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Khi x=-3 thì biểu thức:
\(\Rightarrow B=\left(-3^{2007}+3.\left(-3\right)^{2006}-1\right)^{2007}\)
\(\Rightarrow B=.............\)
máy tính tính cũng không ra nha bạn
Thay \(x=-3\) vào biểu thức B ta được :
\(B=\left(-3^{2007}+3.\left(-3\right)^{2006}-1\right)^{2007}\)
\(=\left(-3^{2007}+3^{2007}-1\right)^{2007}\)
\(=-1^{2007}\)
\(=-1\)
B=\(\left(x^{2007}+3x^{2006}-1\right)^{2007}\)
\(B=\left[\left(-3^{2007}\right)+3\left(-3^{2006}\right)-1\right]^{2007}\)
\(B=\left[\left(-3^{2007}\right)+3.3^{2006}-1\right]^{2007}\)
\(B=\left[\left(-3^{2007}\right)+3^{2007}-1\right]^{2007}\)
\(B=\left(-1\right)^{2007}=\left(-1\right)\)
Ta có: \(B=\left(x^{2007}+3x^{2006}-1\right)^{2007}\)
Thay x = -3 vào B ta có:
\(B=\left(\left(-3\right)^{2007}+3\left(-3\right)^{2006}-1\right)^{2007}\)
=>\(B=\left(\left(-3\right)^{2007}+3\cdot3^{2006}-1\right)^{2007}\)
=>\(B=\left(\left(-3\right)^{2007}+3^{2007}-1\right)^{2007}\)
=>\(B=\left(0-1\right)^{2007}\)
\(=>B=\left(-1\right)^{2007}=-1\)
Ta có:B=\(\left(x^{2007}+3x^{2006}-1\right)^{2007}\)
\(\Rightarrow\)B=\(\left(\left(-3\right)^{2007}+3\times\left(-3\right)^{2006}-1\right)^{2007}\)
B=\(\left(\left(-3\right)\times\left(-3\right)^{2006}+3\times\left(-3\right)^{2006}-1\right)^{2007}\)
B=\(\left(\left(\left(-3\right)+3\right)\times\left(-3\right)^{2006}-1\right)^{2007}\)
B=\(\left(0\times\left(-3\right)^{2006}-1\right)^{2007}\)
B=\(\left(0-1\right)^{2007}\)
B=\(\left(-1\right)^{2007}\)
B=\(\left(-1\right)\)
1. \(A=x^{15}+3x^{14}+5=x^{14}\left(x+3\right)+5\)
Thay \(x+3=0\)vào đa thức ta được:\(A=x^{14}.0+5=5\)
2. \(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)
Thay \(x=-3\)vào đa thức ta được: \(B=\left[x^{2006}\left(-3+3\right)+1\right]^{2017}=\left(x^{2006}.0+1\right)^{2017}=1^{2017}=1\)
3. \(C=21x^4+12x^3-3x^2+24x+15=3x\left(7x^3+4x^2-x+8\right)+15\)
Thay \(7x^3+4x^2-x+8=0\)vào đa thức ta được: \(C=3x.0+15=15\)
4. \(D=-16x^5-28x^4+16x^3-20x^2+32x+2007\)
\(=4x\left(-4x^4-7x^3+4x^2-5x+8\right)+2007\)
Thay \(-4x^4-7x^3+4x^2-5x+8=0\)vào đa thức ta được: \(D=4x.0+2007=2007\)
1. \(A=x^{15}+3x^{14}+5\)
\(A=x^{14}\left(x+3\right)+5\)
\(A=x^{14}+5\)
2. \(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\)
\(B=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)
\(B=\left[x^{2006}.\left(-3+3\right)+1\right]^{2007}\)
\(B=1^{2007}=1\)
3. \(C=21x^4+12x^3-3x^2+24x+15\)
\(C=3x\left(7x^2+4x^2-x+8+5\right)\)
\(C=3x\left(0+5\right)\)
\(C=15x\)
4. \(D=-16x^5-28x^4+16x^3-20x^2+32+2007\)
\(D=4x\left(-4x^4-7x^3+4x^2-5x+8\right)+2007\)
\(D=4x.0+2007\)
\(D=2007\)
1. a) Ta có: M = |x + 15/19| \(\ge\)0 \(\forall\)x
Dấu "=" xảy ra <=> x + 15/19 = 0 <=> x = -15/19
Vậy MinM = 0 <=> x = -15/19
b) Ta có: N = |x - 4/7| - 1/2 \(\ge\)-1/2 \(\forall\)x
Dấu "=" xảy ra <=> x - 4/7 = 0 <=> x = 4/7
Vậy MinN = -1/2 <=> x = 4/7
2a) Ta có: P = -|5/3 - x| \(\le\)0 \(\forall\)x
Dấu "=" xảy ra <=> 5/3 - x = 0 <=> x = 5/3
Vậy MaxP = 0 <=> x = 5/3
b) Ta có: Q = 9 - |x - 1/10| \(\le\)9 \(\forall\)x
Dấu "=" xảy ra <=> x - 1/10 = 0 <=> x = 1/10
Vậy MaxQ = 9 <=> x = 1/10
Tại x= - 3
=> \(B=\left[\left(-3\right)^{2017}+3\left(-3\right)^{2016}-1\right]^{2017}\)
=> \(B=\left[\left(-3\right)^{2017}+3^{2017}-1\right]^{2017}\)
=> \(B=\left(-1\right)^{2017}\)
=> B = - 1
Ta có:
\(B=\left(x^{2007}+3x^{2006}-1\right)^{2007}\)
\(B=\left(\left(-3\right)^{2007}+3\left(-3\right)^{2006}-1\right)^{2007}\)
\(B=\left(\left(-3\right)^{2007}+3\left(3\right)^{2006}-1\right)^{2007}\)
\(B=\left(\left(-3\right)^{2007}+3^1\left(3\right)^{2006}-1\right)^{2007}\)
\(B=\left(\left(-3\right)^{2007}+3^{1+2006}-1\right)^{2007}\)
\(B=\left(\left(-3\right)^{2007}+3^{2007}-1\right)^{2007}\)
\(B=\left(0-1\right)^{2007}\)
\(B=\left(-1\right)^{2007}\)
\(B=1\)