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TQ
0
PL
2 tháng 9 2018
x2 - 25 - (x + 5) = 0
<=> (x - 5)(x + 5) - (x + 5) = 0
<=> (x + 5)( x - 5 - 1) = 0
<=> (x + 5)( x - 6) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x-6=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)
PL
2 tháng 9 2018
(2x - 1)2 - (4x2 - 1) = 0
<=> (2x - 1)(2x - 1 - 2x - 1) = 0
<=> - 4x(2x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)
A
19 tháng 10 2020
a, \(x\left(x+1\right)-x\left(x-5\right)=6\Leftrightarrow x^2+x-x^2+5x=6\)
\(\Leftrightarrow x=1\)
b, \(4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)
c, \(x^2-\frac{1}{4}=0\Leftrightarrow\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\Leftrightarrow x=\pm\frac{1}{2}\)
d, \(5x^2=20x\Leftrightarrow5x^2-20x=0\Leftrightarrow5x\left(x-4\right)=0\Leftrightarrow x=0;4\)
e, \(4x^2-9-x\left(2x-3\right)=0\Leftrightarrow4x^2-9-2x^2=3x\Leftrightarrow2x^2-9-3x=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-3\right)=0\Leftrightarrow x=-\frac{3}{2};3\)
f, \(4x^2-25=\left(2x-5\right)\left(2x+7\right)\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow-2\left(2x+5\right)=0\Leftrightarrow x=-\frac{5}{2}\)
19 tháng 10 2020
a) x( x + 1 ) - x( x - 5 ) = 6
⇔ x2 + x - x2 + 5x = 6
⇔ 6x = 6
⇔ x = 1
b) 4x2 - 4x + 1 = 0
⇔ ( 2x - 1 )2 = 0
⇔ 2x - 1 = 0
⇔ x = 1/2
c) x2 - 1/4 = 0
⇔ ( x - 1/2 )( x + 1/2 ) = 0
⇔ \(\orbr{\begin{cases}x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}}\Leftrightarrow x=\pm\frac{1}{2}\)
d) 5x2 = 20x
⇔ 5x2 - 20x = 0
⇔ 5x( x - 4 ) = 0
⇔ \(\orbr{\begin{cases}5x=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
e) 4x2 - 9 - x( 2x - 3 ) = 0
⇔ ( 2x - 3 )( 2x + 3 ) - x( 2x - 3 ) = 0
⇔ ( 2x - 3 )( 2x + 3 - x ) = 0
⇔ ( 2x - 3 )( x + 3 ) = 0
⇔ \(\orbr{\begin{cases}2x-3=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-3\end{cases}}\)
f) 4x2 - 25 = ( 2x - 5 )( 2x + 7 )
⇔ ( 2x - 5 )( 2x + 5 ) - ( 2x - 5 )( 2x + 7 ) = 0
⇔ ( 2x - 5 )( 2x + 5 - 2x - 7 ) = 0
⇔ ( 2x - 5 )(-2) = 0
⇔ 2x - 5 = 0
⇔ x = 5/2
HH
21 tháng 6 2018
Giải:
1) \(\left(x-6\right)\left(x^2+6x+36\right)-\left(x+4\right)^3=\left(x-2\right)^3+\left(x+5\right)\left(x^2-10x+25\right)-\left(2x^3+6x^2\right)\)
\(\Leftrightarrow x^3-216-\left(x^3+12x^2+48x+64\right)=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow x^3-216-x^3-12x^2-48x-64=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow-280-12x^2-48x=-12x^2+12x+117\)
\(\Leftrightarrow-280-48x-12x-117=0\)
\(\Leftrightarrow-397-60x=0\)
\(\Leftrightarrow-60x=397\)
\(\Leftrightarrow x=-\dfrac{397}{60}\)
Vậy ...
2) \(\left(2x+3\right)^3-\left(2x+5\right)\left(4x^2-10x+25\right)=\left(6x-1\right)^2-\left(x-2\right)\left(x^2+2x+4\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-\left(8x^3+125\right)=36x^2-12x+1-\left(x^3-8\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x^3-125=36x^2-12x+1-x^3+8+x^3\)
\(\Leftrightarrow54x-98=-12x+9\)
\(\Leftrightarrow54x+12x=9+98\)
\(\Leftrightarrow66x=107\)
\(\Leftrightarrow x=\dfrac{107}{66}\)
Vậy ...
2
25 tháng 10 2021
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
2
25 tháng 10 2021
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
NN
6 tháng 12 2017
a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=18:18\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b) \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2-\left(x^2+6x+64\right)=0\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)
\(\Leftrightarrow8^2-x^2-6x-64=0\)
\(\Leftrightarrow64-x^2-6x-64=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow x\left(-x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=-6\)
LH
6 tháng 12 2017
a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=18:18\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b, \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x- 5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2- \left(x^2+6x+64\right)=0\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)
\(\Leftrightarrow8^2-x^2-6x-64=0\)
\(\Leftrightarrow64-x^2-6x-64=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow x\left(-x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=6\)
10 tháng 7 2022
1: \(\Leftrightarrow5x^2+4x-1-2x^2+12x-18=3x^2+5x-2-x^2-8x-16+x^2-x\)
\(\Leftrightarrow3x^2+16x-19=3x^2-4x-18\)
=>20x=1
hay x=1/20
2: \(\Leftrightarrow5x^2-20x-41=x^2-10x+25+4x^2+4x+1-\left(x^2-2x\right)+\left(x-1\right)^2\)
\(\Leftrightarrow5x^2-20x-41=4x^2-4x+26+x^2-2x+1\)
\(\Leftrightarrow-20x-41=-6x+27\)
=>-14x=68
hay x=-34/7
Ta có: \(\left(2x+5\right)\left(4x^2-10x+25\right)-\left(2x+1\right)^2=52\)
<=> \(8x^3+125-4x^2-4x-1=52\)
<=> \(8x^3-4x^4-4x+72=0\)
<=> \(2x^3-x^2-x+18=0\)
<=> \(\left(2x^3+4x^2\right)-\left(5x^2+10x\right)+\left(9x+18\right)=0\)
<=> \(2x^2\left(x+2\right)-5x\left(x+2\right)+9\left(x+2\right)=0\)
<=> \(\left(x+2\right)\left(2x^2-5x+9\right)=0\)
<=> \(\left[{}\begin{matrix}x+2=0\left(1\right)\\2x^2-5x+9=0\left(2\right)\end{matrix}\right.\)
Mặt khác :
\(2x^2-5x+9=2\left(x^2-2,5x+1,5625\right)+5,875\)
= \(2\left(x-1,25\right)^2+5,875\) \(\ge5,875\) > 0
=> PT (2) vô nghiệm
=> x + 2 = 0 => \(x=-2\)
Vậy nghiệm của PT : S = \(\left\{-2\right\}\)