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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!!!
NT
1
FS
2 tháng 3 2019
\(x^4-6x^3+7x^2+6x-8=0\)
\(\Leftrightarrow x^4-4x^3-2x^3+8x^2-x^2+4x+2x-8=0\)
\(\Leftrightarrow x^3\left(x-4\right)-2x^2\left(x-4\right)-x\left(x-4\right)+2\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^3-2x^2-x+2\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left[x^2\left(x-2\right)-\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\in\left\{-1;1;2;4\right\}\)
Vậy S={-1;1;2;4}
2 tháng 5 2019
f, 3x2+4x-4=0
\(\Leftrightarrow\)3x2+6x-2x-4=0
\(\Leftrightarrow\)3x(x+2)-2(x+2)=0
\(\Leftrightarrow\)(x+2)(3x-2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-2\\x=\frac{2}{3}\end{matrix}\right.\left(tm\right)\)
Vậy pt có tập nghiệm S = \(\left\{-2;\frac{2}{3}\right\}\)
24 tháng 2 2020
a) Ta có: \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{\left(2x+1\right)^2\cdot3}{15}-\frac{5\left(x-1\right)^2}{15}-\frac{7x^2-14x-5}{15}=0\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)-7x^2+14x+5=0\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x+3=0\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=\frac{-3}{36}\)
Vậy: \(x=\frac{-3}{36}\)
b) Ta có: \(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\frac{201-x}{99}+\frac{203-x}{97}-\frac{205-x}{95}-3=0\)
\(\Leftrightarrow\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)
\(\Leftrightarrow\frac{201-x+99}{99}+\frac{203-x+97}{97}+\frac{205-x+95}{95}=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\)
nên 300-x=0
\(\Leftrightarrow x=300\)
Vậy: x=300
c) Ta có: \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+1\ge1\ne0\forall x\)(2)
Từ (1) và (2) suy ra x+1=0
hay x=-1
Vậy: x=-1
d) Ta có: \(\left(x-1\right)x\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
Đặt \(x^2+x-1=t\)
\(\Leftrightarrow\left(t+1\right)\left(t-1\right)=24\)
\(\Leftrightarrow t^2-1-24=0\)
\(\Leftrightarrow t^2-25=0\)
\(\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\)
\(\Leftrightarrow\left(x^2+x-1-5\right)\left(x^2+x-1+5\right)=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x^2+3x-2x-6\right)\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{15}{4}\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\right]\)(3)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}\ne0\forall x\)(4)
Từ (3) và (4) suy ra
\(\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
e) Ta có: \(\left(5x-3\right)-\left(4x-7\right)=0\)
\(\Leftrightarrow5x-3-4x+7=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy: x=-4
f) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{1}{3}\right\}\)
g) Ta có: \(x^2+6x-16=0\)
\(\Leftrightarrow x^2-2x+8x-16=0\)
\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-8\right\}\)
h) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;2\right\}\)
i) Ta có: \(x^2+x-2=0\)
\(\Leftrightarrow x^2-x+2x-2=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;-2\right\}\)
k) Ta có: \(3x^2+7x+2=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;\frac{-1}{3}\right\}\)
l) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-2x-10x+5=0\)
\(\Leftrightarrow2x\left(2x-1\right)-5\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
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 \(S=\left\{1;-2;-3\right\}\)
TN
12 tháng 3 2020
\(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\\\Leftrightarrow\left(x+1\right)^2=4\left(x-1\right)^2\\\Leftrightarrow \left(x+1\right)^2-4\left(x-1\right)^2=0\\\Leftrightarrow \left(x+1\right)^2-\left(2x-2\right)^2=0\\\Leftrightarrow \left[\left(x+1\right)+\left(2x-2\right)\right]\left[\left(x+1\right)-\left(2x-2\right)\right] =0\\ \Leftrightarrow\left(x+1+2x-2\right)\left(x+1-2x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(3-x\right)=0\\\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{1}{3};3\right\}\)
\(\left(2x+7\right)^2=9\left(x+2\right)^2\\ \Leftrightarrow\left(2x+7\right)^2-9\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\\ \Leftrightarrow\left[\left(2x+7\right)+\left(3x+6\right)\right]\left[\left(2x+7\right)-\left(3x+6\right)\right]=0\\ \Leftrightarrow\left(2x+7+3x+6\right)\left(2x+7-3x-6\right)=0\\ \Leftrightarrow\left(5x+13\right)\left(1-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+13=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-13}{5}\\x=1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-13}{5};1\right\}\)
\(4\left(2x+7\right)^2=9\left(x+3\right)^2\\\Leftrightarrow 4\left(2x+7\right)^2-9\left(x+3\right)=0\\ \Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\\\Leftrightarrow \left[\left(4x+14\right)+\left(3x+9\right)\right]\left[\left(4x+14\right)-\left(3x+9\right)\right]=0\\\Leftrightarrow \left(4x+14+3x+9\right)\left(4x+14-3x-9\right)=0\\\Leftrightarrow \left(7x+23\right)\left(x+5\right)=0\\\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-23}{7};-5\right\}\)
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 \(S=\left\{1;3\right\}\)
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}
\(6x^2-7x+2=0\)
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 \(\hept{\begin{cases}x+1=0\\x-1=0\end{cases}}\Leftrightarrow x=\pm1\)
\(6x^2-7x+2=0\)
\(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow3x\left(2x-1\right)-2\left(2x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{1}{2}\end{cases}}\)
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 : \(S=\left\{\pm1\right\}\)