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\(A=3x-x^2\)
\(=-\left(x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{9}{4}\right)\)
\(=-\left(\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right)\)
\(=\frac{9}{4}-\left(x-\frac{3}{2}\right)^2\ge\frac{9}{4}\)
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
x2+5.x=0
x.x+5.x=0
x.(x+5)=0
*x=0
*x+5=0
x=0-5
x=-5
Vậy x=0 hoặc x=-5
a) (5x - 12 ) : 4 = 318 : 315
( 5x - 12 ) : 4 = 33
( 5x - 12 ) : 4 = 27
5x - 12 = 108
5x = 120
x = 24
b) 5x3 - 102 = 35
5x3 - 100 = 35
5x3 = 135
x3 = 27
x3 = 33
=> x = 3
c) 65 - 4x+2 = 20180
65 - 4x+2 = 1
4x+2 = 64
4x+2 = 43
x + 2 = 3
=> x = 1
d) 2x+1 - 2x = 32
2x . ( 2 - 1 ) = 32
2x . 1 = 32
2x = 32
2x = 25
=> x = 5
#)Giải :
\(\left|5x-4\right|=\left|x+2\right|\)
\(\Leftrightarrow\left|5x-4\right|-\left|x+2\right|=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-4=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=4\\x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{4}\\x=-2\end{cases}}}\)
\(x^2-5x+4=0\)
\(x^2-x-4x+4=0\)
\(\left(x^2-x\right)-\left(4x-4\right)=0\)
\(x.\left(x-1\right)-4.\left(x-1\right)=0\)
\(\left(x-4\right).\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}}\)
x^2-5x+4=0
(x^2-x) - (4x-4) =0
x(x-1)-4(x-1)=0
(x-4)(x-1)=0
Ta có hai trường hợp:
x-4=0=) x=4
x-1=0=) x=1
5x+xy-4y=9
<=> x(5+y)-4y-20=9-20
<=> x(5+y)-4(y+5)=-11
<=> (x-4)(y+5)=-11
Ta có bảng:
x-4 | 1 | -1 | 11 | -11 |
y+5 | -11 | 11 | -1 | 1 |
x | 5 | 3 | 15 | -7 |
y | -16 | 6 | -6 | -4 |
\(\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
vậy x=1 hoặc x=-3
x2-x-4x+4=0
x(x-1)-4(x-1)=0
(x-1)(x-4)=0
\(\orbr{\begin{cases}x-1=0\\x-4=0\end{cases}}\)
\(\orbr{\begin{cases}x=1\\x=4\end{cases}}\)
x2 - 5x + 4 = 0
=> x2 - x - 4x + 4 = 0
=> (x2 - x) - (4x - 4) = 0
=> x(x - 1) - 4(x - 1) = 0
=> (x - 4)(x - 1) = 0
=> x - 4 = 0 hoặc x - 1 = 0
=> x = 4 hoặc x = 1
vậy_