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16 tháng 1
a: \(x^3+8x=5x^2+4\)
=>\(x^3-5x^2+8x-4=0\)
=>\(x^3-x^2-4x^2+4x+4x-4=0\)
=>\(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)
=>\(\left(x-1\right)\left(x-2\right)^2=0\)
=>\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2: \(x^3+3x^2=x+6\)
=>\(x^3+3x^2-x-6=0\)
=>\(x^3+2x^2+x^2+2x-3x-6=0\)
=>\(x^2\cdot\left(x+2\right)+x\left(x+2\right)-3\left(x+2\right)=0\)
=>\(\left(x+2\right)\left(x^2+x-3\right)=0\)
=>\(\left[{}\begin{matrix}x+2=0\\x^2+x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)
3: ĐKXĐ: x>=0
\(2x+3\sqrt{x}=1\)
=>\(2x+3\sqrt{x}-1=0\)
=>\(x+\dfrac{3}{2}\sqrt{x}-\dfrac{1}{2}=0\)
=>\(\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}=0\)
=>\(\left(\sqrt{x}+\dfrac{3}{4}\right)^2=\dfrac{17}{16}\)
=>\(\left[{}\begin{matrix}\sqrt{x}+\dfrac{3}{4}=-\dfrac{\sqrt{17}}{4}\\\sqrt{x}+\dfrac{3}{4}=\dfrac{\sqrt{17}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{17}-3}{4}\left(nhận\right)\\\sqrt{x}=\dfrac{-\sqrt{17}-3}{4}\left(loại\right)\end{matrix}\right.\)
=>\(x=\dfrac{13-3\sqrt{17}}{8}\left(nhận\right)\)
4: \(x^4+4x^2+1=3x^3+3x\)
=>\(x^4-3x^3+4x^2-3x+1=0\)
=>\(x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)
=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)
=>\(\left(x-1\right)^2\cdot\left(x^2-x+1\right)=0\)
=>(x-1)^2=0
=>x-1=0
=>x=1
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 1
a.
\(x^3+8x=5x^2+4\)
\(\Leftrightarrow x^3-5x^2+8x-4=0\)
\(\Leftrightarrow\left(x^3-4x^2+4x\right)-\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
b.
\(x^3+3x^2-x-6=0\)
\(\Leftrightarrow\left(x^3+x^2-3x\right)+\left(2x^2+2x-6\right)=0\)
\(\Leftrightarrow x\left(x^2+x-3\right)+2\left(x^2+x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)
TT
20 tháng 4 2021
PT 2
\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))
\(\Leftrightarrow\dfrac{3+2x^2-2x-x+3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
\(\Rightarrow2x^2-3x+6=0\)
=> PT vô nghiệm.
12 tháng 9 2023
a: =>|2x-1|=3
=>2x-1=3 hoặc 2x-1=-3
=>2x=-2 hoặc 2x=4
=>x=2 hoặc x=-1
c: \(\Leftrightarrow\left|x-3\right|=11-x\)
=>x<=11 và (x-3)^2=(11-x)^2
=>x<=11 và x^2-6x+9=x^2-22x+121
=>x<=11 và 16x=112
=>x=7
d:
ĐKXĐ: 3x+19>=0
=>x>=-19/3
PT =>x>=-3 và (3x+19)=(x+3)^2=x^2+6x+9
=>x>=-3 và x^2+6x+9-3x-19=0
=>x>=-3 và (x+5)(x-2)=0
=>x=2
e: =>\(\sqrt{x^2+x+5}=x+1\)
=>x>=-1 và x^2+x+5=x^2+2x+1
=>x>=-1 và 2x+1=x+5
=>x=4
23 tháng 8 2021
d) \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow x-3=2\Leftrightarrow x=5\)
e) đk: \(x\ge2\)\(\sqrt{x^2-3x+2}=\sqrt{x-1}\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)f) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow2x-1=x-3\Leftrightarrow x=-2\)
23 tháng 8 2021
c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)
\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)
\(\Leftrightarrow x-4=0\)
hay x=4
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 6 2019
a/ ĐKXĐ:...
\(\Leftrightarrow4x^2-4x\sqrt{2x-1}-3x^2+6x-3=0\)
\(\Leftrightarrow4x\left(x-\sqrt{2x-1}\right)-3\left(x-1\right)^2=0\)
\(\Leftrightarrow\frac{4x\left(x-1\right)^2}{x+\sqrt{2x-1}}-3\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\frac{4x}{x+\sqrt{2x-1}}=3\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4x=3x+3\sqrt{2x-1}\)
\(\Leftrightarrow x=3\sqrt{2x-1}\)
\(\Leftrightarrow x^2-18x+9=0\) \(\Rightarrow9\pm6\sqrt{2}\)
Vậy pt có 3 nghiệm....
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 6 2019
b/ ĐKXĐ:...
\(\Leftrightarrow4x^2-4x\sqrt{4x-3}-x^2+4x-3=0\)
\(\Leftrightarrow4x\left(x-\sqrt{4x-3}\right)-\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\frac{4x\left(x^2-4x+3\right)}{x+\sqrt{4x-3}}-\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\Rightarrow x=...\\\frac{4x}{x+\sqrt{4x-3}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4x=x+\sqrt{4x-3}\)
\(\Leftrightarrow3x=\sqrt{4x-3}\)
\(\Leftrightarrow9x^2-4x+3=0\) (vô nghiệm)
Vậy...
15 tháng 11 2021
\(a,=\dfrac{x^4\left(x-2\right)+2x^2\left(x-2\right)-3\left(x-2\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4+2x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4-x^2+3x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x-1\right)\left(x^2+3\right)}{x+4}\)
\(b,=\dfrac{x^4-3x^2-x^2+3}{x^4-x^2+7x^2-7}=\dfrac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\dfrac{x^2-3}{x^2+7}\\ c,=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}=\dfrac{x^2-1}{x^2+1}\)
KN
1
7 tháng 5 2020
\(4x^4+4x^3+x^2+3x\ge0\)
\(4x^4+4x^2+1-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)\sqrt{\left(x^2-x+1\right)\left(2x^2+1\right)+2x^4+6x^3-2x^3+4x-1}\)
\(\Leftrightarrow\left(2x^2+1\right)^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)\sqrt{\left(x^2-x+1\right)\left(2x^2+1\right)+2x^4+6x^3-2x^3+4x-1}\)
\(2x^2+1=u;\sqrt{4x^4+4x^3+x^2+3x}=v\left(u>0;v>0\right)\)
\(\hept{\begin{cases}u^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)v\\v^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)u\end{cases}\Rightarrow u^2-v^2=\left(x^2-x+1\right)\left(v-u\right)\Leftrightarrow\orbr{\begin{cases}u=v\\u+v+x^2-x+1=0\end{cases}}}\)
- \(u+v+x^2-x+1=0\Leftrightarrow u+v+\left(x-\frac{1}{2}\right)^2=-\frac{3}{4}\)
- \(u=v\Leftrightarrow4x^4+4x^2+1=4x^4+4x^3+x^2+3x\Leftrightarrow\left(x-1\right)^3=-3x^3\Leftrightarrow x-1=-x\sqrt[3]{3}\Leftrightarrow x=\frac{1}{1+\sqrt[3]{3}}\)Đối chiếu điều kiện ta thu được nghiệm duy nhất \(x=\frac{1}{1+\sqrt[3]{3}}\)
28 tháng 1 2019
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
9 tháng 7 2021
f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1