Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(2x^4-x^3+2x^2+1=2x^4-2x^3+2x^2+x^3-x^2+x+x^2-x+1\\ \)
\(=2x^2\left(x^2-x+1\right)+x\left(x^2-x+1\right)+\left(x^2-x+1\right)=\left(x^2-x+1\right)\left(2x^2+x+1\right)\)
Vậy a = 2; b = 1; c = 1.
-7-2x=37-(-24)
-7-2x=61
-2x=61+7
-2x=68
x=68:(-2)
x=-34
k cho mình nhé
\(a,\Rightarrow x+2=-40\\ \Rightarrow x=-42\\ b,\Rightarrow6x-7-2x=5\\ \Rightarrow4x=12\Rightarrow x=3\\ c,\Rightarrow68-56-x=-2\\ \Rightarrow12-x=-2\\ \Rightarrow x=14\)
\(A=3x-x^2\)
\(=-\left(x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{9}{4}\right)\)
\(=-\left(\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right)\)
\(=\frac{9}{4}-\left(x-\frac{3}{2}\right)^2\ge\frac{9}{4}\)
Min A = \(\frac{9}{4}\)khi \(x-\frac{3}{2}=0=>x=\frac{3}{2}\)
\(B=25+2x-x^2\)
\(=-\left(x^2-2x+1-26\right)\)
\(=-\left(\left(x-1\right)^2-26\right)\)
\(=26-\left(x-1\right)^2\ge26\)
Min A = 26 khi \(x-1=0=>x=1\)
\(C=x^2-5x+19\)
\(=x^2-2.x.\frac{5}{2}+\left(\frac{5}{2}\right)^2+\frac{51}{4}\)
\(=\left(x+\frac{5}{2}\right)^2+\frac{51}{4}\ge\frac{51}{4}\)
Min C = \(\frac{51}{4}\)khi \(x+\frac{5}{2}=0=>x=\frac{-5}{2}\)
@@@ nha các bạn . Thanks
Ta có : (2x - 1)3 = 125
=> (2x - 1)3 = 53
=> 2x - 1 = 5
=> 2x = 5 + 1
=> 2x = 6
=> x = 6 : 2
=> x = 3
( 2x - 1 )3 = 125
ta có :
(2x - 1)3 = 53
=> 2x - 1 = 5
2x = 5 + 1
2x = 6
x = 6 : 2
x = 3
VẬY x = 3
2x + 1 + 7. 2x + 3 = 232
2x . 2 + 7. 2x . 23 = 232
2x . 2 + 7 . 8 . 2x = 232
2x . 2 + 56 . 2x = 232
2x . ( 2 + 56 ) = 232
2x . 58 = 232
2x = 232 : 58
2x = 4
2x = 22
=> x = 2
2x x 2 + 7 x 2x x 23=232
2x(2+7 x 23)=232
2x x 58 =232
2x=232:58=4=22
=>x=2