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MC
12 tháng 7 2021
\(x\left(3x-5\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\3x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{5}{3}\right\}\)
AD
12 tháng 7 2021
a) \(x\left(3x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}}\)
b) \(3x^2-27=0\)
\(\Leftrightarrow3x^2=27\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow x=\pm3\)
c) \(\left(x-5\right)^2=x-5\)
\(\Leftrightarrow x^2-10x+25-x+5=0\)
\(\Leftrightarrow x^2-11x+30=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}}\)
d) \(2\left(x+7\right)-x^2-7x=0\)
\(\Leftrightarrow2x+14-x^2-7x=0\)
\(\Leftrightarrow-x^2-5x+14=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=2\end{cases}}}\)
e)\(7x\left(x-3\right)+2.3x=0\)
\(\Leftrightarrow7x^2-21x+6x=0\)
\(\Leftrightarrow7x^2-15x=0\)
\(\Leftrightarrow x\left(7x-15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\7x-15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{15}{7}\end{cases}}}\)
#H
27 tháng 7 2018
I don't now
sorry
...................
nha
KT
27 tháng 7 2018
a) \(\left(3x-1\right)^2+2\left(3x-1\right)\left(2x+1\right)+\left(2x+1\right)^2=0\)
\(\Leftrightarrow\)\(\left[\left(3x-1\right)+\left(2x-1\right)\right]^2=0\)
\(\Leftrightarrow\)\(\left(5x-2\right)^2=0\)
\(\Leftrightarrow\)\(5x-2=0\)
\(\Leftrightarrow\)\(x=\frac{2}{5}\)
Vậy...
b) \(\left(7x+2\right)^2+\left(7x-2\right)^2-2\left(7x+2\right)\left(7x-2\right)=0\)
\(\Leftrightarrow\)\(\left[\left(7x+2\right)-\left(7x-2\right)\right]^2=0\)
\(\Leftrightarrow\)\(4^2=0\) vô lí
Vậy pt vô nghiệm
N
1
13 tháng 7 2021
(3x+2).(x+1)=3x.(5+x)
\(\Rightarrow\)\(3x^2+3x+2x+2=15x+3x^2\)
\(\Rightarrow3x^2+5x+2=15x+3x^2\)
\(\Rightarrow5x-15x+2=3x^2-3x^2\)
\(\Rightarrow-10x+2=0\)
\(-10x=-2\)
\(x=\frac{1}{5}\)
H
2
NN
8 tháng 12 2019
(Ko chép lại đề)
\(a.\Rightarrow\orbr{\begin{cases}3x+5=0\\4-3x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{4}{3}\end{cases}}\)
\(b.\left(3x-2\right)\left(x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\x-7=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=7\end{cases}}\)
\(c.7x^2=28\)
\(x^2=4\)
\(x^2=2^2\)
\(x=\pm2\)
\(d.\left(2x+1\right)\left(1+x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\1+x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-1\end{cases}}\)
AD
12 tháng 8 2021
a)\(\left(3x+5\right)\left(4-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+5=0\\4-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{4}{3}\end{cases}}}\)
b)\(3x\left(x-7\right)-2\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\3x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=\frac{2}{3}\end{cases}}}\)
c) \(7x^2-28=0\)
\(\Leftrightarrow7x^2=28\)
\(\Leftrightarrow7x^2=7.4\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
d)\(\left(2x+1\right)+x\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(1+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\1+x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-1\end{cases}}}\)
#H
21 tháng 7 2020
a) ( x - 3 )2 - 4 = 0
<=> ( x - 3 )2 = 4
<=> \(\orbr{\begin{cases}\left(x-3\right)^2=2^2\\\left(x-3\right)^2=\left(-2\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy S = { 5 ; 1 }
b) x2 - 9 = 0
<=> x2 = 9
<=> \(\orbr{\begin{cases}x^2=3^2\\x^2=\left(-3\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy S = { 3 ; -3 }
c) x( x - 2x ) - x2 - 8 = 0
<=> x2 - 2x2 - x2 - 8 = 0
<=> -2x2 - 8 = 0
<=> -2x2 = 8
<=> x2 = -4 ( vô lí )
<=> x = \(\varnothing\)
Vậy S = { \(\varnothing\)}
21 tháng 7 2020
d) 2x( x - 1 ) - 2x2 + x - 5 = 0
<=> 2x2 - 2x - 2x2 + x - 5 = 0
<=> -x - 5 = 0
<=> -x = 5
<=> x = -5
Vậy S = { -5 }
e) x( x - 3 ) - ( x + 1 )( x - 2 ) = 0
<=> x2 - 3x - ( x2 - x - 2 ) = 0
<=> x2 - 3x - x2 + x + 2 = 0
<=> - 2x + 2 = 0
<=> -2x = -2
<=> x = 1
Vậy S = { 1 }
f) x( 3x - 1 ) - 3x2 - 7x = 0
<=> 3x2 - x - 3x2 - 7x = 0
<=> -8x = 0
<=> x = 0
Vậy S = { 0 }
27 tháng 9 2020
a) \(3x^3-12x=0\)
=> \(3x\left(x^2-4\right)=0\)
=> \(\orbr{\begin{cases}3x=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
=> \(x^2\left(x-3\right)+\left(-4x+12\right)=0\)
=> \(x^2\left(x-3\right)-4x+12=0\)
=> \(x^2\left(x-3\right)-4\left(x-3\right)=0\)
=> \(\left(x-3\right)\left(x^2-4\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) \(\left(3x-1\right)^2-\left(2x-3\right)^2=0\)
=> \(\left[3x-1-\left(2x-3\right)\right]\left(3x-1+2x-3\right)=0\)
=> \(\left(3x-1-2x+3\right)\left(3x-1+2x-3\right)=0\)
=> \(\left(x+2\right)\left(5x-4\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{4}{5}\end{cases}}\)
d) \(x^2-4x-21=0\)
=> \(x^2+3x-7x-21=0\)
=> \(x\left(x+3\right)-7\left(x+3\right)=0\)
=> (x + 3)(x - 7) = 0 => x = -3 hoặc x = 7
e) 3x2 - 7x - 10 = 0
=> 3x2 + 3x - 10x - 10 = 0
=> 3x(x + 1) - 10(x + 1) = 0
=> (x + 1)(3x - 10) = 0
=> x = -1 hoặc x = 10/3
F
27 tháng 9 2020
a) \(3x^3-12x=0\)
\(\Leftrightarrow3x\left(x^2-4\right)=0\)
\(\Leftrightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow x\in\left\{-2;0;2\right\}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow x\in\left\{-2;2;3\right\}\)
c) \(\left(3x-1\right)^2-\left(2x-3\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(5x-4\right)=0\)
\(\Leftrightarrow x\in\left\{-2;\frac{4}{5}\right\}\)
XO
27 tháng 9 2020
Ta có : 3x3 - 12x = 0
=> 3x(x2 - 4) = 0
=> x(x - 2)(x + 2) = 0
=> \(x\in\left\{0;2;-2\right\}\)
b) x2(x - 3) + 12 - 4x = 0
=> x2(x - 3) - 4(x - 3) = 0
=> (x2 - 4)(x - 3) = 0
=> \(\orbr{\begin{cases}x^2-4=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=4\\x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm2\\x=3\end{cases}}\)
Vậy \(x\in\left\{-2;2;3\right\}\)
c) (3x - 1)2 - (2x - 3)2 = 0
=> (3x - 1 - 2x + 3)(3x - 1 + 2x - 3) = 0
=> (x + 2)(5x - 4) = 0
=> \(\orbr{\begin{cases}x+2=0\\5x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=0,8\end{cases}}\)
Vậy \(x\in\left\{-2;0,8\right\}\)
d) x2 - 4x - 21 = 0
=> x2 - 7x + 3x - 21 = 0
=> x(x - 7) + 3(x - 7) = 0
=> (x + 3)(x - 7) = 0
=> \(\orbr{\begin{cases}x+3=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=7\end{cases}}\)
Vậy \(x\in\left\{-3;7\right\}\)
e) 3x2 - 7x - 10 = 0
=> 3x2 + 3x - 10x - 10 = 0
=> 3x(x + 1) - 10(x + 1) = 0
=> (3x - 10)(x + 1) = 0
=> \(\orbr{\begin{cases}3x-10=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=-1\end{cases}}\)
Vậy \(x\in\left\{\frac{10}{3};-1\right\}\)
\(\Rightarrow7x^3+7x^2-4x^2-4x+x+1=0\\ \Rightarrow\left(x+1\right)\left(7x^2-4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\7x^2-4x+1=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Rightarrow7\left(x^2-2\cdot\dfrac{2}{7}x+\dfrac{4}{49}\right)+\dfrac{3}{7}=0\\ \Rightarrow7\left(x-\dfrac{2}{7}\right)^2+\dfrac{3}{7}=0\left(\text{vô lí}\right)\)
Vậy x=-1
\(x^2\left(7x-3\right)=0\)