TM
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NT
1
1 tháng 8 2018
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
\(=>4.\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)
\(=>4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(=>4x-13=11\)\(=>4x=11+13=24\)
\(=>x=24:4=6\)
CHúc bạn Hk tốt!!!
N
2 tháng 11 2018
\(x^2-3x+2.\left(x-3\right)=0\)
\(x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(x.\left(x-3\right)-3x+9=0\)
\(x.\left(x-3\right)-3.\left(x-3\right)=0\)
\(\left(x-3\right)^2=0=>x=3\)
2 tháng 11 2018
a,\(x^2-3x+2\left(x-3\right)=0.\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
PN
12 tháng 8 2020
không ai trả lời
a,\(2\left(3x-1\right)-5\left(x-3\right)-9\left(2x-4\right)=24\)
\(< =>6x-2-5x+15-18x+36=24\)
\(< =>-29x+49=24< =>29x=25< =>x=\frac{25}{29}\)
b,\(2x^2+4\left(x^2-1\right)=2x\left(3x+1\right)\)
\(< =>2x^2+4x^2-4=6x^2+2x\)
\(< =>2x=-4< =>x=-\frac{4}{2}=-2\)
c, \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=4\)
\(< =>10x-6x^2+6x^2-10x-3x+21=4\)
\(< =>-3x=4-21=-17< =>x=\frac{17}{3}\)
d, \(5x\left(x+1\right)-4x\left(x+2\right)=1-x\)
\(< =>5x^2+5x-4x^2-8x=1-x\)
\(< =>x^2-3x+x-1=0\)
\(< =>x^2-2x-1=0\)
\(< =>\left(x-1\right)^2=2\)
\(< =>\orbr{\begin{cases}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{cases}}\)
20 tháng 7 2018
Tìm x, biết:
1) 2x ( x - 5) - x ( 2x - 4 ) = 15
<=> 2x2 - 10x - 2x2 + 4x - 15 = 0
<=> -6x - 15 = 0
<=> -6x = 15
<=> x = -15/6
2) ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6
<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0
<=> -4x = -16
<=> x = 4
3) 4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x
<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0
<=> x + 4 = 0
<=> x = -4
4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5
<=> 2x2 + x + 6x + 3 - 2x2 - 4x + 5 = 0
<=> 3x + 8 = 0
<=> 3x = -8
<=> x = -8/3
5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0
<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0
.......
6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)
<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0
<=> -2x + 40 = 0
<=> -2x = -40
<=> x = 20
Còn lại tương tự ....
TN
4
CU
14 tháng 7 2019
\(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)
\(\Rightarrow4x^2+4x+1-4\left(x^2+4x+4\right)=9\)
\(\Rightarrow4x^2+4x+1-4x^2-16x-16-9=0\)
\(\Rightarrow-12x-24=0\)
\(\Rightarrow-12x=24\)
\(\Rightarrow x=-2\)
14 tháng 7 2019
Ta có (2x+1)\(^2\)-4(x+2)\(^2\)=9
\(\Leftrightarrow\)4x\(^2\)+4x+1-[4(x\(^2\)+4x+4)]=9
\(\Leftrightarrow\)4x\(^2\)+4x+1-4x\(^2\)-16x-16=9
\(\Leftrightarrow\)-12x-15=9
\(\Leftrightarrow\)-12x=24
\(\Leftrightarrow\)x=-2
Vậy x=-2
24 tháng 6 2017
a)\(2x\left(x+1\right)-3-2x=5\)
\(\Leftrightarrow2x^2+2x-3-2x=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4=\left(-2\right)^2=2^2\)
\(\Rightarrow x=2;-2\)
b)\(2x\left(3x+1\right)+\left(4-2x\right)=7\)
\(\Leftrightarrow6x^2+2x+4-2x=7\)
\(\Leftrightarrow6x^2+4=7\)
\(\Leftrightarrow6x^2=3\)
\(\Leftrightarrow x^2=\frac{1}{2}=-\sqrt{\frac{1}{2}}=\sqrt{\frac{1}{2}}\)
c)\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x-1\right)^2=6\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x^2-2x+1\right)=6\)
\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)
\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)
\(\Leftrightarrow3x^2+15x=0\)
\(\Leftrightarrow3x\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x+5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
28 tháng 10 2019
\(3\left(2x-1\right)-5\left(x-3\right)=0\)
\(6x-3-5x+15=0\)
\(x+12=0\)
\(x=-12\)
KN
28 tháng 10 2019
a) \(3\left(2x-1\right)-5\left(x-3\right)=0\)
\(\Leftrightarrow6x-3-5x+15=0\)
\(\Leftrightarrow x+12=0\)
\(\Leftrightarrow x=-12\)
b) \(\left(x-3\right)\left(x-7\right)-x^2=1\)
\(\Leftrightarrow x^2-10x+21-x^2=1\)
\(\Leftrightarrow-10x=-20\Leftrightarrow x=2\)
c) \(5\left(x+3\right)-2x\left(3+x\right)=0\)
\(\Leftrightarrow5x+15-6x-2x^2=0\)
\(\Leftrightarrow-2x^2-x+15=0\)
Ta có: \(\Delta=1^2+4.2.5=41\)
pt có 2 nghiệm
\(x_1=\frac{1+\sqrt{41}}{-4}\);\(x_2=\frac{1-\sqrt{41}}{-4}\)
\(\left(2x+1\right)^2-4.\left(x+2\right)=9\)
\(\Leftrightarrow4x^2+4x+1-4x-8=9\)
\(\Leftrightarrow4x^2=16\)
\(\Leftrightarrow x^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
vậy...............
\(\left(2x+1\right)^2-4\left(x+2\right)=9\)
\(\Leftrightarrow4x^2+4x+1-4x-8=9\)
\(\Leftrightarrow4x^2-7-9=0\)
\(\Leftrightarrow4x^2-16=0\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)