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8 tháng 2 2020
Bài 2 :
a, Ta có : \(\left(x+4\right)\left(x-1\right)=0\)
=> \(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
b, Ta có : \(\left(3x-2\right)\left(4x-7\right)=0\)
=> \(\left[{}\begin{matrix}3x-2=0\\4x-7=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}3x=2\\4x=7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{7}{4}\end{matrix}\right.\)
c, Ta có : \(\left(x+5\right)\left(x^2+1\right)=0\)
=> \(\left[{}\begin{matrix}x+5=0\\x^2+1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-5\\x^2+1=0\left(VL\right)\end{matrix}\right.\)
d, Ta có : \(x\left(x-1\right)\left(x^2+4\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\x-1=0\\x^2+4=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=1\\x^2+4=0\left(VL\right)\end{matrix}\right.\)
e, Ta có : \(\left(3x+2\right)\left(x+\frac{1}{2}\right)=0\)
=> \(\left[{}\begin{matrix}3x+2=0\\x+\frac{1}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{1}{2}\end{matrix}\right.\)
f, Ta có : \(\left(x+2\right)\left(x+3\right)\left(x^2+7\right)=0\)
=> \(\left[{}\begin{matrix}x+2=0\\x-3=0\\x^2+7=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-2\\x=3\\x^2+7=0\left(VL\right)\end{matrix}\right.\)
8 tháng 2 2020
Bài 1 :
a, Ta có : \(1-\frac{x+3}{4}-\frac{x-2}{6}=0\)
=> \(\frac{12}{12}-\frac{3\left(x+3\right)}{12}-\frac{2\left(x-2\right)}{12}=0\)
=> \(12-3\left(x+3\right)-2\left(x-2\right)=0\)
=> \(12-3x-9-2x+4=0\)
=> \(-5x=-7\)
=> \(x=\frac{7}{5}\)
TT
23 tháng 4 2019
a. \(x^2-x-6=0\)
\(\Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
b. \(x^2+8x-20=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(10x-20\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+10\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)
c. \(x^4+4x^2-5=0\)
\(\Leftrightarrow\left(x^4+4x^2+4\right)-9=0\)
\(\Leftrightarrow\left(x^2+2\right)^2-3^2=0\)
\(\Leftrightarrow\left(x^2+2+3\right)\left(x^2+2-3\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-5\left(vo.nghiem\right)\\x=1\\x=-1\end{matrix}\right.\)
d. \(x^3-19x-30=0\)
\(\Leftrightarrow\left(x^3-5x^2\right)+\left(5x^2-25x\right)+\left(6x-30\right)=0\)
\(\Leftrightarrow x^2\left(x-5\right)+5x\left(x-5\right)+6\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[\left(x^2+2x\right)+\left(3x+6\right)\right]=0\)
\(\Leftrightarrow\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\\x=-3\end{matrix}\right.\)
HT
1
18 tháng 4 2019
\(192-\left(x^2-1\right)\left(x^2+4x+3\right)=0\)
\(\Leftrightarrow192-\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow192-\left[\left(x-1\right)\left(x+3\right)\right]\left[\left(x+1\right)\left(x+1\right)\right]=0\)
\(\Leftrightarrow192-\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)
Đặt \(x^2+2x-3=a\)
\(pt\Leftrightarrow192-a\left(a+4\right)=0\)
\(\Leftrightarrow192-a^2-4a=0\)
\(\Leftrightarrow-a^2-16a+12a+192=0\)
\(\Leftrightarrow-a\left(a+16\right)+12\left(a+16\right)=0\)
\(\Leftrightarrow\left(a+16\right)\left(-a+12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-16\\a=12\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+2x-3=-16\\x^2+2x-3=12\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+2x+13=0\\x^2+2x-15=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+2x+1+12=0\\x^2+5x-3x-15=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^2=-12\\x\left(x+5\right)-3\left(x+5\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x\in\varnothing\\\left(x+5\right)\left(x-3\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)
Vậy.....
KS
2
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 2 2020
\(\Leftrightarrow x^3+x^2-2x+5x^2+5x-10=0\)
\(\Leftrightarrow x\left(x^2+x-2\right)+5\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+2\right)\left(x-1\right)=0\)
b/ \(\Leftrightarrow x^3+5x^2+6x-x^2-5x-6=0\)
\(\Leftrightarrow x\left(x^2+5x+6\right)-\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)
NH
10 tháng 2 2020
\(x^3+6x^2+3x-10=0\)
\(\Leftrightarrow x^3-x^2+7x^2-7x+10x-10=0\)
\(\Leftrightarrow x^2\left(x-1\right)+7x\left(x-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+5x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-5\end{matrix}\right.\)
Vậy \(S=\left\{1;-2;-5\right\}\)
\(x^3+4x^2+x-6=0\)
\(\Leftrightarrow x^3-x^2+5x^2-5x+6x-6=0\)
\(\Leftrightarrow x^2\left(x-1\right)+5x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+3x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{1;-2;-3\right\}\)
TB
9 tháng 6 2017
a)
\(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2-x+3x-1=0\)
\(\Leftrightarrow x\left(3x-1\right)+\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\)
b)
\(x^2-5x+6=0\)
\(\Leftrightarrow x^2-3x-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
DH
9 tháng 6 2017
a, \(3x^2+2x-1=0\)
\(\Rightarrow3x^2-x+3x-1=0\)
\(\Rightarrow\left(3x^2-x\right)+\left(3x-1\right)=0\)
\(\Rightarrow x.\left(3x-1\right)+\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right).\left(x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x=1\\x=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\)
Vậy......
b, \(x^2-5x+6=0\)
\(\Rightarrow x^2-3x-2x+6=0\)
\(\Rightarrow\left(x^2-3x\right)-\left(2x-6\right)=0\)
\(\Rightarrow x.\left(x-3\right)-2.\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right).\left(x-2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy......
Chúc bạn học tốt!!!
HT
2
KN
23 tháng 1 2020
\(6x^2-7x+2=0\)
Ta có \(\Delta=7^2-4.6.2=1,\sqrt{\Delta}=1\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7+1}{12}=\frac{2}{3}\\x=\frac{7-1}{12}=\frac{1}{2}\end{cases}}\)
\(x^6-1=0\)
\(\Leftrightarrow\left(x^3+1\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)\left(x-1\right)\left(x^2+x+1\right)=0\)
Dễ thấy \(\hept{\begin{cases}x^2-x+1>0\forall x\\x^2+x+1>0\forall x\end{cases}}\)nên \(\hept{\begin{cases}x+1=0\\x-1=0\end{cases}}\Leftrightarrow x=\pm1\)
MN
23 tháng 1 2020
\(6x^2-7x+2=0\)
\(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow3x\left(2x-1\right)-2\left(2x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{1}{2}\end{cases}}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{2}{3};\frac{1}{2}\right\}\)
\(x^6-1=0\)
\(\Leftrightarrow x^6=1\)
\(\Leftrightarrow x=\pm1\)
Vậy tập nghiệm của pt là : \(S=\left\{\pm1\right\}\)
31 tháng 3 2017
\(\dfrac{x+1}{x-1}+\dfrac{1}{x+1}=0\\ < =>\dfrac{\left(x+1\right)^2}{x^2-1}+\dfrac{x-1}{x^2-1}=0->\left(1\right)\\ ĐKXĐ:x^2-1\ne0< =>\left[{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.< =>\left[{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
\(\left(1\right)=>\dfrac{\left(x+1\right)^2}{x^2-1}+\dfrac{x-1}{x^2-1}=0\\ =>\left(x+1\right)^2+\left(x-1\right)=0\\ < =>x^2+2x+1+x-1=0\\ < =>x^2+3x=0\\ < =>x\left(x+3\right)=0\\ =>\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=-3\left(TMĐK\right)\end{matrix}\right.\)
Vậy: Tập nghiệm của pt là S= {-3;0}
KT
31 tháng 3 2017
\(\dfrac{x}{x-3}+\dfrac{6x}{9-x^2}=0\) (ĐKXĐ: \(x\ne\pm3\))
\(\Leftrightarrow\dfrac{-x\left(3+x\right)+6x}{9-x^2}=0\)
\(\Rightarrow-3x-x^2+6x=0\\ \Leftrightarrow x\left(-x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\-x+3=0\Leftrightarrow x=3\left(loại\right)\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={0}
DD
6
HY
10 tháng 2 2020
\(a.2x^2+7x-9=0\\ \Leftrightarrow2\left(x^2+\frac{7}{2}x-\frac{9}{2}\right)=0\\\Leftrightarrow x^2+\frac{7}{2}x-\frac{9}{2}=0\\ \Leftrightarrow x^2+\frac{9}{2}x-x-\frac{9}{2}=0\\\Leftrightarrow x\left(x+\frac{9}{2}\right)-\left(x+\frac{9}{2}\right)=0\\\Leftrightarrow \left(x-1\right)\left(x+\frac{9}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+\frac{9}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{9}{2}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{1;-\frac{9}{2}\right\}\)
HY
10 tháng 2 2020
\(b.x^2-4x+3=0\\\Leftrightarrow x^2-x-3x+3=0\\ \Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-3\right)=0\\\Rightarrow \left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{1;3\right\}\)
PN
1
4 tháng 3 2018
\(x^2-2x+y^2-8y+17=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-8y+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-4\right)^2=0\)
Vì \(\left(x-1\right)^2\ge0\)
\(\left(y-4\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2=\left(y-4\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-4\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\)
Vậy phương trình có nghiệm \(\left(x,y\right)=\left(1;4\right)\)
