tìm x, biết
x^2 -2x=0
(3x-1)^2-16=0
x^2-25x=0
(4x-1)^2-9=0
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12 tháng 12 2021
\(a,PT\Leftrightarrow3x^2+3x-2x^2-4x=-1-x\Leftrightarrow x^2=-1\left(\text{vô nghiệm}\right)\)
Vậy: ...
\(b,PT\Leftrightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\Leftrightarrow\left(x-2019\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy: ...
\(c,PT\Leftrightarrow\left(x-4-6\right)\left(x-4+6\right)=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
Vậy: ...
\(d,PT\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)
Vậy: ...
\(e,PT\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
Vậy: ...
\(f,PT\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\Leftrightarrow x=\pm\dfrac{3}{5}\)
Vậy: ...
12 tháng 12 2021
câu c sao tính ra vậy đc vậy k hiểu giải thích hộ e đi 36 đâu mất òi
5 tháng 11 2023
2: \(3x\left(x-4\right)+2x-8=0\)
=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
3: 4x(x-3)+x2-9=0
=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(4x+x+3\right)=0\)
=>\(\left(x-3\right)\left(5x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)
4: \(x\left(x-1\right)-x^2+3x=0\)
=>\(x^2-x-x^2+3x=0\)
=>2x=0
=>x=0
5: \(x\left(2x-1\right)-2x^2+5x=16\)
=>\(2x^2-x-2x^2+5x=16\)
=>4x=16
=>x=4
PQ
2
26 tháng 8 2021
a) 9-64x^2=0
=> 64x^2 = 8
=> \(x^2=\frac{8}{64}=\frac{1}{8}\)
=> \(x=\frac{1}{\sqrt{8}}\)
b ) 25x^2 - 3 = 0
=> 25x^2 = 3
=> \(x^2=\frac{3}{25}\)
=> \(x=\frac{\sqrt{3}}{5}\)
C) 7 - 16x^2 =0
=> 16x^2 = 7
=> \(x^2=\frac{7}{16}\)
=> \(x=\frac{\sqrt{7}}{4}\)
d) 4x^2 - (x-4)^2 = 0
=> 4x^2 - x^2 + 8x - 16 =0
=> 3x^2 + 8x -16 = 0
=> ( 3x^2 + 12x ) - ( 4x +16 ) = 0
=> 3x( x + 4 ) - 4( x + 4 ) = 0
=>( x + 4 )( 3x - 4 ) = 0
=> \(\orbr{\begin{cases}x+4=0\\3x-4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-4\\x=\frac{4}{3}\end{cases}}\)
e) ( 3x + 4 )^2 - ( 2x - 5 )^2 = 0
=> ( 3x + 4 + 2x - 5 )( 3x + 4 - 2x + 5 ) = 0
=> ( 5x -1 ) ( x + 9 ) = 0
=> \(\orbr{\begin{cases}5x-1=0\\x+9=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{5}\\x=-9\end{cases}}\)
QA
26 tháng 8 2021
Trả lời:
a, \(9-64x^2=0\)
\(\Leftrightarrow\left(3-8x\right)\left(3+8x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3-8x=0\\3+8x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{8}\\x=-\frac{3}{8}\end{cases}}}\)
Vậy x = 3/8; x = - 3/8 là nghiệm của pt.
b, \(25x^2-3=0\)
\(\Leftrightarrow\left(5x-\sqrt{3}\right)\left(5x+\sqrt{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-\sqrt{3}=0\\5x+\sqrt{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}}{5}\\x=-\frac{\sqrt{3}}{5}\end{cases}}}\)
Vậy \(x=\pm\frac{\sqrt{3}}{5}\)
c, \(7-16x^2=0\)
\(\Leftrightarrow\left(\sqrt{7}-4x\right)\left(\sqrt{7}+4x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{7}-4x=0\\\sqrt{7}+4x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{7}}{4}\\x=-\frac{\sqrt{7}}{4}\end{cases}}}\)
Vậy \(x=\pm\frac{\sqrt{7}}{4}\)
d, \(4x^2-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(2x-x+4\right)\left(2x+x-4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\3x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{4}{3}\end{cases}}}\)
Vậy x = - 4; x = 4/3 là nghiệm của pt.
e, \(\left(3x+4\right)^2-\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(3x+4-2x+5\right)\left(3x+4+2x-5\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-9\\x=\frac{1}{5}\end{cases}}}\)
Vậy x = - 9; x = 1/5 là nghiệm của pt.
4 tháng 1 2022
\(a,\Leftrightarrow\left(4-5x\right)\left(4+5x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1-2\right)\left(x+1+2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(3x+1-2x\right)\left(3x+1+2x\right)=0\\ \Leftrightarrow\left(x+1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{5}\end{matrix}\right.\\ d,Sửa:\left(4x+1\right)^2-\left(x-2\right)^2=0\\ \Leftrightarrow\left(4x+1-x+2\right)\left(4x+1+x-2\right)=0\\ \Leftrightarrow\left(3x+3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{5}\end{matrix}\right.\\ e,\Leftrightarrow\left(2x+1-x-3\right)\left(2x+1+x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
HH
1 tháng 7 2018
a/ \(25x^2-9=0\)
<=> \(\left(5x-3\right)\left(5x+3\right)=0\)
<=> \(\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}5x=3\\5x=-3\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)
b/ \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)
<=> \(x^2+8x+16-x^2+8x-9=16\)
<=> \(16x+7=16\)
<=> \(16x=9\)
<=> \(x=\frac{9}{16}\)
1 tháng 7 2018
a) \(25x^2-9=0\)
\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=3\\5x=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}}\)
Vậy S = {3/5 ; -3/5}
b) \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)
\(\Leftrightarrow\left(x+4\right)^2-4^2-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+8\right)-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow x^2+8x-x^2-8x+9=0\)
\(\Leftrightarrow9=0\left(vl\right)\)
Vậy S = \(\varnothing\)
30 tháng 12 2016
b) ( 2x - 3 ) - ( 3 - 2x )( x - 1 ) = 0
<=> ( 2x - 3 ) + ( 2x - 3 )( x - 1 ) = 0
<=> ( 2x - 3 )( 1 + x - 1 ) = 0
<=> x( 2x - 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)
Vậy .....
30 tháng 12 2016
a, 25x^2 - 1 - (5x -1)(x+2)=0
=> (5x)^2 - 1 + (5x-1)(x+2) = 0
=> (5x-1)(5x+1) + (5x-1)(x+2) = 0
=> (5x-1)(5x+1+x+2) = 0
=> (5x-1)(6x+3) = 0
=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)
a) \(x^2-2x=0\)
\(x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
b) \(\left(3x-1\right)^2-16=0\)
\(\left(3x-1\right)^2-4^2=0\)
\(\left(3x-1-4\right)\left(3x-1+4\right)=0\)
\(\left(3x-5\right)\left(3x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-5=0\\3x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
c) \(x^2-25x=0\)
\(x\left(x-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)
d) \(\left(4x-1\right)^2-9=0\)
\(\left(4x-1\right)^2-3^2=0\)
\(\left(4x-1-3\right)\left(4x-1+3\right)=0\)
\(\left(4x-4\right)\left(4x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4x-4=0\\4x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{2}\end{cases}}}\)
a) \(x^2-2x=0\)
\(x.\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
vậy..
b) \(\left(3x-1\right)^2-16=0\)
\(\left(3x-1\right)^2=16\)
\(\left(3x-1\right)^2=4^2=\left(-4\right)^2\)
\(\Rightarrow\orbr{\begin{cases}3x-1=4\\3x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
vậy ...
c) \(x^2-25x=0\)
\(x.\left(x-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)
vậy ....
d) \(\left(4x-1\right)^2-9=0\)
\(\left(4x-1\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow\orbr{\begin{cases}4x-1=3\\4x-1=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)
vậy ...