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29 tháng 11 2023
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
NT
5
11 tháng 9 2017
a) \(x^3\)\(-\)\(\frac{1}{4}x\)\(=\)\(0\)
\(x\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2=0,5^2\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=+-0,5\end{cases}}\)
Vậy .............................
b) \(\left(2x-1\right)^2\)\(-\)\(\left(x+3\right)^2\)\(=\)\(0\)
\(\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)
\(\left(3x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-4=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=4\end{cases}}\)\(\orbr{\begin{cases}x=\frac{-2}{3}\\x=4\end{cases}}\)
Vậy ................................
c) \(x^2\)\(\left(x-3\right)\)\(+\)\(12\)\(-\)\(4x\)\(=\)\(0\)
\(x^2\)\(\left(x-3\right)\)\(-\)\(4\)\(\left(x-3\right)\)\(=\)\(0\)
\(\left(x^2-4\right)\left(x-3\right)\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x-3=0\end{cases}-4=0}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x=3\end{cases}=2^2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=+-2\\x=3\end{cases}}\)
11 tháng 9 2017
a)\(x^3-\frac{1}{4}x=0\)
\(\Leftrightarrow x\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
27 tháng 3 2020
a) Ta có: \(x^2-9x+20=0\)
\(\Leftrightarrow x^2-5x-4x+20=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
Vậy: x∈{4;5}
b) Ta có: \(x^3-4x^2+5x=0\)
\(\Leftrightarrow x\left(x^2-4x+5\right)=0\)(1)
Ta có: \(x^2-4x+5\)
\(=x^2-4x+4+1=\left(x-2\right)^2+1\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2+1\ge1>0\forall x\)
hay \(x^2-4x+5>0\forall x\)(2)
Từ (1) và (2) suy ra x=0
Vậy: x=0
c) Sửa đề: \(x^2-2x-15=0\)
Ta có: \(x^2-2x-15=0\)
\(\Leftrightarrow x^2+3x-5x-15=0\)
\(\Leftrightarrow x\left(x+3\right)-5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
Vậy: x∈{-3;5}
d) Ta có: \(\left(x^2-1\right)^2=4x+1\)
\(\Leftrightarrow x^4-2x^2+1-4x-1=0\)
\(\Leftrightarrow x^4-2x^2-4x=0\)
\(\Leftrightarrow x\left(x^3-2x-4\right)=0\)
\(\Leftrightarrow x\left(x^3+2x^2+2x-2x^2-4x-4\right)=0\)
\(\Leftrightarrow x\cdot\left[x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\right]=0\)
\(\Leftrightarrow x\cdot\left(x^2+2x+2\right)\cdot\left(x-2\right)=0\)(3)
Ta có: \(x^2+2x+2\)
\(=x^2+2x+1+1=\left(x+1\right)^2+1\)
Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^2+1\ge1>0\forall x\)
hay \(x^2+2x+2>0\forall x\)(4)
Từ (3) và (4) suy ra
\(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy: x∈{0;2}
11 tháng 7 2021
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(A=\dfrac{1}{x+2}-\dfrac{x^3-4x}{x^2+4}\cdot\left(\dfrac{1}{x^2+4x+4}+\dfrac{1}{4-x^2}\right)\)
\(=\dfrac{1}{x+2}-\dfrac{x\left(x+2\right)\left(x-2\right)}{x^2+4}\cdot\dfrac{x-2-x-2}{\left(x+2\right)^2\left(x-2\right)}\)
\(=\dfrac{1}{x+2}-\dfrac{-4x}{\left(x+2\right)\left(x^2+4\right)}\)
\(=\dfrac{x^2+4+4x}{\left(x+2\right)\left(x^2+4\right)}\)
\(=\dfrac{x+2}{x^2+4}\)
b) Để A>0 thì x+2>0
hay x>-2 và \(x\ne2\)
Để A<0 thì x+2<0
hay x<-2
Để A=0 thì x+2=0
hay x=-2(loại)
6 tháng 6 2018
Bài 1. a) 4x - 3 = 0
⇔ x = \(\dfrac{3}{4}\)
KL.....
b) - x + 2 = 6
⇔ x = - 4
KL...
c) -5 + 4x = 10
⇔ 4x = 15
⇔ x = \(\dfrac{15}{4}\)
KL....
d) 4x - 5 = 6
⇔ 4x = 11
⇔ x = \(\dfrac{11}{4}\)
KL....
h) 1 - 2x = 3
⇔ -2x = 2
⇔ x = -1
KL...
Bài 2. a) ( x - 2)( 4 + 3x ) = 0
⇔ x = 2 hoặc x = \(\dfrac{-4}{3}\)
KL......
b) ( 4x - 1)3x = 0
⇔ x = 0 hoặc x = \(\dfrac{1}{4}\)
KL.....
c) ( x - 5)( 1 + 2x) = 0
⇔ x = 5 hoặc x = \(\dfrac{-1}{2}\)
KL.....
d) 3x( x + 2) = 0
⇔ x = 0 hoặc x = -2
KL.....
6 tháng 6 2018
Bài 3.a) 3( x - 4) - 2( x - 1) ≥ 0
⇔ x - 10 ≥ 0
⇔ x ≥ 10
b) 3 - 2( 2x + 3) ≤ 9x - 4
⇔ - 4x - 3 ≤ 9x - 4
⇔ 13x ≥1
⇔ x ≥ \(\dfrac{1}{13}\)
N
1
4 tháng 8 2018
\(4x^2+4x-3=0\)
\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)
\(\left(2x+1\right)^2-2^2=0\)
\(\left(2x+1-2\right).\left(2x+1+2\right)=0\)
\(\left(2x-1\right).\left(2x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)
\(x^4-3x^3-x+3=0\)
\(x^3.\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right).\left(x^3-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(x^2.\left(x-1\right)-4x^2+8x-4=0\)
\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)
\(x^2.\left(x-1\right)-\left(2x-2\right)^2=0\)
\(x^2.\left(x-1\right)-4.\left(x-1\right)^2=0\)
\(\left(x-1\right).\left[x^2-4.\left(x-1\right)\right]=0\)
\(\left(x-1\right).\left[x^2-2.x.2+2^2\right]=0\)
\(\left(x-1\right).\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy \(\begin{cases}x=1\\x=2\end{cases}\)
Tham khảo nhé~
N
0
15 tháng 1 2017
a) \(^{x^3}\) - 7x+6=0
\(\Leftrightarrow\) \(^{x^3}\) - x-6x+6=0
\(\Leftrightarrow\) \(\left(x^3-x\right)\) - \(\left(6x-6\right)\) =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
\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left[x-6\left(x+1\right)\right]\) =0
\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left(6-5x\right)\) =0
\(\Leftrightarrow\) \(\left[\begin{matrix}x-1=0\\6-5x=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\5x=-6\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\x=-\frac{6}{5}\end{matrix}\right.\)
Những câu sau dùng phương pháp phân tích đa thức thành nhân tử nhé!
AK
1
PA
17 tháng 10 2016
\(x^2+4x+3=0\)
\(x^2+x+3x+3=0\)
\(x\left(x+1\right)+3\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+3\right)=0\)
\(\left[\begin{array}{nghiempt}x+1=0\\x+3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=-1\\x=-3\end{array}\right.\)
\(4x^2+4x-3=0\)
\(4x^2-2x+6x-3=0\)
\(2x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(2x+3\right)=0\)
\(\left[\begin{array}{nghiempt}2x-1=0\\2x+3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=1\\2x=-3\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-\frac{3}{2}\end{array}\right.\)
\(x^2-x-12=0\)
\(x^2-4x+3x-12=0\)
\(x\left(x-4\right)+3\left(x-4\right)=0\)
\(\left(x-4\right)\left(x+3\right)=0\)
\(\left[\begin{array}{nghiempt}x-4=0\\x+3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=4\\x=-3\end{array}\right.\)
\(x^2-25-\left(x-5\right)=0\)
\(\left(x-5\right)\left(x+5\right)-\left(x-5\right)=0\)
\(\left(x-5\right)\left(x+5-1\right)=0\)
\(\left(x-5\right)\left(x+4\right)=0\)
\(\left[\begin{array}{nghiempt}x-5=0\\x+4=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=5\\x=-4\end{array}\right.\)
\(x^2\left(x^2+1\right)-x^2-1=0\)
\(x^2\left(x^2+1\right)-\left(x^2+1\right)=0\)
\(\left(x^2+1\right)\left(x^2-1\right)=0\)
\(\left(x^2+1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+1=0\end{array}\right.\) (vì \(x^2+1\ge1>0\))
\(\left[\begin{array}{nghiempt}x=1\\x=-1\end{array}\right.\)
13 tháng 2 2022
a: =>-2x=-8
hay x=4
b: =>7x=-21
hay x=-3
c: =>0,25x=-1,5
hay x=-6
d: =>5,3x=6,36
hay x=6/5
e: =>-4x=-12
hay x=3
f: =>-10x=-10
hay x=1
g: =>2x+2-3-2x=0
=>-1=0(vô lý)
h: =>3-3x+4x-3=0
=>x=0
DT
13 tháng 2 2022
a,
\(3-x=x-5\\ \Leftrightarrow3x-x+5=0\Leftrightarrow2x+5=0\)
\(\Rightarrow x=-\dfrac{5}{2}\)
b, \(\Rightarrow x=-\dfrac{21}{7}=-3\)
c, \(\Leftrightarrow x=\left(0-1,5\right):0,25=-6\)
\(\frac{x^3-1}{4x}=0\)
Ta có : \(4x>0\)
=> x3 -1 = 0
x3 = 1
x = 1
hok tốt .
\(a)\frac{x^3-1}{4x}=0\)
\(\Rightarrow x(x^2-\frac{1}{4})=0\Rightarrow x(x-\frac{1}{2})(x+\frac{1}{2})=0\)
VT là tích của 3 số , VT là 0 nên có :
\(x=0\)hoặc \((x-\frac{1}{2})=0\)hoặc \((x+\frac{1}{2})=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=\frac{-1}{2}\end{cases}}\)
Chúc bạn học tốt