Giải phương trình:
(x^2 + 3x + 2)(x^2 + 7x + 12)=24
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FW
2
OM
21 tháng 4 2017
(x^2+3x+2)(x^2+7X+12)=24
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
đặt \(x^2+5x+5=a\)=> ta có phương trình \(\Leftrightarrow\left(a-1\right)\left(a+1\right)=24\)
\(\Leftrightarrow a^2-1=24\)\(\Leftrightarrow a^2=25\Leftrightarrow a=\orbr{\begin{cases}5\\-5\end{cases}}\)
+)\(x^2+5x+5=5\)\(\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow x=\orbr{\begin{cases}5\\0\end{cases}}\)
+) \(x^2+5x+5=-5\)\(\Leftrightarrow x^2+5x+10=0\)\(\Rightarrowđenta=5^2-4.10=-15< 0\Rightarrow ptvonghiem\)
vậy \(x=\orbr{\begin{cases}0\\5\end{cases}}\)
17 tháng 5 2017
( x^2 + 3x + 2 )( x^2 + 7x + 12 ) = 24
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
Đặt x2 + 5x + 5 = a = ta có : \(\Leftrightarrow\left(a-1\right)\left(a+1\right)=24\)
\(\Leftrightarrow a^2-1=24\Leftrightarrow a^2=25\Leftrightarrow a=\orbr{\begin{cases}5\\-5\end{cases}}\)
+)\(x^2+5x+5=5\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow x=\orbr{\begin{cases}5\\0\end{cases}}\)
+)\(x^2+5x+5=-5\Leftrightarrow x^2+5x+10=0\)
\(\Rightarrowđenta=5^2-4.10=-15< 0\Rightarrow ptvonghiem\)
\(Vay.x=\orbr{\begin{cases}5\\0\end{cases}}\)
NV
29 tháng 6 2016
(x2 + 3x + 2)(x2 + 7x + 12) = 24
=> (x + 1)(x + 2)(x + 3)(x + 4) = 24
=> (x2 + 5x + 4)(x2 + 5x + 6) = 24
Đặt a = x2 + 5x + 4 ta được:
a.(a + 2) = 24 => a2 + 2a - 24 = 0 => (a - 4)(a + 6) = 0 => a = 4 hoặc a = -6
+ Với a = 4 => x2 + 5x + 4 = 4 => x2 + 5x = 0 => x(x + 5) = 0 => x = 0 hoặc x = -5
+ Với a = -6 => x2 + 5x + 4 = -6 => x2 + 5x + 10 = 0, mà x2 + 5x + 10 > 0 => vô nghiệm
Vậy x = 0 , x = -5
CC
4
29 tháng 12 2017
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
\(\Leftrightarrow\left(x^2+5x+5\right)^2-1=24\)
\(\Leftrightarrow\left(x^2+5x+5\right)^2=25\)
Mà \(x^2+5x+5>0\forall x\)
\(\Rightarrow x^2+5x+5=5\Rightarrow x\left(x+5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy pt có tập nghiệm S={0,-5}
29 tháng 12 2017
pt <=> (x+1).(x+2).(x+3).(x+4) = 24
<=> [(x+1).(x+4)].[(x+2).(x+3)] = 24
<=> (x^2+5x+4).(x^2+5x+6) = 24
<=> (x^2+5x+5)^2-1 = 24
<=> (x^2+5x+5) = 25
=> x^2+5x+5 = 5 [ vì x^2+5x+5 = (x+2,5)^2-0,25 >= -0,25 > -5 ]
=> x=0 hoặc x=-5
Vậy pt có tập nghiệm S = {-5;0}
k mk nha
TM
11 tháng 8 2021
1/ \(7x-5=13-5x\)
\(\Leftrightarrow12x=18\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)
==========
2/ \(19+3x=5-18x\)
\(\Leftrightarrow21x=-14\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
Vậy: \(S=\left\{-\dfrac{2}{3}\right\}\)
==========
3/ \(x^2+2x-4=-12+3x+x^2\)
\(\Leftrightarrow-x=-8\)
\(\Leftrightarrow x=8\)
Vậy: \(S=\left\{8\right\}\)
===========
4/ \(-\left(x+5\right)=3\left(x-5\right)\)
\(\Leftrightarrow-x-5=3x-15\)
\(\Leftrightarrow-4x=-10\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy: \(S=\left\{\dfrac{5}{2}\right\}\)
==========
5/ \(3\left(x+4\right)=\left(-x+4\right)\)
\(\Leftrightarrow3x+12=-x+4\)
\(\Leftrightarrow4x=-8\)
\(\Leftrightarrow x=-2\)
Vậy: \(S=\left\{-2\right\}\)
[----------]
NT
11 tháng 8 2021
1. \(7x-5=13-5x\) \(\Leftrightarrow12x=18\Leftrightarrow x=\dfrac{3}{2}\)
2. \(19+3x=5-18x\Leftrightarrow21x=-14\Leftrightarrow x=-\dfrac{2}{3}\)
3. \(x^2+2x-4=-12+3x+x^2\Leftrightarrow-x=-8\Leftrightarrow x=8\)
4. \(-\left(x+5\right)=3\left(x-5\right)\Leftrightarrow-x-5=3x-15\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{5}{2}\)
5. \(3\left(x+4\right)=-x+4\Leftrightarrow3x+12=-x+4\Leftrightarrow4x=-8\Leftrightarrow x=-2\)
12 tháng 5 2023
1:
a: =>3x=6
=>x=2
b: =>4x=16
=>x=4
c: =>4x-6=9-x
=>5x=15
=>x=3
d: =>7x-12=x+6
=>6x=18
=>x=3
2:
a: =>2x<=-8
=>x<=-4
b: =>x+5<0
=>x<-5
c: =>2x>8
=>x>4
15 tháng 7 2021
a) \(\left(3x-2\right)\left(3x-1\right)=\left(3x+1\right)^2\)
<=> \(9x^2-9x+2=9x^2+6x+1\)
<=> \(15x=1\) <=> \(x=\frac{1}{15}\)
b) \(\left(4x-1\right)\left(x+1\right)=\left(2x-3\right)^2\)
<=> \(4x^2+3x-1=4x^2-12x+9\)
<=> \(15x^2=10\) <=> \(x=\frac{2}{3}\)
c) \(\left(5x+1\right)^2=\left(7x-3\right)\left(7x+2\right)\) <=> \(25x^2+10x+1=49x^2-7x-6\)
<=> \(24x^2-17x-7=0\) <=> \(24x^2-24x+7x-7=0\)
<=> \(\left(24x+7\right)\left(x-1\right)=0\) <=> \(\orbr{\begin{cases}x=-\frac{7}{24}\\x=1\end{cases}}\)
15 tháng 7 2021
d) (4 - 3x)(4 + 3x) = (9x - 3)(1 - x)
<=> 16 - 9x2 = 12x - 9x2 - 3
<=> 12x = 19
<=> x = 19/12
e) x(x + 1)(x + 2)(x + 3) = 24
<=> (x2 + 3x)(x2 + 3x + 2) = 24
<=> (x2 + 3x)2 + 2(x2 + 3x) - 24 = 0
<=> (x2 + 3x)2 + 6(x2 + 3x) - 4(x2 + 3x) - 24 = 0
<=> (x2 + 3x + 6)(x2 + 3x - 4) = 0
<=> \(\orbr{\begin{cases}x^2+3x+6=0\\x^2+3x-4=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x+\frac{3}{2}\right)^2+\frac{15}{4}=0\left(vn\right)\\\left(x+4\right)\left(x-1\right)=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)
g) (7x - 2)2 = (7x - 3)(7x + 2)
<=> 49x2 - 28x + 4 = 49x2 - 7x - 6
<=> 21x = 10 <=> x = 10/21
LL
1
DH
6 tháng 9 2017
\(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Rightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)-24=0\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)
\(\Rightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24=0\)
\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
Đặt \(x^2+5x+4=t\Rightarrow x^2+5x+6=t+2\) ta được:
\(t\left(t+2\right)-24=0\Rightarrow t^2+2t-24=0\)
\(\Rightarrow t^2-4t+6t-24=0\Rightarrow\left(t-4\right)\left(t-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t-4=0\\t-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}t=4\\t=6\end{matrix}\right.\)
Vì \(t=x^2+5x+4\) nên
\(\left[{}\begin{matrix}x^2+5x+4=4\\x^2+5x+6=6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\left(x+5\right)=0\\x\left(x+5\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy................
Chúc bạn học tốt!!!
10 tháng 7 2019
\(a,\left(3x-2\right)\left(3x-1\right)=\left(3x+1\right)^2\)
\(9x^2-3x-6x+2=9x^2+6x+1\)
\(-9x+2-6x-1=0\)
\(-15x+1=0\)
\(-15x=-1\)
\(x=\frac{1}{15}\)
PT
3
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 2 2020
a/ \(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)
b/ ĐKXĐ; ...
\(\Leftrightarrow\frac{x^2}{x^2+4x+4}+12x+5=3x^2+6x+2\)
\(\Leftrightarrow\frac{x^2+\left(12x+5\right)\left(x^2+4x+4\right)}{x^2+4x+4}=3x^2+6x+2\)
\(\Leftrightarrow\frac{\left(4x+10\right)\left(3x^2+6x+2\right)}{x^2+4x+4}=3x^2+6x+2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2+6x+2=0\\\frac{4x+10}{x^2+4x+4}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x^2+6x+2=0\\x^2=6\end{matrix}\right.\)
8 tháng 2 2020
a) Ta có :
(x2 + 3x + 2)(x2 + 7x + 12) = 24
⇔ ( x + 1 ) ( x + 2 ) (x + 3 ) ( x + 4 ) = 24
⇔ ( x + 1 ) ( x + 4 ) ( x + 2 ) ( x + 3 ) - 24 = 0
⇔ ( x2 + 5x + 4 ) ( x2 + 5x + 6 ) - 24 = 0
Đặt t = x2 + 5x + 4, ta có :
t ( t + 2 ) - 24 = 0
⇔ t2 + 2t +1 - 25 = 0
⇔ ( t + 1 )2 - 52 = 0
⇔ ( t - 4 ) ( t + 6 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}t-4=0\\t+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+4=4\\x^2+5x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+5x=0\\x^2+5x+10=0\end{matrix}\right.\)
Sau đó tìm x bạn tự làm nha
Ý b) là - 3 à !?
(x+1)(x+2)(x+3)(x+4)=24
(x+1)(x+4)(x+2)(x+3)=24
(x\(^2\)+5x+4)(x2 +5x+6)=24
Đặt x2+5x+5=t
\(\Rightarrow\)(t+1)(t-1)=24
\(\Rightarrow\) t2 -1=24
\(\Rightarrow\) t2-25=0
\(\Rightarrow\) (t-5)(t+5)=0
\(\Rightarrow\) (x2+5x)(x2+5x+10)=0
\(\Rightarrow\) x(x+5)(x+5)2=0
\(\Rightarrow\) x(x+5)3=0
\(\Rightarrow\) x=0 hoặc (x+5)3=0
Vậy x=0 hoặc x= -5