giải pt | 3x-2 | =3-2x
giải pt: x^5 + 2x^4 +3x^3 + 3x^2 + 2x +1=0
giải pt: x^4 + 3x^3 - 2x^2 +x - 3=0
ta có : x^5+2x^4+3x^3+3x^2+2x+1=0
\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0
\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0
\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0
\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0
\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0
VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)
\(\Rightarrow\)x+1=0
\(\Rightarrow\)x=-1
CÒN CÂU B TỰ LÀM (02042006)
b: x^4+3x^3-2x^2+x-3=0
=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0
=>(x-1)(x^3+4x^2+2x+3)=0
=>x-1=0
=>x=1
giải pt:
a) x^5 + 2x^4 + 3x^3 + 3x^2 + 2x +1=0
b) x^4 + 3x^3 - 2x^2 + x - 3 = 0
a) \(x^5+2x^4+3x^3+3x^2+2x+1=0\)
\(\Leftrightarrow x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0\)
\(\Leftrightarrow x^4\left(x+1\right)+x^3\left(x+1\right)+2x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+2x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+x^2+x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2+1\right)=0\)
Dễ thấy \(x^2+x+1>0\forall x;x^2+1>0\forall x\)
\(\Rightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy....
b) \(x^4+3x^3-2x^2+x-3=0\)
\(\Leftrightarrow x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0\)
\(\Leftrightarrow x^3\left(x-1\right)+4x^2\left(x-1\right)+2x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+4x^2+2x+3\right)=0\)
...
\(\Leftrightarrow x=1\)
p/s: có bác nào giải đc pt \(x^3+4x^2+2x+3=0\)thì giúp nhé :))
giải pt 2x^3+3x^2-8x+3
GIẢI CÁC PT SAU:
\(\sqrt{5x+10}=8-x\)
\(\sqrt{4x^2+x-12}=3x-5\)
\(\sqrt{x^2-2x+6}=2x-3\)
\(\sqrt{3x^2-2x+6}+3-2x=0\)
giải pt (3x/x^2-x+3)-(2x/x^2-3x+3)=1
help me m cần gấp ạ
Đặt x/(x^2-3x+3) = t ta được
\(3t-2t=1\Leftrightarrow t=1\)
Theo cách đặt \(x=x^2-3x+3\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\Leftrightarrow x=3;x=1\)
Giải PT: \(\sqrt{2x^4-3x^2+1}+\sqrt{2x^4-x^2}=4x-3\)
Do vế trái dương nên pt chỉ có nghiệm khi \(x\ge\dfrac{3}{4}\), kết hợp điều kiện \(2x^4-3x^2+1\ge0\Rightarrow x\ge1\)
Khi đó:
\(4x-3=\sqrt{2x^4-3x^2+1}+\sqrt{2x^4-x^2}\ge\sqrt{2x^4-3x^2+1+2x^4-x^2}\)
\(\Rightarrow4x-3\ge\sqrt{4x^4-4x^2+1}\)
\(\Rightarrow4x-3\ge\left|2x^2-1\right|=2x^2-1\)
\(\Rightarrow2x^2-4x+2\le0\)
\(\Rightarrow2\left(x-1\right)^2\le0\)
\(\Rightarrow x=1\)
giải pt sau:
\(2x^2+3x+3=5\sqrt{2x^2+3x+9}\)
\(2x^2+3x+3=5\sqrt{2x^2+3x+9}\)
\(\Leftrightarrow2x^2+3x+3=5\sqrt{2x^2+3x+3+6}\)(*)
Đặt \(2x^2+3x+3=a\)
(*) \(\Leftrightarrow a=5\sqrt{a+6}\)
\(\Leftrightarrow a^2=25\left(a+6\right)\)
\(\Leftrightarrow a^2-25a-150=0\)
\(\Leftrightarrow\left(a-30\right)\left(a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=30\\a=-5\end{matrix}\right.\)
Trả lại biến cũ: \(2x^2+3x+3=30\Leftrightarrow2x^2+3x-27=0\)\(\Leftrightarrow\left(x-3\right)\left(2x+9\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{9}{2}\end{matrix}\right.\)
\(2x^2+3x+3=-5\Leftrightarrow2x^2+3x+8=0\)\(\Leftrightarrow\left(x\sqrt{2}+\frac{3\sqrt{2}}{4}\right)^2=-\frac{55}{8}\left(L\right)\)
Đat: \(2x^2+3x+3=a\)
\(\Rightarrow a=5\sqrt{a+6}\Leftrightarrow a^2=25a+150\Leftrightarrow a^2-25a-150=0\Leftrightarrow\left(a-12,5\right)^2=6,25\Leftrightarrow\left[{}\begin{matrix}a=10\\a=15\end{matrix}\right.\) \(+,a=10\Leftrightarrow x^2+3x+3=10\Leftrightarrow\left(x+\frac{3}{2}\right)^2=9,25\Leftrightarrow x=\pm\sqrt{9,25}-\frac{3}{2}\)
\(+,a=15\Leftrightarrow x^2+3x+2,25=14,25\Leftrightarrow\left(x+\frac{3}{2}\right)^2=14,25\Leftrightarrow x=\pm\sqrt{14,25}-\frac{3}{2}\)
\(5) (3x -1)^2 - (x +3)(2x-1) = 7(x + 1)(x -2) -3x\)
Giải pt
3.Giải pt : 2x2 + 3x2 + 3x +1 = 0
\(2x^2+3x^2+3x+1=0\\ \Leftrightarrow5x^2+3x+1=0\)
Ta có: $a=5,b=3,c=1$
\(\Delta=b^2-4ac=3^2-4.5.1=9-20=-11\)
\(\sqrt{\Delta}=-11< 0\)
Vậy phương trình vô nghiệm