Giải phuong trình: \(x^4+4x^3-6x^2-x-10=0\)
giải phuong trình x^4+2x^3+5x^2+4x-12=0
Giải bất phương trình sau : a/ 2x ^ 2 + 6x - 8 < 0 x ^ 2 + 5x + 4 >=\ 2) Giải phương trình sau : a/ sqrt(2x ^ 2 - 4x - 2) = sqrt(x ^ 2 - x - 2) c/ sqrt(2x ^ 2 - 4x + 2) = sqrt(x ^ 2 - x - 3) b/ x ^ 2 + 5x + 4 < 0 d/ 2x ^ 2 + 6x - 8 > 0 b/ sqrt(- x ^ 2 - 5x + 2) = sqrt(x ^ 2 - 2x - 3) d/ sqrt(- x ^ 2 + 6x - 4) = sqrt(x ^ 2 - 2x - 7)
2:
a: =>2x^2-4x-2=x^2-x-2
=>x^2-3x=0
=>x=0(loại) hoặc x=3
b: =>(x+1)(x+4)<0
=>-4<x<-1
d: =>x^2-2x-7=-x^2+6x-4
=>2x^2-8x-3=0
=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)
Giải các phương trình sau:
g/ x(x + 3)(x – 3) – (x + 2)(x2 – 2x + 4) = 0
h/ (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
i/ (x + 2)(3 – 4x) = x2 + 4x + 4
k/ x(2x – 7) – 4x + 14 = 0
m/ x2 + 6x – 16 = 0
n/ 2x2 + 5x – 3 = 0
\(m,x^2+6x-16=0\)
\(\Leftrightarrow x^2-2x+8x-16=0\)
\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+8\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=2\end{matrix}\right.\)
\(n,2x^2+5x-3=0\)
\(\Leftrightarrow2x^2-x+6x-3=0\)
\(\Leftrightarrow x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(k,x\left(2x-7\right)-4x+14=0\)
\(\Leftrightarrow2x^2-4x-7x+14=0\)
\(\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)
Giải các phương trình sau:
a \(x^2+3x+4=0\)
b \(3x^3-x+2=0\)
c \(x^4-4x^3-9x^2+8x+4=0\)
d \(x^4+4x^3+6x^2-5x-8=0\)
a: Ta có: \(x^2+3x+4=0\)
\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)
Do đó: Phương trình vô nghiệm
giải phương trình:
a, x^3 - 5x^2 + 6x - 4=0
b,x^3 - 7x + 6=0
c, x^3 + 8x^2 + 17x + 10=0
d,x(x+1)(x+2)(x+3) = -1
e, (4x+1)(12x-1)(3x+2)(x+1)=4
giải phương trình
a)x*4+x*3-4x*2+x+1=0
b)x*4+x*3-10x*2+x+1=0
c)x*4+3x*3-14x*2-6x+4=0
giải phương trình :
a, \(\sqrt{x-3}+\sqrt[3]{x^2+1}+x^2+x-2=0\)
b,\(4x^2+\sqrt{2x+3}=8x+1\)
c, \(2x^2-6x+10-5\left(x-2\right)\sqrt{x+1=0}\)
a.
ĐKXĐ: \(x\ge3\)
(Tốt nhất bạn kiểm tra lại đề cái căn đầu tiên của \(\sqrt{x-3}\) là căn bậc 2 hay căn bậc 3). Vì nhìn ĐKXĐ thì thấy căn bậc 2 là không hợp lý rồi đó
Pt tương đương:
\(\sqrt{x-3}+\sqrt[3]{x^2+1}+\left(x+1\right)\left(x-2\right)=0\)
Do \(x\ge3\Rightarrow x-2>0\Rightarrow\left(x+1\right)\left(x-2\right)>0\)
\(\Rightarrow\sqrt{x-3}+\sqrt[3]{x^2+1}+\left(x+1\right)\left(x-2\right)>0\)
Pt vô nghiệm
b.
ĐKXĐ: \(x\ge-\dfrac{3}{2}\)
Pt: \(2x+3-\sqrt{2x+3}-\left(4x^2-6x+2\right)=0\)
Đặt \(\sqrt{2x+3}=t\ge0\) ta được:
\(t^2-t-\left(4x^2-6x+2\right)=0\)
\(\Delta=1+4\left(4x^2-6x+2\right)=\left(4x-3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t_1=\dfrac{1+4x-3}{2}=2x-1\\t_2=\dfrac{1-4x+3}{2}=2-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+3}=2x-1\left(x\ge\dfrac{1}{2}\right)\\\sqrt{2x+3}=2-2x\left(x\le1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4x^2-4x+1\left(x\ge\dfrac{1}{2}\right)\\2x+3=4x^2-8x+4\left(x\le1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{17}}{4}\\x=\dfrac{5-\sqrt{21}}{4}\end{matrix}\right.\)
c.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2\left(x^2-4x+4\right)+2\left(x+1\right)-5\left(x-2\right)\sqrt{x+1}=0\)
\(\Leftrightarrow2\left(x-2\right)^2-5\left(x-2\right)\sqrt{x+1}+2\left(x+1\right)=0\)
Đặt \(\left\{{}\begin{matrix}x-2=a\\\sqrt{x+1}=b\end{matrix}\right.\) ta được:
\(2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=x-2\left(x\ge2\right)\\\sqrt{x+1}=2x-4\left(x\ge2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-4x+4\\x+1=4x^2-16x+16\end{matrix}\right.\) (\(x\ge2\))
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\) (đã loại nghiệm)
Giải phương trình:
1. \(x^4-6x^2-12x-8=0\)
2. \(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
3. \(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
4. \(2x^2.\sqrt{-4x^4+4x^2+3}=4x^4+1\)
5. \(x^2+4x+3=\sqrt{\dfrac{x}{8}+\dfrac{1}{2}}\)
6. \(\left\{{}\begin{matrix}4x^3+xy^2=3x-y\\4xy+y^2=2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}\sqrt{x^2-3y}\left(2x+y+1\right)+2x+y-5=0\\5x^2+y^2+4xy-3y-5=0\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\sqrt{2x^2+2}+\left(x^2+1\right)^2+2y-10=0\\\left(x^2+1\right)^2+x^2y\left(y-4\right)=0\end{matrix}\right.\)
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)
\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)
Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)
\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)
Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:
\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)
\(\Leftrightarrow10b+40=3\left(b+8\right)b\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)
TH1: \(b=2\Leftrightarrow...\)
TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)
Giải phương trình sau bằng cách đặt biến phụ
x^4+4x^3+6x^2 +4x+1=0
Mình ko biết đặt biến phụ nên mình sẽ giải bừa :>
\(x^4+4x^3+6x^2+4x+1=0\)
\(\Leftrightarrow x^4+2x^3+x^2+2x^3+4x^2+2x+x^2+2x+1=0\)
\(\Leftrightarrow x^2\left(x^2+2x+1\right)+2x\left(x^2+2x+1\right)+\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^4=0\Leftrightarrow x=-1\)
Thấy ngay x= 0 không phải là nghiệm của pt. Chia 2 vế của pt cho x2 ta được:
\(x^2+4x+6+4.\frac{1}{x}+\frac{1}{x^2}=0\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)+4\left(x+\frac{1}{x}\right)+6=0\left(1\right)\)
Đặt \(x+\frac{1}{x}=t\Rightarrow\left(x+\frac{1}{x}\right)^2=t^2\Rightarrow x^2+\frac{1}{x^2}=t^2-2\) Khi đó ta có:
\(\left(1\right)\Leftrightarrow t^2-2+4t+6=0\)
\(\Leftrightarrow t=-2\Leftrightarrow x+\frac{1}{x}=-2\Leftrightarrow x^2+2x+1=0\Leftrightarrow x=-1\)
Vậy pt có 1 nghiệm x = -1