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11 tháng 8 2019
a) 3x4 - 13x3 + 16x2 - 13x + 3 = 0
(x - 3)(3x - 1)(x2 - x + 1) = 0
nhưng vì x2 - x + 1 # 0 nên:
x - 3 = 0 hoặc 3x - 1 = 0
x = 0 + 3 3x = 0 + 1
x = 3 3x = 1
x = 1/3
b) 6x4 + 5x3 - 38x2 + 5x + 6 = 0
(x - 2)(x + 3)(3x + 1)(2x - 1) = 0
x - 2 = 0 hoặc x + 3 = 0 hoặc 3x + 1 = 0 hoặc 2x - 1 = 0
x = 0 + 2 x = 0 - 3 3x = 0 - 1 2x = 0 + 1
x = 2 x = -3 3x = -1 2x = 1
x = -1/3 x = 1/2
7 tháng 7 2019
a, \(16x^2-5=0\)
\(\Rightarrow16x^2=5\)
\(\Rightarrow x^2=\frac{5}{16}\)
\(\Rightarrow x=\sqrt{\frac{5}{16}}\Rightarrow x=\frac{\sqrt{5}}{4}\)
b, \(2\sqrt{x-3}=4\)
\(\Rightarrow\sqrt{x-3}=4:2\)
\(\Rightarrow\sqrt{x-3}=2\)
\(\Rightarrow x-3=4\)
\(\Rightarrow x=4+3\)
\(\Rightarrow x=7\)
c, \(\sqrt{4x^2-4x+1}=3\)
\(\Rightarrow\sqrt{\left(2x-1\right)^2}=3\)
\(\Rightarrow2x-1=3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
d, \(\sqrt{x+3}\ge5\)
\(\Rightarrow x+3\ge25\)
\(\Rightarrow x\ge22\)
e, \(\sqrt{3x-1}< 2\)
\(\Rightarrow3x-1< 4\)
\(\Rightarrow3x< 5\)
\(\Rightarrow x< \frac{5}{3}\)
g, \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)
\(\Rightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)
\(\Rightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)
\(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)
\(\Rightarrow\sqrt{x-3}=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
7 tháng 7 2019
a) \(16x^2-5=0\)
\(\Leftrightarrow16x^2=5\)
\(\Leftrightarrow x^2=\frac{5}{16}\)
\(\Leftrightarrow x=\pm\sqrt{\frac{5}{16}}\)
b) \(2\sqrt{x-3}=4\)
\(\Leftrightarrow\sqrt{x-3}=2\)
\(\Leftrightarrow x-3=4\)
\(\Leftrightarrow x=7\)
c) \(\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=3\)
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
d) \(\sqrt{x+3}\ge5\)
\(\Leftrightarrow x+3\ge25\)
\(\Leftrightarrow x\ge22\)
e) \(\sqrt{3x-1}< 2\)
\(\Leftrightarrow3x-1< 4\)
\(\Leftrightarrow3x< 5\)
\(\Leftrightarrow x< \frac{5}{3}\)
g) \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)
Vì \(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)
\(\Leftrightarrow\sqrt{x-3}=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
24 tháng 4 2020
\(2x^2-7x+3=0\Leftrightarrow2x^2-x-6x+3=0\Leftrightarrow x\left(2x-1\right)-3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-3=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}}\)
\(2x^2-7x+3=0\Leftrightarrow x=\frac{-1\pm\sqrt{119}t}{12}\)
hoặc bn cho là vô nghiệm cx đc
\(16x^2+24x+9=0\Leftrightarrow\left(4x+3\right)^2=0\Leftrightarrow4x+3=0\Leftrightarrow x=-\frac{3}{4}\)
OD
3
TD
7 tháng 1 2017
a. \(x^4-10x^3+25x^2-36=0\)
=> \(x^3\left(x-3\right)-7x^2\left(x-3\right)+4x\left(x-3\right)+12\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x^3-7x^2+4x+12\right)=0\)
=>\(\left(x-3\right)\left[x^2\left(x-2\right)-5x\left(x-2\right)-6\left(x-2\right)\right]=0\)=> \(\left(x-3\right)\left(x-2\right)\left(x^2-5x-6\right)=0\)
=> \(\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(x-6\right)=0\)
=>\(\left[\begin{matrix}x=3\\x=2\\x=-1\\x=6\end{matrix}\right.\)
b) \(x^4\) - \(^{9x^2}\) - 24x - 16 = 0
=> \(x^3\left(x-4\right)+4x^2\left(x-4\right)+7x\left(x-4\right)+4\left(x-4\right)=0\)=>\(\left(x-4\right)\left(x^3+4x^2+7x+4\right)=0\)
=> \(\left(x-4\right)\left[x^2\left(x+1\right)+3x\left(x+1\right)+4\left(x+1\right)\right]=0\)=>\(\left(x-4\right)\left(x+1\right)\left(x^2+3x+4\right)=0\)
=> \(\left(x-4\right)\left(x+1\right)=0\) (vì x^2 + 3x + 4> 0)
=>\(\left[\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
TD
7 tháng 1 2017
a,pt\(\Leftrightarrow\left(x^4-10x^3+25x\right)-36=0\)\(\Leftrightarrow\left(x^2-5x\right)^2-36=0\)
\(\Leftrightarrow\left(x^2-5x-6\right)\left(x^2-5x+6\right)=0\)\(\Leftrightarrow\left[\begin{matrix}x^2-5x-6=0\\x^2-5x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}\left(x+1\right)\left(x-6\right)=0\\\left(x-2\right)\left(x-3\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-1,x=6\\x=2,x=3\end{matrix}\right.\)
vậy pt có 4 nghiệm x=(-1,6,2,3)
