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TN
28 tháng 7 2016
a)(2x-3)2=(x+5)2
=>4x2-12x+9=x2+10x+25
=>3x2-22x-16=0
=>3x2+2x-24x-16=0
=>x(3x+2)-8(3x+2)=0
=>(x-8)(3x+2)=0
=>x=8 hoặc x=-2/3
b)X2.(x-1)-4x2+8x-4=0
=>x2(x-1)-4x2+4x+4x-4=0
=>x2(x-1)-4x(x-1)-4(x-1)=0
=>x2(x-1)-(4x-4)(x-1)=0
=>(x2-4x+4)(x-1)=0
=>(x-2)2(x-1)=0
=>x=2 hoặc x=1
c) 4x2- 25 - (2x- 5) . ( 2x+7)=0
=>4x2-25-(4x2+14x-10x-35)=0
=>4x2-25-4x2-14x+10x+35=0
=>-4x+10=0
=>-4x=-10 <=>x=5/2
d) x3+27+(x+3).(x-9)=0
=>x3+33+(x+3)(x-9)=0
=>(x+3)(x2-3x+9)+(x+3)(x-9)=0
=>(x2-3x+9+x-9)(x+3)=0
=>(x2-2x)(x+3)=0
=>x(x-2)(x+3)=0
=>x=0 hoặc x=2 hoặc x=-3
e) (x-2).(x+5)- x2+4=0
=>(x-2)(x+5)-(x-2)(x+2)=0
=>(x-2)(x+5-x-2)=0
=>3(x-2)=0 <=>x=2
28 tháng 7 2016
Sau khi khai triển hằng đẳng thức và thực hiện chuyển vế bạn sẽ đk kết quả như này!(\(\left(2x-3\right)^2=\left(x+5\right)^2=3x^2-22x-14\)
DT
4
NN
8 tháng 12 2019
\(2x\left(x^2-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-25=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)
\(2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\left(2x+1\right)\left(3x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x-5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{3}\end{cases}}\)
NN
8 tháng 12 2019
\(9\left(3x-2\right)-x\left(2-3x\right)=0\)
\(9\left(3x-2\right)+x\left(3x-2\right)=0\)
\(\left(9+x\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}9+x=0\\3x-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-9\\x=\frac{2}{3}\end{cases}}\)
\(\left(2x-1\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
ND
1
19 tháng 10 2016
a, x^2(x^2+4)-x^2-4=0
=) x^2(x^2+4)-(x^2+4)=0
=) (x^2+4)(x^2-1)=0
=)(x^2+4)(x-1)(x+1)=0
=) x^2+4=0 =)x^2=4 =)x=2
x-1=0 =)x=1
x+1=0 =)x=-1
b,x^2-25-x-5=0
=)(x+5)(x-5)-(x+5)=0
=)(x+5)(x-6)=0
=)x+5=0 =) x=-5
x-6=0 =) x=6
KK
8 tháng 12 2018
a.(2x - 5)(3x + 4) - x(6x - 5) = 4
⇔ 6x2 +8x -15x-20-6x2+5x=4
⇔-2x=24
⇔ x=-12
vậy x=12
b.(x - 2)2 + x(x - 2) = 0
⇔(x-2)(x-2+x)=0
⇔(x-2) (2x-2)=0
⇔ (x-2)2(x-2)=0
⇔(x-2)2.2=0
⇔(x-2)2=0
⇔x-2=0
⇔x=2
vậy x=2
c.(x3 + 4x2 - x - 4) : (x + 4) = 0
⇔[(x3+4x2)-(x+4)] :(x+4)=0
⇔ [x2(x+4)-(x+4)] :(x+4)=0
⇔ (x+4)(x2-1):(x+4)=0
⇔(x-1)(x+1)=0
⇔ \(\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
2T
17 tháng 8 2019
\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
2T
17 tháng 8 2019
\(x^5-9x=0\)
\(\Leftrightarrow x\left(x^4-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^4-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt[4]{9}\end{cases}}\)
LB
1
D
3 tháng 9 2018
a ) \(x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
b ) \(\left(2x-1\right)^2-4x^2-1=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2-1=0\)
\(\Leftrightarrow-4x=0\)
\(\Leftrightarrow x=0\)
c ) \(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow x^2\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-1\\x=1\end{matrix}\right.\)
Chúc bạn học tốt !!
CN
1
14 tháng 3 2020
a) x2 + 10x + 25 - 4x2 - 20x = 0
<=> 3x2 + 10x - 25 = 0
<=> (x + 5)(3x - 5) = 0 <=> \(\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}\)
Vậy S = \(\left\{-5;\frac{5}{3}\right\}\)
b. (4x - 5)2 - 2(4x - 5)(4x + 5) = 0
<=> (4x - 5)[(4x - 5) - 2(4x + 5)] = 0
<=> (4x - 5)(4x - 5 - 8x - 10) = 0
<=> (4x - 5)(-4x - 15) = 0 <=> \(\orbr{\begin{cases}x=\frac{5}{4}\\x=-\frac{15}{4}\end{cases}}\)
Vậy S = \(\left\{-\frac{15}{4};\frac{5}{4}\right\}\)
NT
1
KB
22 tháng 10 2018
a ) \(x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
b ) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(2x+1\right)\right]=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
c ) \(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow x^2\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x^2+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x^2=-4\left(VL\right)\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
16 tháng 11 2022
a: \(\Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
=>(x-2)(x^2+4x+6)=0
=>x-2=0
=>x=2
b: =>(2x-5)(2x+5)-(2x-5)(2x+7)=0
=>(2x-5)(2x+5-2x-7)=0
=>2x-5=0
=>x=5/2
c: =>(x+3)(x^2-3x+9+x-9)=0
=>(x+3)(x^2-2x)=0
=>\(x\in\left\{0;2;-3\right\}\)
24 tháng 4 2020
\(2x^3+5x^2+3x=0\\ < =>x\left(2x^2+5x+3\right)=0\\ < =>x\left[2x\left(x+1\right)+3\left(x+1\right)\right]=0\\< =>x\left(2x+3\right)\left(x+1\right)=0\\ < =>\left[{}\begin{matrix}x=0\\2x+3=0\\x+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=\frac{-3}{2}\\-1\end{matrix}\right.\)
\(\left(x+5\right)\left(x-3\right)+x^2-25=0\\ < =>\left(x+5\right)\left(x+3\right)+\left(x-5\right)\left(x+5\right)=0\\ < =>\left(x+5\right)\left(x-3+x-5\right)=0\\ < =>\left(x+5\right)\left(2x-8\right)=0\\ < =>\left[{}\begin{matrix}x+5=0\\2x-8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-5\\x=4\end{matrix}\right.\)
\(x\left(x-2\right)-3x+6=0\\ < =>x\left(x-2\right)-3\left(x-2\right)=0\\ < =>\left(x-2\right)\left(x-3\right)=0\\< =>\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
20 tháng 2 2021
@Mốc
2x3 + 5x2 + 3x = 0
⇔ x.(2x2 + 5x + 3) = 0
⇔ x.(x + 1).(2x + 3) = 0
TH1: x = 0
TH2: x + 1 = 0
⇔ x = - 1
TH3: 2x + 3 = 0
⇔ x = \(\dfrac{-3}{2}\)
Vậy S = {0;- 1;\(\dfrac{-3}{2}\)}
(x + 5).(x - 3) + x2 - 25 = 0
⇔ (x + 5).(x - 3) + (x - 5).(x + 5) = 0
⇔ (x + 5).(x - 3 + x - 5) = 0
⇔ (x + 5).(2x - 8) = 0
TH1: x + 5 = 0
⇔ x = - 5
TH2: 2x - 8 = 0
⇔ x = 4
Vậy S = {- 5; 4}
x.(x - 2) - 3x + 6 = 0
⇔ x.(x - 2) - 3.(x - 2) = 0
⇔ (x - 2).(x - 3) = 0
TH1: x - 2 = 0
⇔ x = 2.
TH2: x - 3 = 0
⇔ x = 3
Vậy S = {2;3}
#chucbanhoctot:)
\(x^2-25+x\left(x+5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5\right)+x\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x-5+x\right)=0\\ \Leftrightarrow\left(x+5\right)\left(2x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{5}{2}\end{matrix}\right.\)