^2+36=36\)
\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2=0\)
\(\Leftrightarrow8x=-13\)
\(\Leftrightarrow x=-\frac{13}{8}\)
Vậy pt có nghiệm duy nhất \(x=-\frac{13}{8}\)
a, Ta có : \(\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\)
=> \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)
=> \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)
=> \(\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\)
=> \(\left(x-1\right)\left(4x-3\right)=0\)
=> \(\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{1,\frac{3}{4}\right\}\)
b, Ta có : \(5\left(x^2+3x\right)-9\left(3x+3\right)=x^2-36\)
=> \(5x^2+15x-27x-27=x^2-36\)
=> \(5x^2+15x-27x-27-x^2+36=0\)
=> \(4x^2-12x+9=0\)
=> \(\left(2x-3\right)^2=0\)
=> \(x=\frac{3}{2}\)
Vậy phương trình có tập nghiệm là \(S=\left\{\frac{3}{2}\right\}\)
\(a.\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{1;\frac{3}{4}\right\}\)
\(b.5\left(x^2+3x\right)-9\left(3x+3\right)=x^2-36\\ \Leftrightarrow5x^2+15x-27x-27=x^2-36\\ \Leftrightarrow5x^2+15x-27x-27-x^2+36=0\\ \Leftrightarrow4x^2-12x+9=0\\ \Leftrightarrow\left(2x-3\right)^2=0\\ \Leftrightarrow x=\frac{3}{2}\)
Vậy pt có tập nghiệm \(S=\left\{\frac{3}{2}\right\}\)
Chúc bạn học tốt!!!!!!!!!!!