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7 tháng 7 2016
a) \(5x\left(x-1\right)=x-1\)
\(\Rightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right).\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
Vậy \(x=1\) hoặc \(x=\frac{1}{5}\)
b) \(2\left(x+5\right)-x^2-5x\)
\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)\)
\(\Rightarrow\left(x+5\right).\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
Vậy \(x=-5\)hoặc \(x=2\)
HH
25 tháng 6 2018
a/ \(\left(x+2\right)^2-9=0\)
<=> \(\left(x+2-3\right)\left(x+2+3\right)=0\)
<=> \(\left(x-1\right)\left(x+5\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
b/ \(x^2-2x+1=25\)
<=> \(\left(x-1\right)^2=25\)
<=> \(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)
<=> \(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
TK
6
26 tháng 8 2020
a) 5x( x - 1 ) = x - 1
<=> 5x2 - 5x = x - 1
<=> 5x2 - 5x - x + 1 = 0
<=> 5x2 - 6x + 1 = 0
<=> 5x2 - 5x - x + 1 = 0
<=> 5x( x - 1 ) - 1( x - 1 ) = 0
<=> ( x - 1 )( 5x - 1 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
b) 2( x + 5 ) - x2 - 5x = 0
<=> 2x + 10 - x2 - 5x = 0
<=> -x2 - 3x + 10 = 0
<=> -x2 - 5x + 2x + 10 = 0
<=> -x( x + 5 ) + 2( x + 5 ) = 0
<=> ( x + 5 )( 2 - x ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
c) x2 - 2x - 3 = 0
<=> x2 + x - 3x - 3 = 0
<=> x( x + 1 ) - 3( x + 1 ) = 0
<=> ( x + 1 )( x - 3 ) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
d) 2x2 + 5x - 3 = 0
<=> 2x2 - x + 6x - 3 = 0
,<=> x( 2x - 1 ) + 3( 2x - 1 ) = 0
<=> ( 2x - 1 )( x + 3 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)
PN
26 tháng 8 2020
a) 5x ( x - 1 ) = x - 1 <=> 5x2 - 5x - x + 1 = 0
<=> 5x2 - 6x + 1 = 0 <=> 5x2 - x - ( 5x - 1 ) = 0
<=> x ( 5x - 1 ) - ( 5x - 1 ) = 0 <=> ( x - 1 )( 5x - 1 ) = 0
<=> x = 1 hoặc x = 1/5
b) 2 ( x + 5 ) - x2 - 5x = 0 <=> 2 ( x + 5 ) - x ( x + 5 ) = 0
<=> ( 2 - x ) ( x + 5 ) = 0 <=> x = 2 hoặc x = -5
c) x2 - 2x - 3 = 0 <=> x2 + x - 3x - 3 = 0
<=> x ( x + 1 ) - 3 ( x + 1 ) = 0 <=> ( x - 3 ) ( x + 1 ) = 0
<=> x = 3 hoặc x = -1
d) 2x2 + 5x - 3 = 0
Ta có : delta = 52 - 4.2.3 = 25 - 24 = 1
Khi đó : x = -1 hoặc x = 3/2
MA
1
B
30 tháng 10 2020
a) \(5x\left(x+4\right)-x\left(5x+1\right)=0\)
\(\Leftrightarrow x\left[5\left(x+4\right)-5x-1\right]=0\)
\(\Leftrightarrow x\left(5x+20-5x-1\right)=0\Leftrightarrow x=0\)
b) \(3x\left(5-x\right)+4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(4-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)
c) \(x\left(x-3\right)+4x-12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
d) \(x^2-36=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
e) \(x^2+3x+1=2\)
\(\Leftrightarrow x^2+3x+1-2=0\)
\(\Leftrightarrow x^2+3x-1=0\)
\(\Leftrightarrow x^2+3x+\frac{3}{2}-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}+\frac{\sqrt{5}}{\sqrt{2}}\right)\left(x+\frac{3}{2}-\frac{\sqrt{5}}{\sqrt{2}}\right)=0\)
Còn lại ........... Tự lm nất nha
CM
2
10 tháng 8 2016
a) 5.(x - 1) = x - 1
=> 5.(x - 1) - (x - 1) = 0
=> (x - 1).(5 - 1) = 0
=> (x - 1).4 = 0
=> x - 1 = 0
=> x = 0 + 1 = 1
b) 2.(x + 5) - x2 - 5x = 0
=> 2.(x + 5) - x.(x + 5) = 0
=> (x + 5).(2 - x) = 0
=> \(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)=> \(\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
A
1 tháng 9 2020
Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
25 tháng 7 2019
a. (5x-1)2 - (5x-4) (5x-4) +7
= (5x-1)2 - (5x-4)2 + 7
=[(5x-1)+(5x-4)] [(5x-1)-(5x-4)] +7 ( đoạn này bỏ cx đc)
=(10x-5) .3+7
=30x-15+7
=30x-8
SA
5
NC
19 tháng 8 2020
Bài làm:
a) \(x+5x^2=0\)
\(\Leftrightarrow x\left(1+5x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+5x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}\)
b) \(x\left(x-1\right)=x-1\)
\(\Leftrightarrow x^2-x-x+1=0\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
c) \(5x\left(x-1\right)=1-x\)
\(\Leftrightarrow5x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{5}\end{cases}}\)
d) \(\left(3x-4\right)^2-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(2x-5\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\4x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{4}\end{cases}}\)
PN
19 tháng 8 2020
\(a,x+5x^2=0< =>x\left(5x+1\right)=0\)
\(< =>\orbr{\begin{cases}x=0\\5x+1=0\end{cases}< =>\orbr{\begin{cases}x=0\\5x=-1\end{cases}< =>\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}}}\)
\(b,x\left(x-1\right)=x-1< =>x^2-x=x-1\)
\(< =>x^2-x-x+1=0< =>x\left(x-1\right)-\left(x-1\right)=0\)
\(< =>\left(x-1\right)\left(x-1\right)=0< =>x=1\)
\(c,5x\left(x-1\right)=1-x< =>5x^2-5x=1-x\)
\(< =>5x^2-5x+x-1=0< =>5x^2-4x-1=0\)
\(< =>5x^2-5x+x-1=0< =>5x\left(x-1\right)+x-1=0\)
\(< =>\left(5x+1\right)\left(x-1\right)=0< =>\orbr{\begin{cases}5x+1=0\\x-1=0\end{cases}}\)
\(< =>\orbr{\begin{cases}5x=-1\\x=1\end{cases}< =>\orbr{\begin{cases}x=-\frac{1}{5}\\x=1\end{cases}}}\)
\(d,\left(3x-4\right)^2-\left(x+1\right)^2=0\)
\(< =>9x^2-24x+16-x^2-2x-1=0\)
\(< =>8x^2-26x+15=0< =>8\left(x^2-\frac{13}{4}x+\frac{169}{64}\right)-\frac{2082}{64}=0\)
\(< =>\left(x-\frac{13}{8}\right)^2=\frac{2082}{512}=\frac{2082}{16\sqrt{2}}\)
\(< =>\orbr{\begin{cases}x-\frac{13}{8}=\frac{\sqrt{2082}}{4\sqrt[4]{2}}\\x-\frac{13}{8}=-\frac{\sqrt{2082}}{4\sqrt[4]{2}}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{13}{8}+\frac{\sqrt{2082}}{4\sqrt[4]{2}}\\x=\frac{13}{8}-\frac{\sqrt{2082}}{4\sqrt[4]{2}}\end{cases}}\)(nghiệm vô tỉ)
23 tháng 7 2016
1/a ) = (x+y)3 -(x+y)
= (x+y)[(x+y)2+1]
c) = 5(x2-xy+y2)-20z2
=5(x-y)2-20z2
= 5 [ (x-y)2- 4z2 ]
=5(x-y-4z)(x-y+4z)
TN
23 tháng 7 2016
Bài 1:
a) x3-x+3x2y+3xy2+y3-y
=x3+2x2y-x2+xy2-xy+x2y+2xy2-xy+y3-y2+x2+2xy-x+y2-y
=x(x2+2xy-x+y2-y)+y(x2+2xy-x+y2-y)+(x2+2xy-x+y2-y)
=(x2+2xy-x+y2-y)(x+y+1)
=[x(x+y-1)+y(x+y-1)](x+y+1)
=(x+y-1)(x+y)(x+y+1)
c) 5x2-10xy+5y2-20z2
=-5(2xy-y2+4z2-2)
Bài 2:
5x(x-1)=x-1
=>5x2-6x+1=0
=>5x2-x-5x+1
=>x(5x-1)-(5x-1)
=>(x-1)(5x-1)=0
=>x=1 hoặc x=1/5
b) 2(x+5)-x2-5x=0
=>2(x+5)-x(x+5)=0
=>(2-x)(x+5)=0
=>x=2 hoặc x=-5
sửa lại:
a) \(5x\left(x-1\right)=x-1\)
=> \(5x\left(x-1\right)-\left(x-1\right)=0\)
=> \(\left(x-1\right)\left(5x-1\right)=0\)
=> x - 1 = 0 hoặc 5x - 1 = 0
=> x = 1 hoặc 5x = 1 => x = 1/5
Vậy x = 1 hoặc x = 1/5
b) \(2\left(5+x\right)-x^2-5x=0\)
=> \(2\left(5+x\right)-\left(x^2+5x\right)=0\)
=> \(2\left(5+x\right)-x\left(x+5\right)=0\)
=> \(\left(x+5\right)\left(2-x\right)=0\)
=> x + 5 = 0 hoặc 2 - x = 0
=> x = -5 hoặc x = 2
a, 5x(x-1)=x-1
<=>5x(x-1)-(x-1)=0
<=>(5x-1)(x-1)=0
<=>5x-1=0 hoặc x-1=0
<=>x=1/5 hoặc x=1
b,2(5+x)-x2-5x=0
<=>2(x+5)-x(x+5)=0
<=>(2-x)(x+5)=0
<=>2-x=0 hoặc x+5=0
<=>x=2 hoặc x=-5