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31 tháng 12 2017
a. \(2x\left(x+5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2+10x-3x-2x^2=26\Leftrightarrow7x=26\Leftrightarrow x=\dfrac{26}{7}\)
Vậy \(x=\dfrac{26}{7}\)
b. \(5x\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
c. \(2\left(x+5\right)-x^2-5x=0\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
d. \(\left(2x-3\right)^2-\left(x+5\right)^2=0\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
e. \(3x^3-48x=0\Leftrightarrow3x\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}3x=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)
f. \(x^3+x^2-4x=4\Leftrightarrow x^3+x^2-4x-4=0\Leftrightarrow\left(x^2-4x+4\right)+\left(x^3-8\right)=0\Leftrightarrow\left(x-2\right)^2+\left(x-2\right)\left(x^2+2x+4\right)=0\Leftrightarrow\left(x-2\right)\left(x-2+x^2+2x+4\right)=0\left(x-2\right)\left(x^2+3x+2\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+x+2x+2\right)=0\Leftrightarrow\left(x-2\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]=0\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\\x=-2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)
g. \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+3-x\right)=0\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
h. \(x^2-4x+8=2x-1\Leftrightarrow x^2-4x+8-2x+1=0\Leftrightarrow x^2-6x+9=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy \(x=3\)
__________________________Chúc bạn học tốt____________________________
12 tháng 10 2020
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
5 tháng 12 2017
1)⇔x2+1x-3x+3=0
⇔x(x+1)-3(x+1)=0
⇔(x+1)(x-3)=0
⇔x+1=0 hoặc x-3=0
⇔x=-1 hoặc x=3
5 tháng 12 2017
4)⇔x(1+5x)=0
⇔x=0 hoặc 1+5x=0
⇔x=0 hoặc 5x=-1
⇔x=0 hoặc x=-0.2
VM
21 tháng 2 2020
Mình giải kĩ lại câu cuối nha.
\(\left(3x+5\right).\left(x^2+x+1\right)=0\)
+ Vì \(x^2+x+1>0\) \(\forall x.\)
\(\Rightarrow x^2+x+1\ne0.\)
\(\Leftrightarrow3x+5=0\)
\(\Leftrightarrow3x=0-5\)
\(\Leftrightarrow3x=-5\)
\(\Leftrightarrow x=\left(-5\right):3\)
\(\Leftrightarrow x=-\frac{5}{3}\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-\frac{5}{3}\right\}.\)
Chúc bạn học tốt!
28 tháng 9 2018
\(2x\left(x-3\right)-x+3=0\)
<=> \(2x\left(x-3\right)-\left(x-3\right)=0\)
<=> \(\left(x-3\right)\left(2x-1\right)=0\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
19 tháng 2 2019
1) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)
2) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)
3) \(\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)
AL
12 tháng 1 2017
1. \(\left(x-4\right)^2-25=0\)
<=> (x-4+5).(x-4-5) = 0
<=> (x+1)(x-9) = 0
<=> \(\left[\begin{matrix}x+1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = {-1;9}
2. \(\left(2x-1\right)^2+\left(2-x\right)\left(2x-1\right)=0\)
<=> (2x-1)(2x-1+2-x) = 0
<=> (2x-1)(x+1) = 0
<=> \(\left[\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}2x=1\\x=-1\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0.5\\x=-1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = {-1 ; 0,5}
3. \(x^2+6x+9=4x^2\)
<=> \(\left(x+3\right)^2-4x^2=0\)
<=> (x+3+2x)(x+3-2x) = 0
<=> (3x+3)(3-x) = 0
<=> \(\left[\begin{matrix}3x+3=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}3x=-3\\x=3\end{matrix}\right.\Leftrightarrow}\left[\begin{matrix}x=-1\\x=3\end{matrix}\right.\) Vậy phương trình có tập nghiệm S = {-1 ; 3}
4. (2x-5)(x+11) = (5-2x)(2x+1)
<=> (2x-5)(x+11) = - (2x-5)(2x+1)
<=> x + 11 = -2x - 1
<=> x+2x = -12
<=> 3x = -12
<=> x = -4
Vậy phương trình có một nghiệm duy nhất là x = -4
5. \(2x^2+5x+3=0\)
<=> \(2x^2+2x+3x+3=0\)
<=> \(2x\left(x+1\right)+3\left(x+1\right)=0\)
<=> \(\left(x+1\right)\left(2x+3\right)=0\)
<=> \(\left[\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\Leftrightarrow}\left[\begin{matrix}x=-1\\x=\frac{-3}{2}\end{matrix}\right.\) Vậy phương trình có tập nghiệm S = { -1 ; -3/2 }
DN
12 tháng 1 2017
1) (x-4)^2-25=0
<=> (x-4+5)(x-4-5)=0
\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
2) (2x-1)2+(2-x)(2x-1)=0
<=> (2x-1)(2+2-x)=0
<=> \(\left[\begin{matrix}x=\frac{1}{2}\\x=4\end{matrix}\right.\)
3) x^2+6x+9=4x^2
<=> 3x^2 -6x-9=0
<=> x^2 -2x -3=0
<=> x^2 -3x+x-3=0
<=> x(x-3)+(x-3)=0
<=> (x-3)(x+1)=0
=>\(\left[\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
4) (2x-5)(x+11)=(5-2x)(2x+1)
-(5-2x)(x+11)-(5-2x)(2x+1)=0
(5-2x)(x+11+2x+1)=0
=>\(\left[\begin{matrix}x=\frac{5}{2}\\x=-4\end{matrix}\right.\)
5)2x^2+5x+3=0
2x^2+2x+3x+3=0
2x(x+1)+3(x+1)=0
(x+1)(2x+3)=0
=>\(\left[\begin{matrix}x=-1\\x=\frac{-3}{2}\end{matrix}\right.\)
CM
4 tháng 12 2018
a) \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
b) \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\left(2x+5\right)^2=\left(x+2\right)^2\)
\(\Rightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\)
\(\Rightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\)
\(\Rightarrow\left(x+3\right)\left(3x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\3x+7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\3x=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
d) \(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
e) \(2x^3+6x^2=x^2+3x\)
\(\Rightarrow2x^3+6x^2-x^2-3x=0\)
\(\Rightarrow2x^3+5x^2-3x=0\)
\(\Rightarrow x\left(2x^2+5x-3\right)=0\)
\(\Rightarrow2x^2+5x-3=0\)
\(\Rightarrow2x^2-6x+x-3=0\)
\(\Rightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
f) \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)-2x^2\)
\(\Rightarrow\left(x^2-1\right)\left(x+2\right)-\left(x^3-8\right)-2x^2=0\)
\(\Rightarrow x^3+2x^2-x+2-x^3+8-2x^2=0\)
\(\Rightarrow-x+10=0\)
\(\Rightarrow x=10\)
Tập nghiệm: \(S=\left\{1;-\frac{3}{2}\right\}\)