giai phuong trinh x(x+1)(x+2)(x+3)=24
Giai phuong trinh
(x+1)2(x-2)+(x+1)2(x+2)=-24
ta có pt
<=>\(\left(x+1\right)^2\left(x-2+x+2\right)=-24\Leftrightarrow2x\left(x+1\right)^2=-24\Leftrightarrow x\left(x^2+2x+1\right)=-12\)
<=>\(x^3+2x^2+x+12=0\Leftrightarrow x^3+3x^2-x^2-3x+4x+12=0\Leftrightarrow\left(x+3\right)\left(x^2-x+4\right)=0\)
đến đây thì dễ rồi nhé ^_^
giai phuong trinh :
(x+1)^2 * (x+2)+(x+1)^2 *(x-2)=-24
(x+1)^2*(x+2)+(x+1)^2*(x-2)=-24
<=>(x+1)^2*(x+2+x-2)=-24
<=>2x(x+1)^2=-24
<=>x(x^2+2x+1)=-12
<=>x^3+2x^2+x+12=0
<=>(x+3)(x^2-x+4)=0
<=>(x+3)((x+1/2)^2+15/4)=0
<=> x+3=0 hay (x+1/2)^2+15/4=0
<=>x=-3 hay (x+1/2)^2=-15/4(sai)
vậy x=-3
(x+1)^2*(x+2)+(x+1)^2*(x-2)=-24
<=>(x+1)^2*(x+2+x-2)=-24
<=>2x(x+1)^2=-24
<=>x(x^2+2x+1)=-12
<=>x^3+2x^2+x+12=0
<=>(x+3)(x^2-x+4)=0
<=>(x+3)((x+1/2)^2+15/4)=0
<=> x+3=0 hay (x+1/2)^2+15/4=0
<=>x=-3 hay (x+1/2)^2=-15/4(sai)
vậy x=-3
giup minh bai nay:
giai phuong trinh:
x(x+1)(x-1)(x+2)=24
2x(8x-1)^2(3x+2)(2x+1)=3
thanks!
a) [x(x+1].[(x-1)(x+2)]=24
(x2+x)(x2+x+2)=24
Dat x2+x=a , ta dc: a(a+2)=24
=> a2+2a-24=0
=> (a-4)(a+6)=0
=> a=4 hoac a=-6
Thay vao roi tu tim x nha
b)
giai phuong trinh
a , x(x+1)(x-1)(x+2)= 24
b , (x-4)(x-5)(x-6)(x-7) = 1680
b) \(\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)=1680\)
\(\Leftrightarrow\left(x-4\right)\left(x-7\right)\left(x-5\right)\left(x-6\right)=1680\)
\(\Leftrightarrow\left(x^2-11x+28\right)\left(x^2-11x+28+2\right)-1680=0\)
\(\Leftrightarrow\left(x^2-11x+28\right)^2+2\left(x^2-11x+28\right)+1-1681=0\)
\(\Leftrightarrow\left(x^2-11x+28+1\right)^2-41^2=0\)
\(\Leftrightarrow\left(x^2-11x+29-41\right)\left(x^2-11x+29+41\right)=0\)
\(\Leftrightarrow\left(x^2-11x-12\right)\left(x^2-11x+70\right)=0\)
Th1: \(x^2-11x-12=0\Leftrightarrow x^2+x-12x-12=0\Leftrightarrow\left(x-12\right)\left(x+1\right)=0\)
\(\Leftrightarrow x-12=0\Leftrightarrow x=12\) hoặc \(x+1=0\Leftrightarrow x=-1\)
Th2:\(x^2-11x+70=0\Leftrightarrow x^2-2.x.\frac{11}{2}+\left(\frac{11}{2}\right)^2+\frac{159}{4}=0\Leftrightarrow\left(x-\frac{11}{2}\right)^2+\frac{159}{4}=0\)
Vì\(\left(x-\frac{11}{2}\right)^2\ge0\Rightarrow\left(x+\frac{11}{2}\right)^2+\frac{159}{4}\ge\frac{159}{4}\)
Mà ta có \(\left(x+\frac{11}{2}\right)^2+\frac{159}{4}=0\) Nên k có giá trị của x
Vậy tập nghiệm của phương trình là \(S=\left\{12;-1\right\}\)
a) x=-3,
x=2;
x = -(căn bậc hai(3)*căn bậc hai(5)*i+1)/2;
x = (căn bậc hai(3)*căn bậc hai(5)*i-1)/2;
giai phuong trinh 3∛x-3 +4∛8x-24 -1/3∛27x-81 =-20
a, (3x-7)^2-4(x+1)^2=0
b,(x+1)^2.(x+2)+(x+1)^2.(x-2)=-24
giai phuong trinh nha
cho phuong trinh (an x):(m-1)x +m^2 -1=0(1) a,giai phuong trinh (1) voi m=2 b, tim gia tri cua m sao cho phuong trinh (1) nhan x=3 lam nghiem
mk dang can gap .cam on truoc a
giai phuong trinh (x-1)(x-4) + (x-3)(x-2)=2
x2 - 5x + 4 + x2 - 5x + 6 = 2
<=> 2x2 - 10x + 8 = 0
<=> x2 - 5x + 4 = 0
<=> x = 1 hoặc x = 4
X^2-4x-x+4+x^2-2x-3x+6=2 rút gọn và chuyển vế : 2x^2-10x+8=0 bấm máy tính ; x=4 và x=1
\(\left(x-1\right)\left(x-4\right)+\left(x-3\right)\left(x-2\right)=2\)
\(< =>x^2-4x-x+4+x^2-2x-3x+6=2\)
\(< =>2x^2-10x+10-2=0\)
\(< =>2x^2-10x+8=0\)
\(< =>x^2-5x+4=0\)
\(< =>x^2-x-4\left(x-1\right)=0\)
\(< =>\left(x-4\right)\left(x-1\right)=0\)
\(< =>\orbr{\begin{cases}x=1\\x=4\end{cases}}\)
Cho phuong trinh : x+m=\(\sqrt{x+1}\) (1)
1/giai phuong trinh (1) khi m=1
2/giai va bien luan phuong trinh (1)theo m
1; Khi m=1 thì pt sẽ là \(\sqrt{x+1}=x+1\)
=>(x+1)^2=(x+1)
=>x(x+1)=0
=>x=0hoặc x=-1
2: \(\Leftrightarrow x+1=\left(x+m\right)^2\)
=>x^2+2mx+m^2-x-1=0
=>x^2+x(2m-1)+m^2-1=0
Δ=(2m-1)^2-4(m^2-1)
=4m^2-4m+1-4m^2+4
=-4m+5
Để pt có 2 nghiệm pb thì -4m+5>0
=>-4m>-5
=>m<5/4
Để pt có nghiệm kép thì 5-4m=0
=>m=5/4
Để pt vô nghiệm thì -4m+5<0
=>m>5/4