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9 tháng 6 2017
e, Ta có: \(\Delta\)'\(=\left(-6\right)^2-4.5=16>0\)
Suy ra \(\sqrt{\Delta'}=\sqrt{16}=4\)
Vậy phương trình đã cho có 2 nghiệm phân biệt
\(x_1=\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{6-4}{4}=\dfrac{1}{2}\)
\(x_2=\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{6+4}{4}=\dfrac{5}{2}\)
Vậy phương trình đã cho có 2 nghiệm là 1/2;5/2
f,Ta có : a+-b+c=2-5+3=0
Do đó phương trình đã cho có 2 nghiệm \(x_1\)=-1 hoặc \(x_2=\dfrac{-c}{a}=-\dfrac{3}{2}\)
g,Ta có: a+b+c=1+1-2=0
Do phương trình đã cho có 2 nghiệm \(x_1\)=1 hoặc \(x_2=\dfrac{c}{a}=-2\)
h,Ta có a+b+c=1-4+3=0
Do đó phương trình đã cho có 2 nghiệm \(x_1=1\) hoặc \(x_2=\dfrac{c}{a}=3\)
DH
9 tháng 6 2017
g, \(x^2+x-2=0\)
\(\Rightarrow x^2-x+2x-2=0\)
\(\Rightarrow\left(x^2-x\right)+\left(2x-2\right)=0\)
\(\Rightarrow x.\left(x-1\right)+2.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right).\left(x+2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy......
h, \(x^2-4x+3=0\)
\(\Rightarrow x^2-3x-x+3=0\)
\(\Rightarrow\left(x^2-3x\right)-\left(x-3\right)=0\)
\(\Rightarrow x.\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right).\left(x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy......
Chúc bạn học tốt!!!
18 tháng 3 2020
b) Ta có: \(x^3-7x+6=0\)
\(\Leftrightarrow x^3-6x-x+6=0\)
\(\Leftrightarrow x\left(x^2-1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x+1\right)-6\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+3x-2x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\\x=2\end{matrix}\right.\)
Vậy: x∈{1;-3;2}
c) Ta có: \(x^4-4x^3+12x-9=0\)
\(\Leftrightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)
\(\Leftrightarrow x^2\left(x^2-4x+3\right)-3\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\left(x^2-4x+3\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\\x^2-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=\pm\sqrt{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{3;1;\pm\sqrt{3}\right\}\)
d) Ta có: \(x^5-5x^3+4x=0\)
\(\Leftrightarrow x^5-x^3-4x^3+4x=0\)
\(\Leftrightarrow x^3\left(x^2-1\right)-4x\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^3-4x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\cdot x\left(x^2-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=0\\x=\pm2\end{matrix}\right.\)
Vậy: x∈{-2;-1;0;1;2}
e) Ta có: \(x^4-4x^3+3x^2+4x-4=0\)
\(\Leftrightarrow x^4-4x^3+4x^2-x^2+4x-4=0\)
\(\Leftrightarrow x^2\left(x^2-4x+4\right)-\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x=1\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
Vậy: x∈{-1;1;2}
HM
1
HZ
5 tháng 8 2017
a, x2 - 2x + 3 > 0
Xét : VT = x2 - 2x + 1 + 2 = ( x - 1 )2 + 2 .
Có : ( x - 1 )2 \(\ge\) 0 với mọi x \(\Rightarrow\) ( x - 1 )2 + 2 > 0 với mọi x hay
VT > 0 .
Vậy BĐT x2 - 2x + 3 > 0 đúng .
Các câu còn lại tương tự .
Chúc bn học tốt !!!!!!!!
1 tháng 9 2020
a) x3 + 3x2 + 3x + 1 = 64
=> (x + 1)3 = 64
=> (x + 1)3 = 43
=> x + 1 = 4 => x = 3
b) x3 + 6x2 + 9x = 4x
=> x3 + 6x2 + 9x - 4x = 0
=> x3 + 6x2 + 5x = 0
=> x3 + 5x2 + x2 + 5x = 0
=> x2(x + 5) + x(x + 5) = 0
=> (x + 5)(x2 + x) = 0
=> (x + 5)x(x + 1) = 0
=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)
c) 4(x - 2)2 = (x + 2)2
=> 4(x2 - 4x + 4) = x2 + 4x + 4
=> 4x2 - 16x + 16 = x2 + 4x + 4
=> 4x2 - 16x + 16 - x2 - 4x - 4 = 0
=> 3x2 - 20x + 12 = 0
=> 3x2 - 18x - 2x + 12 = 0
=> 3x(x - 6) - 2(x - 6) = 0
=> (x - 6)(3x - 2) = 0
=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)
d) x4 - 16x2 = 0
=> x2(x2 - 16) = 0
=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
e) x4 - 4x3 + x2 - 4x = 0
=> x4 + x2 - 4x3 - 4x = 0
=> x2(x2 + 1) - 4x(x2 + 1) = 0
=> (x2 - 4x)(x2 + 1) = 0
=> x(x - 4)(x2 + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x2 + 1 \(\ge\)1 > 0 \(\forall\)x)
f) x3 + x = 0 => x(x2 + 1) = 0 => x = 0 (vì x2 + 1 \(\ge1>0\forall\)x)
25 tháng 6 2018
a ) \(x^3-x=0\)
\(\Leftrightarrow x\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy..................
b ) \(x^3+5x^2+4x+20=0\)
\(\Leftrightarrow x^2\left(x+5\right)+4\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x+5\right)=0\) . Vì \(x^2+4>0\)
\(\Leftrightarrow x=-5\)
c) \(x^2-25+3\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+3\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5+3x-15\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(4x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\4x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{10}{4}\end{matrix}\right.\)
Vậy......................
d ) Có nhầm đề không ?
HH
25 tháng 6 2018
Giải:
a) \(x^3-x=0\)
\(\Leftrightarrow x\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy ...
b) \(x^3+5x^2+4x+20=0\)
\(\Leftrightarrow x^2\left(x+5\right)+4\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x+5=0\left(x^2+4>0\right)\)
\(\Leftrightarrow x=-5\)
Vậy ...
c) \(x^2-25+3\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+3\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5+3x-15\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(4x-10\right)=0\)
\(\Leftrightarrow2\left(x-5\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy ...
d) \(2+4\sqrt{2}x+4x^2=0\)
\(\Leftrightarrow\left(\sqrt{2}\right)^2+2\sqrt{2}.2x+\left(2x\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{2}+2x\right)^2=0\)
\(\Leftrightarrow\sqrt{2}+2x=0\)
\(\Leftrightarrow x=-\dfrac{\sqrt{2}}{2}\)
Vậy ...
5 tháng 11 2017
\(a,x^4-4x^3+x^2-4x=0\)
\(\Rightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)
\(\Rightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x^2+x\right)=0\)
\(\Rightarrow x\left(x-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-1\end{matrix}\right.\)
\(b,x^3-5x^2+4x-20=0\)
\(\Rightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)
\(\Rightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x^2+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\)
\(\Rightarrow x=5\)
AN
5 tháng 11 2017
a) \(x^4-4x^3+x^2-4x=0\)
\(\Leftrightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)
\(\Leftrightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^3+x\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x^2+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x^2=-1\left(loai\right)\end{matrix}\right.\)
Vậy x=0; x=4
b) \(x^3-5x^2+4x-20=0\)
\(\Leftrightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)
\(\Leftrightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=-4\left(loai\right)\end{matrix}\right.\)
Vậy x=5
6 tháng 4 2020
1/ \(3x^2-2x=0\Leftrightarrow x\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy .....................
2/ \(4x^2-4x+\frac{1}{2}=0\)
\(\Delta=b'^2-ac=2^2-\frac{4.1}{2}=2\)
\(\Rightarrow\left[{}\begin{matrix}x_1=\frac{2+\sqrt{2}}{4}\\x_2=\frac{2-\sqrt{2}}{4}\end{matrix}\right.\)
Vậy ........................
3/ \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)
<=> \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)
<=> \(\left(x-1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy ......................
6 tháng 4 2020
a) Ta có: \(3x^2+2x=0\)
\(\Leftrightarrow x\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{-2}{3}\right\}\)
b) Ta có: \(4x^2-4x+\frac{1}{2}=0\)
\(\Leftrightarrow4x^2-4x+1-\frac{1}{2}=0\)
\(\Leftrightarrow\left(2x-1\right)^2-\frac{1}{2}=0\)
\(\Leftrightarrow\left(2x-1\right)^2=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{\frac{1}{2}}\\2x-1=-\sqrt{\frac{1}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\sqrt{\frac{1}{2}}+1\\2x=-\sqrt{\frac{1}{2}}+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2+\sqrt{2}}{4}\\x=\frac{2-\sqrt{2}}{4}\end{matrix}\right.\)
Vậy: \(x=\frac{2\pm\sqrt{2}}{4}\)
15 tháng 10 2018
a. \(\left(2x-1\right)^2-4x^2+1=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
\(\Leftrightarrow2-4x=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
b/ \(6x^3-24x=0\)
\(\Leftrightarrow6x\left(x^2-4\right)=0\)
\(\Leftrightarrow6x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy ...
c/ \(2x\left(x-3\right)-4x+12=0\)
\(\Leftrightarrow2x\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow2\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy ...
d/ \(x^3-5x^2+x-5=0\)
\(\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)=0\)
Mà \(x^2+1>0\)
\(\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy..
\(x^3-4x=0\)
\(\Rightarrow x\left(x^2-4\right)=0\)
\(\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\)x = 0 , x - 2 = 0 hoặc x + 2 = 0
\(\Rightarrow\)\(x=0,x=2,x=-2\)
\(x^3-4x=0\)
\(\Leftrightarrow x\left(x^2-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
Vậy \(x\in\left\{-2;0;2\right\}\)