LN
6
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
HH
24 tháng 6 2018
a/ \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
<=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
<=> \(\left(2x+3\right)^2-4x^2+1=22\)
<=> \(\left(2x+3-2x\right)\left(2x+3+2x\right)=21\)
<=> \(3\left(4x+3\right)=21\)
<=> \(4x+3=7\)
<=> \(4x=4\)
<=> \(x=1\)
BD
24 tháng 6 2018
......................?
mik ko biết
mong bn thông cảm
nha ................
25 tháng 8 2019
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=0+4\)
\(\left(x-3\right)^2=4\)
\(\left(x-3\right)^2=\pm4\)
\(\left(x-3\right)^2=\pm2^2\)
\(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(4x^2+12x+9-4x^2+1=22\)
\(12x+10=22\)
\(12x=22-10\)
\(12x=12\)
\(x=1\)
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
\(16x^2-9-16x^2+40x-25=16\)
\(-34+40x=16\)
\(40x=16+34\)
\(40x=50\)
\(x=\frac{50}{40}=\frac{5}{4}\)
d) \(x^3-9x^2+27x-27=-8\)
\(x^3-9x^2+27x-27+8=0\)
\(x^3-9x^2+27x-19=0\)
\(\left(x^2-8x+19\right)\left(x-1\right)=0\)
Vì \(\left(x^2-8x+19\right)>0\) nên:
\(x-1=0\)
\(x=1\)
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
\(x^3+2x^2+x+x^2+2x+1-x^2-3x^2=2\)
\(3x+1=2\)
\(3x=2-1\)
\(3x=1\)
\(x=\frac{1}{3}\)
TG
7 tháng 8 2020
a) \(\left(2x+1\right)\left(1-2x\right)+\left(2x-1\right)^2=22\)
\(\Rightarrow\left(1+2x\right)\left(1-2x\right)+\left[\left(2x\right)^2-2.2x+1^2\right]=22\)
\(\Rightarrow1^2-\left(2x\right)^2+\left(4x^2-4x+1\right)=22\)
\(\Rightarrow1-4x^2+4x^2-4x+1=22\)
\(\Rightarrow2-4x=22\)
\(\Rightarrow-4x=22-2=20\)
\(\Rightarrow x=20:\left(-4\right)=-5\)
b/ \(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2.\left(x+1\right)^2=0\)
\(\Rightarrow\left(x^2-2.x.5+5^2\right)+\left(x^2-3^2\right)+2.\left(x^2+2.x.1+1^2\right)=0\)
\(\Rightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)
\(\Rightarrow x^2-10x+25+x^2-9-2x^2-4x-2=0\)
\(\Rightarrow-14x+14=0\)
\(\Rightarrow-14x=0-14=-14\)
\(\Rightarrow x=\left(-14\right):\left(-14\right)=1\)
7 tháng 8 2020
b/\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)
\(\Leftrightarrow x^2-10x+25+x^2-3^2-2\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow x^2-10x+25+x^2-9-2x^2-4x-2=0\)
\(\Leftrightarrow14x=14\Leftrightarrow x=1\)
c/\(\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(4x^2-9\right)=0\)
\(\Leftrightarrow4x^2+12x+9+4x^2-12x+9-8x^2+18=0\)
\(\Leftrightarrow0x=-36\Leftrightarrow x=0\)
a/\(\left(2x+1\right).\left(1-2x\right)+\left(2x-1\right)^2=22\Leftrightarrow2x-4x^2+1-2x+4x^2-4x+1=22\Leftrightarrow-4x=20\Leftrightarrow x=-5\)
4 tháng 7 2018
a) \(\left(x+3\right)^2-\left(2x+1\right).\left(2x-1\right)=22\)
\(\Leftrightarrow x^2+6x+9-\left(4x^2-1\right)=22\)
\(\Leftrightarrow x^2+6x+9-4x^2+1=22\)
\(\Leftrightarrow-3x^2+6x-12=0\)
\(\Leftrightarrow x^2-2x+4=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+3=0\)
\(\Leftrightarrow\left(x-1\right)^2+3=0\)(vô lý)
b) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)
\(\Leftrightarrow16x^2-9-\left(16x^2-40x+25\right)=46\)
\(\Leftrightarrow16x^2-9-16x^2+40x-25-46=0\)
\(\Leftrightarrow40x-80=0\)
\(\Leftrightarrow x=2\)
NM
25 tháng 8 2019
b) ( 2x+3)^2 - (2x+1)(2x-1) =22
=> 4x2+12x+9-4x2+1=22
=> 12x=12
=>x=1
c) (4x+3)(4x-3) -(4x-5)^2 =16
=>16x2-9-16x2+40x-25=16
=>40x=50
=>x=4/5
NB
25 tháng 8 2019
a)\(\left(x-13\right)^2-4=0\\\left(x-13\right)^2=4\\ \left(x-13\right)^2=2^2\\ \Rightarrow\left\{{}\begin{matrix}x-13=2\\x-13=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}15\\-11\end{matrix}\right.\)
vậy...
NN
0
Ta có : (2x + 3)2 - (2x + 1)(2x - 1) = 22
=> 4x2 + 12x + 9 - 4x2 + 1 = 22
=> 12x + 10 = 22
=> 12x = 12
=> x = 1
Vậy x = 1
\(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(\Leftrightarrow\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
\(\Leftrightarrow\left(2x+3\right)^2-4x^2+1=22\)
\(\Leftrightarrow\left(2x+3-2x\right)\left(2x+3+2x\right)=21\)
\(\Leftrightarrow3.\left(4x+3\right)=21\)
\(\Leftrightarrow4x+3=7\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)