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a) P(x) = 7x2 . (x2 – 5x + 2 ) – 5x .(x3 – 7x2 + 3x)
= 7x2 . x2 + 7x2 . (-5x) + 7x2 . 2 – [5x. x3 + 5x . (-7x2) + 5x . 3x]
= 7. (x2 . x2) + [7.(-5)] . (x2 . x) + (7.2).x2 – {5. (x.x3) + [5.(-7)]. (x.x2) + (5.3).(x.x)}
= 7x4 + (-35). x3 + 14x2 – [ 5x4 + (-35)x3 + 15x2 ]
= 7x4 + (-35). x3 + 14x2 - 5x4 + 35x3 - 15x2
= (7x4 – 5x4) + [(-35). x3 + 35x3 ] + (14x2 - 15x2 )
= 2x4 + 0 - x2
= 2x4 – x2
b) Thay x = \( - \dfrac{1}{2}\) vào P(x), ta được:
P(\( - \dfrac{1}{2}\)) = 2. (\( - \dfrac{1}{2}\))4 – (\( - \dfrac{1}{2}\))2 \))
\(\begin{array}{l} = 2.\dfrac{1}{{16}} - \dfrac{1}{4} \\ = \dfrac{1}{8} - \dfrac{{2}}{8} \\ = \dfrac{-1}{8} \end{array}\)
Vì | 5x - 10 | = 5x - 10 hoặc -( 5x + 10 )
\(\rightarrow\)TH1:
Ta có : | 5x - 10 | - x + 6
= 5x - 10 -x + 6
= ( 5 - 1 ) . x - ( 10 - 6 )
= 4x -4 = 4.( x - 1 )
\(\rightarrow\)TH2 :
Ta có : | 5x - 10 | - x + 6
= - ( 5x - 10 ) - x + 6
= - 5x + 10 - x + 6
= ( - 5 - 1 ) . x + ( 10 + 6 )
= -6x + 16
Bài 2:
3x + 2(5 - x) = 0
<=> 3x + 10 - 2x = 0
<=> x + 10 = 0
<=> x = 0 - 10
<=> x = -10
=> x = -10
Bài 3:
6(3q + 4q) - 8(5p - q) + (p - q)
= 6.3p + 6.4q - 8.5p - (-8).q + p - q
= 18p + 24q - 40p + 8q + p - q
= (18p - 40p + p) + (24q + 8q - q)
= -21p + 31q
a)*TH1: 2x+1>0 .Suy ra: |2x+1|=2x+1. Suy ra A=5x-2-2x-1=5x-2x-2-1=3x-3
*TH2: 2x+1<0. Suy ra: |2x+1|=-2x-1. Suy ra: A= 5x-2+2x+1=5x+2x-2+1=7x-1
b) Ta có: A>0.Suy ra: 5x-2>|2x+1|. Suy ra: 5x-2>0. Suy ra:5x>2. Suy ra x>2/5.
Vậy, nếu x>2/5 thì A>0.
\(^{10^{n+1}-6.10^n}\)=\(^{10^n-10^n-6.10^n}\)
=\(^{4.10^n}\)
=>rút gọn thành 4.10^n
|5x - 10 |-x - 6 TH1 : 5x - 10 > 0 \(\Leftrightarrow\) 5x >10 \(\Leftrightarrow\)x >2 \(\Rightarrow\)|5x - 10 |-x - 6 = 5x - 10 -x -6 = 4x -16 = 4 ( x-4 ) TH2 : 5x - 10 < 0 \(\Leftrightarrow\) 5x < 10 \(\Leftrightarrow\) x< 2 \(\Rightarrow\)|5x -10|-x-6 = -5x +10-x-6 = -6x +4