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7 tháng 3 2021
b) Ta có: \(x^3+4x+5=0\)
\(\Leftrightarrow x^3-x+5x+5=0\)
\(\Leftrightarrow x\left(x^2-1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+5\right)=0\)
mà \(x^2-x+5>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
18 tháng 3 2021
a)x2-(x+3)(3x+1)=9
⇔(x-3)(x+3)-(x+3)(3x+1)=0
⇔x+3=0 hoặc 3x+1=0
1.x+3=0 ⇔x=-3
2.3x+1=0⇔x=-1/3
phương trình có 2 nghiệm x=-3 và x=-1/3
OM
6 tháng 5 2020
Giải phương trình
a, x2 - (x-3)(3x+1) = 9
\(\Leftrightarrow\) x2 - 3x2 + 8x +3 = 9
\(\Leftrightarrow\) -2x2 + 8x - 6 = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
b, (x+14)3 - (x+12)3 =1352
\(\Leftrightarrow\) (x+14-x-12)[(x+14)2 + (x+14)(x+12) + (x+12)2 ] = 1352
\(\Leftrightarrow\) 6(x2 + 28x + 196 + x2 + 26x + 168 + x2 +24x +144) =1352
\(\Leftrightarrow\) 18x2 +468x + 3048 = 1352
Pt nghiệm vô tỉ
DK
6 tháng 5 2020
a) \(x^2-\left(x-3\right)\left(3x+1\right)=9\)
\(\Leftrightarrow x^2-9-\left(x-3\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3-3x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy nghiệm của pt x = 3 hoặc x = 1
MA
27 tháng 9 2016
a) \(\frac{36\left(x-2\right)}{32-16x}=\frac{36\left(x-2\right)}{16\left(2-x\right)}=-\frac{36\left(2-x\right)}{16\left(2-x\right)}=-\frac{36}{16}=-\frac{9}{4}\)
b) \(\frac{3x^2-12x+12}{x^4-8x}=\frac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}=\frac{3x-6}{x^3+2x^2+4x}\)
c) \(\frac{7x^2+14x+7}{3x^2+3x}=\frac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}=\frac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\frac{7\left(x+1\right)}{3x}=\frac{7x+7}{3x}\)
d) \(\frac{x^4-5x^2+4}{x^4-10x^2+9}=\frac{x^4-x^2-4x^2+4}{x^4-x^2-9x^2+9}=\frac{x^2\left(x^2-1\right)-4\left(x^2-1\right)}{x^2\left(x^2-1\right)-9\left(x^2-1\right)}=\frac{\left(x^2-4\right)\left(x^2-1\right)}{\left(x^2-9\right)\left(x^2-1\right)}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\)
e) \(\cdot\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}=\frac{\left(x^3+1\right)\left(x+1\right)}{x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)}=\frac{\left(x+1\right)^2\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}=\frac{\left(x+1\right)^2}{x^2+1}=\frac{x^2+2x+1}{x^2+1}\)
13 tháng 6 2018
1/ (x+1)(-3)+5(x-4)=-3
\(\Leftrightarrow\)-3x - 3 + 5x - 20= -3
\(\Leftrightarrow\)2x - 23=-3
\(\Leftrightarrow\)x=10
2/3(5x-1) -x (x+1)+x2=14
\(\Leftrightarrow\)15x - 3 - x2 -x + x2=14
\(\Leftrightarrow\)14x=17
\(\Leftrightarrow\)x=17/14
3/2(x-1)-x(3-x)=x2
\(\Leftrightarrow\)2x - 2 - 3x + x2=x2
\(\Leftrightarrow\)2x-3x+x2-x2=2
\(\Leftrightarrow\)x= -2
4/ 3x(x+5)-2(x+5)=3x2
\(\Leftrightarrow\)3x2 + 15x - 2x - 10=3x2
\(\Leftrightarrow\)13x = 10
\(\Leftrightarrow\)x=10/13
5/ 4x(x+2)+x(4-x)=3x2+12
\(\Leftrightarrow\)4x2 + 8x + 4x - x2 = 3x2 + 12
\(\Leftrightarrow\)12x=12
\(\Leftrightarrow\)x=1
10 tháng 1 2018
2. \(x\left(x+2\right)\left(x+3\right)\left(x+5\right)=280\)
\(\Leftrightarrow x\left(x+5\right)\left(x+2\right)\left(x+3\right)=280\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+6\right)=280\)
Đặt \(x^2+5x+3=t\)
\(\Rightarrow\left(t-3\right)\left(t+3\right)=280\)
\(\Leftrightarrow t^2-9=280\)
\(\Leftrightarrow t^2=289\Leftrightarrow\left[{}\begin{matrix}t=17\\t=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+3=17\\x^2+5x+3=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x-14=0\\x^2+5x+20=0\end{matrix}\right.\)
\(\Leftrightarrow x^2+5x-14=0\text{(vì }x^2+5x+20=\left(x+\dfrac{5}{2}\right)^2+\dfrac{55}{4}>0\forall x\text{)}\)
\(\Leftrightarrow x^2-2x+7x-14=0\)
\(\Leftrightarrow x\left(x-2\right)+7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\)
\(\Leftrightarrow\) x - 2 = 0 hoặc x + 7 = 0
\(\Leftrightarrow\) x = 2 hoặc x = - 7
Vậy x = 2 hoặc x = -7.
10 tháng 1 2018
3. \(\left(x+3\right)\left(x+4\right)\left(x+5\right)=x\)
\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\left(x+5\right)-x=0\)
\(\Leftrightarrow x^3+12x^2+47x+60-x=0\)
\(\Leftrightarrow x^3+12x^2+46x+60=0\)
\(\Leftrightarrow x^3+6x^2+6x^2+36x+10x+60=0\)
\(\Leftrightarrow x^2\left(x+6\right)+6x\left(x+6\right)+10\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x^2+6x+10\right)=0\)
\(\Leftrightarrow x+6=0\text{(vì }x^2+6x+10=\left(x+3\right)^2+1>0\forall x\text{)}\)
\(\Leftrightarrow x=-6\)
Vậy x = -6.
23 tháng 3 2020
\(\text{a) 2(x+3)-3(x-1)=2}\)
\(2x+6-3x+3=2\)
\(2x-3x=2-3-6\)
\(-x=-7\)
\(x=7\)
\(\text{b) 7-(x-2)=5(2x-3)}\)
\(7-x+2=10x-15\)
\(-x-10x=-15-2-7\)
\(-11x=-24\)
\(x=-24:\left(-11\right)\)
\(x=\frac{24}{11}\)
\(\text{c) 32-4(0,5y-5)=3y+2}\)
\(32-2y+20=3y+2\)
\(-2y-3y=2-20-32\)
\(-y=-50\)
\(y=50\)
\(\text{d) 3(x-1)-x=2x-3}\)
\(3x-3-x=2x-3\)
\(3x-x-2x=-3+3\)
\(0=0\)( vô nghiệm )
23 tháng 3 2020
a) 2(x + 3) - 3(x - 1) = 2
<=> 2x + 6 - 3x + 3 = 2
<=> -x + 9 = 2
<=> -x = -2 - 9
<=> -x = -7
<=> x = 7
b) 7 - (x - 2) = 5(2x - 3)
<=> 7 - x + 2 = 10x - 15
<=> 9 - x = 10x - 15
<=> 9 - x - 10 = -15
<=> 9 - 11x = -15
<=> -11x = -15 - 9
<=> -11x = -24
<=> x = 24/11
c) 32 - 4(0,5y - 5) = 3y + 2
<=> 32 - 2y + 20 = 3y + 2
<=> 52 - 2y = 3y + 2
<=> 52 - 2y - 3y = 2
<=> 52 - 5y = 2
<=> -5y = 2 - 52
<=> -5y = -50
<=> y = 10
Cảm ơn bạn nhé
3: Đặt x+3=a
Ta có: (x+3)(x+4)(x+5)=x
⇔a(a+1)(a+2)=a-3
⇔\(a^3+3a^2+2a-a+3=0\)
\(\Leftrightarrow a^3+3a^2+a+3=0\)
\(\Leftrightarrow a^2\left(a+3\right)+\left(a+3\right)=0\)
\(\Leftrightarrow\left(a+3\right)\left(a^2+1\right)=0\)(1)
Ta có: \(a^2\ge0\forall a\)
\(\Rightarrow a^2+1\ge1>0\forall a\)(2)
Từ (1) và (2) suy ra a+3=0
hay \(x+6=0\)
⇔x=-6
Vậy: x=-6