x^5-5x+5x^3-5x^2+5x-1 với x=4
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\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)
\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)
\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)
\(=-10x^3+19x^2+74x+1\)
\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)
\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)
\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)
\(=-5x^4-11x^3+24x^2+12x+7\)
\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)
\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)
\(=-2x^2-27x+57\)
\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)
\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)
\(=-x^3+4x^2+22x+5\)
\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)
\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)
\(=-9x^3-55x^2+4x+35\)
\(g,\left(x-1\right)^2-\left(x+2\right)^2\)
\(=x^2-2x+1-x^2-4x-4\)
\(=-6x-3\)
x = 4
=> x + 1 = 5
Khi đó A = x5 - 5x4 + 5x3 - 5x2 + 5x - 1
= x5 - (x + 1)x4 + (x + 1)x3 - (x + 1)x2 + (x + 1)x - 1
= x5 - x5 - x4 + x4 + x3 - x3 - x2 + x2 + x - 1
= x - 1
= 4 - 1 = 3
Thay \(x=4\)vào biểu thức A ta có:
\(A=4^5-5.4^4+5.4^3-5.4^2+5.4-1\)
\(=1024-5.256+5.64-5.16+20-1\)
\(=1024-1280+320-80+20-1\)
\(=3\)
Vậy giá trị của biểu thức A khi x =4 là 3
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
Lời giải:
Tại $x=4$ thì:
\(A=5(x^5-x^4+x^3-x^2+x-1)-1\)
\(=(x+1)(x^5-x^4+x^3-x^2+x-1)-1=x^6+1-1=x^6\)
\(=4^6=4096\)
Tại \(x=4\)thì:
\(x^5-5x^4+5x^3-5x^2+5x-1\)
= \(x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-1\)
= \(x^5-x^4-x^4+x^4+x^3-x^3-x^2+x^2+x-1\)
= \(3\)
mình chỉnh lại đề nhé:
Do: \(x=4\)\(\Rightarrow\)\(x+1=5\)
\(x^5-5x^4+5x^3-5x^2+5x-1\)
\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-1\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-1\)
\(=x-1\)
\(=4-1=3\)