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RN
2
XO
1 tháng 10 2020
a) (2x - 3)2 = (x + 5)2
=> 4x2 - 12x + 9 = x2 + 10x + 25
=> 4x2 - 12x + 9 - (x2 + 10x + 25) = 0
=> 3x2 - 22x - 16 = 0
=> 3x2 - 24x + 2x - 16 = 0
=> 3x(x - 8) + 2(x - 8) = 0
=> (3x + 2)(x - 8) = 0
=> \(\orbr{\begin{cases}3x+2=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=8\end{cases}}\)
b) x2(x - 1) - 4x2 + 8x - 4 = 0
=> x2(x - 1) - (2x - 2)2 = 0
=> x2(x - 1) - [2(x- 1)]2 = 0
=> x2(x - 1) - 4(x - 1)2 = 0
=> (x - 1)(x2 - 4(x - 1) = 0
=> (x - 1)(x2 - 4x + 4) = 0
=> (x - 1)(x - 2)2 = 0
=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
c) x2 + 7x + 12 = 0
=> x2 + 3x + 4x + 12 = 0
=> x(x + 3) + 4(x + 3) = 0
=> (x + 4)(x + 3) = 0
=> \(\orbr{\begin{cases}x+4=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-4\\x=-3\end{cases}}\)
d) x2 + 3x - 18 = 0
=> x2 + 6x - 3x - 18 = 0
=> x(x + 6) - 3(x + 6) = 0
=> (x - 3)(x + 6) = 0
=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)
e) x(x + 6) - 7x - 42 = 0
=> x(x + 6) - 7(x + 6) = 0
=> (x - 7)(x + 6) = 0
=> \(\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)
1 tháng 10 2020
1. ( 2x - 3 )2 = ( x + 5 )2
<=> ( 2x - 3 )2 - ( x + 5 )2 = 0
<=> [ ( 2x - 3 ) - ( x + 5 ) ][ ( 2x - 3 ) + ( x + 5 ) ] = 0
<=> ( 2x - 3 - x - 5 )( 2x - 3 + x + 5 ) = 0
<=> ( x - 8 )( 3x + 2 ) = 0
<=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)
2. x2( x - 1 ) - 4x2 + 8x - 4 = 0
<=> x2( x - 1 ) - ( 4x2 - 8x + 4 ) = 0
<=> x2( x - 1 ) - 4( x2 - 2x + 1 ) = 0
<=> x2( x - 1 ) - 4( x - 1 )2 = 0
<=> ( x - 1 )[ x2 - 4( x - 1 ) ] = 0
<=> ( x - 1 )( x2 - 4x + 4 ) = 0
<=> ( x - 1 )( x - 2 )2 = 0
<=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
3. x2 + 7x + 12 = 0
<=> x2 + 3x + 4x + 12 = 0
<=> x( x + 3 ) + 4( x + 3 ) = 0
<=> ( x + 3 )( x + 4 ) = 0
<=> \(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)
4. x2 + 3x - 18 = 0
<=> x2 - 3x + 6x - 18 = 0
<=> x( x - 3 ) + 6( x - 3 ) = 0
<=> ( x - 3 )( x + 6 ) = 0
<=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)
5. x( x + 6 ) - 7x - 42 = 0
<=> x( x + 6 ) - 7( x + 6 ) = 0
<=> ( x + 6 )( x - 7 ) = 0
<=> \(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}\)
7 tháng 11 2021
e) \(\left(9x^2-49\right)+\left(3x+7\right)\left(7x+3\right)=0\)
\(\Rightarrow\text{[}\left(3x\right)^2-7^2\text{]}+\left(3x+7\right)\left(7x+3\right)=0\)
\(\Rightarrow\left(3x-7\right)\left(3x+7\right)+\left(3x+7\right)\left(7x+3\right)=0\)
\(\Rightarrow\left(3x+7\right)\text{[}\left(3x-7\right)+\left(7x+3\right)\text{]}=0\)
\(\Rightarrow\left(3x+7\right)\left(3x-7+7x+3\right)=0\)
\(\Rightarrow\left(3x+7\right)\left(10x-4\right)=0\)
=> 2 TH
*3x+7=0 *10x-4=0
=>3x=-7 =>10x=4
=>x=-7/3 =>x=4/10=2/5
vậy x=-7/3 hoặc x=2/5
g) \(\left(x-4\right)^2=\left(2x-1\right)^2\)
\(\Rightarrow\left(x-4\right)^2-\left(2x-1\right)^2=0\)
\(\Rightarrow\left(x-4-2x+1\right)\left(x-4+2x-1\right)=0\)
\(\Rightarrow\left(-x-3\right)\left(3x-5\right)=0\)
\(\Rightarrow-\left(x+3\right)\left(3x-5\right)=0\)
=> 2 TH
*-(x+3)=0 *3x-5=0
=>-x=-3 =>3x=5
=x=3 =>x=5/3
h)\(x^2-x^2+x-1=0\)
\(\Rightarrow0+x-1=0\)
\(\Rightarrow x-1=0\)
=>x=0+1
=>x=1
vậy x=1
k, x(x+ 16) - 7x - 42 = 0
=>x^2+16x-7x-42=0
=>x^2+9x-42=0
vì x^2>0
do đó x^2+9x-42>0
nên o có gt nào của x t/m y/cầu đề bài
m)x^2+7x+12=0
=>x^2+3x++4x+12=0
=>x(x+3)+4(x+3)=0
=>(x+4).(x+3)=0
=>2 TH
=> *x+4=0
=>x=-4
vậy x=-4
*x+3=0
=>x=-3
vậy x=-3
n)x^2-7x+12=0
=>x^2-4x-3x+12=0
=>x(x-4)-3(x-4)=0
=>(x-3).(x-4)=0
=>2 TH
*x-3=0=>x=0+3=>x=3
*x-4=0=>x=0+4=>x=4
vậy x=3 hoặc x=4
KA
7 tháng 11 2021
a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1
b)(x+1)(x+2)(x+5)−x2(x+8)=27⇔x2+2x+x+2(x+5)−x3−8x2=27⇔x2(x+5)+2x(x+5)+x(x+5)+2(x+5)−x3−8x2=27⇔x3+5x2+2x2+10x+x2+5x+2x+10−x3−8x2=27⇔17x+10=27⇔17x=17⇒x=1
23 tháng 6 2017
\(\left(x+1\right)^2=x+1\)
\(\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+1-1\right)=0\)
\(\left(x+1\right)x=0\)
\(\orbr{\begin{cases}x+1=0\\x=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)vậy.....
\(x\left(x-5\right)^2-4x+20=0\)
\(x\left(x-5\right)^2-4\left(x-5\right)=0\)
\(\left(x-5\right)\left[x\left(x-5\right)-4\right]=0\)
\(\left(x-5\right)\left(x^2-5x-4\right)=0\)
\(\orbr{\begin{cases}x-5=0\\x^2-5x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-0,7015621187\end{cases}}}\)vậy.........
\(x\left(x+6\right)-7x-42=0\)
\(x\left(x+6\right)-7\left(x+6\right)=0\)
\(\left(x+6\right)\left(x-7\right)=0\)
\(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}}\) vậy....
\(x^3-5x^2+x-5=0\)
\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\orbr{\begin{cases}x-5=0\\x^2+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x^2=-1\Rightarrow x\in\Phi\end{cases}}}\)vậy........
\(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x-2\right)\left(x^3+10x\right)=0\)
\(\orbr{\begin{cases}x-2=0\\x^3+10x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)vậy..............
nhớ chọn mk nha
3 tháng 1 2020
a. x(x-5)-4x+20=0
\(\Leftrightarrow\)x(x-5)-4(x-5)=0
\(\Leftrightarrow\)(x-4)(x-5)=0
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}}\)
b, x(x+6)-7x-42=0
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\\ \Leftrightarrow\left(x-7\right)\left(x+6\right)=0\\ \Leftrightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}}\)
3, x^3-5x^2+x-5=0
\(\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-5\right)=0\\ \Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow x-5=0\\ \Leftrightarrow x=5\)
ET
19 tháng 10 2018
1
a) x^2+2x-5 b) x^2+x+7 9 (dư 8)
2
x=2; x = -(3*căn bậc hai(7)*i+1)/2;x = (3*căn bậc hai(7)*i-1)/2;
3
a=2
26 tháng 6 2018
a) Qui đồng rồi khử mẫu ta được:
3(3x+2)-(3x+1)=2x.6+5.2
<=> 9x+6-3x-1 = 12x+10
<=> 9x-3x-12x = 10-6+1
<=> -6x = 5
<=> x = -5/6
Vậy ....
b) ĐKXĐ: \(x\ne\pm2\)
Qui đồng rồi khử mẫu ta được:
(x+1)(x+2)+(x-1)(x-2) = 2(x2+2)
<=> x2+3x+2+x2-3x+2 = 2x2+4
<=> x2+x2-2x2+3x-3x = 4-2-2
<=> 0x = 0
<=> x vô số nghiệm
Vậy x vô số nghiệm với x khác 2 và x khác -2
c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)
\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)
Vậy ......
d) (x+1)2-4(x2-2x+1) = 0
<=> x2+2x+1-4x2+8x-4 = 0
<=> -3x2+10x-3 = 0
giải phương trình
31 tháng 7 2018
1,x^2-(x+1)(x-1)=0
x^2-x^2+1+0
1=0(vô lý)
2,5x^3+3x^2+3x+1=4x^2
x^3+3x^2+3x+1=0
(x+1)=0
x=-1
3,x^3+x^2=0
x^2(x+1)=0
x=0 or x=-1
4,2x^3-12x^2+18x=0
x^3-6x^2+9x=0
x(x^2-6x+9)=0
x(x-3)^2=0
x=0 or x=3
5,5x^2-4(x^2-2x+1)+20=0
5x^2-4x^2+8x-4+20=0
x^2+8x+16=0
(x+4)^2=0
x=-4
6,5x(x-3)+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
5x-7=0 or x-3=0
x=7/5 or x=3
7,2x^3-50x=0
2x(x^2-25)=0
2x(x-5)(x+5)=0
x=0 or x=5 or x=-5
8,(4x-1)^2-9(x+3)^2=0
(4x-1)^2-3^2*(x+3)^2=0
(4x-1)^2-(3x+9)^2=0
(4x-1-3x-9)(4x-1+3x+9)=0
(x-10)(7x+8)=0
x=10 or x=-8/7
9,3(x-2)^2-x+2=0
3*(x-2)*(x-2)-(x-2)=0
(3x-6)(x-2)-(x-2)=0
(x-2)(3x-6-1)=0
(x-2)(3x-7)=0
x=2 or x=7/3
10,9x^2+6x-8=0
9x^2+12x-6x-8=0
3x(3x-2)+4(3x-2)=0
(3x+4)(3x-2)=0
3x+4=0 or 3x-2=0
x=-4/3 or x=2/3
20 tháng 7 2022
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
\(x^6-7x^3-8=0\)
\(\Leftrightarrow\left(x^3+1\right)\left(x^3-8\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)\left(x-2\right)\left(x^2+2x+4\right)=0\)(*)
Ta có: \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)(loại)
\(x^2+2x+4=\left(x+1\right)^2+3>0\)(loại)
Nên (*)\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\x=2\end{array}\right.\)(thỏa mãn)
Vậy phương trình đã cho có tập nghiệm là S={-1;2}
làm bài khác mk vs