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a; \(P=x^3+1+x-\left(x^3-1\right)+2017\)
\(=x^3+1+x-x^3+1+2017\)
=x+2019=-2017+2019=2
b: \(Q=64x^3-80x-64x^3-1=-80x-1=-16-1=-17\)
(x - 1)2 - (x + 2)(x - 5) + (4x - 1)2 = (-x - 1)(2 - 16x)
=> x2 - 2x + 1 - x2 + 5x - 2x + 10 + 16x2 - 8x + 1 = -2x + 16x2 -2 + 16x
=> 16x2 - 7x + 12 = 14x - 2 + 16x2
=> -21x = -14
=> 21x = 14
=> x = 2/3
\(\left(x-1\right)^2-\left(x+2\right)\left(x-5\right)+\left(4x-1\right)^2=\left(-x-1\right)\left(2-16x\right)\)
\(\Rightarrow x^2-2x+1-x^2+3x+10+16x^2-8x+1-16x^2-14x+2=0\)
\(\Rightarrow\left(x^2-x^2+16x^2-16x^2\right)+\left(-2x+3x-8x-14x\right)+\left(1+10+1+2\right)=0\)
\(\Rightarrow-21x+14=0\)
\(\Rightarrow-21x=-14\)
\(\Rightarrow x=\frac{2}{3}\)
a)\(P=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+2018\)
\(=\left(x^3+1\right)+x-\left(x^3-1\right)+2018=1+\left(x+2019\right)\)
Mà x=-2019 nên x+2019=0
\(\Rightarrow P=1\)
Vậy P=1 tại x=-2019
b)\(Q=16x\left(4x^2-5\right)-\left(4x+1\right)\left(16x^2-4x+1\right)\)
\(=64x^3-16.5x-\left(64x^3+1\right)=64x^3-64x^3-1-16.5x=-1-16.5x\)
Mà x=1/5 nên 5x=1 từ đó suy ra Q=-1-16=-17
Vậy Q=-17 tại x=1/5
\(P=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)\)\(P=\left(x^3+1^3\right)+x-\left(x^3-1^3\right)\)
\(P=1^3+1^3-2017\)
\(P=-2015\)
Tương tự
( 4x - 1 )3 + ( 3 - 4x )( 9 + 12x + 16x2 ) = ( 8x - 1 )( 8x + 1 ) - ( 3x - 5 )
<=> 64x3 - 48x2 + 12x - 1 + [ 33 - ( 4x )3 ] = ( 8x )2 - 1 - 3x + 5
<=> 64x3 - 48x2 + 12x - 1 + 27 - 64x3 = 64x2 - 3x + 4
<=> -48x2 + 12x + 26 = 64x2 - 3x + 4
<=> -48x2 + 12x + 26 - 64x2 + 3x - 4 = 0
<=> -112x2 + 15x + 22 = 0 (*)
\(\Delta=b^2-4ac=15^2-4\cdot\left(-112\right)\cdot22=225+9856=10081\)
\(\Delta>0\)nên (*) có hai nghiệm phân biệt
\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{\sqrt{10081}-15}{-224}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-15-\sqrt{10081}}{-224}\end{cases}}\)
Lớp 8 sao nghiệm xấu thế -..-
\(16x^3-16x^4+4x-8x^2-1=0\)
<=> \(-16x^4-4x^2+16x^3+4x-4x^2-1=0\)
<=> \(-4x^2\left(4x+1\right)+4x\left(4x^2+1\right)-\left(4x^2+1\right)=0\)
<=> \(-\left(4x^2+1\right)\left(4x^2-4x+1\right)=0\)
<=> \(-\left(4x^2+1\right)\left(2x-1\right)^2=0\)
<=> \(2x-1=0\) (do 4x2 + 1 > 0 )
<=> \(x=\frac{1}{2}\)
1.
PT \(\Leftrightarrow (x+2)(x-3)(x-4)(x+6)=16x^2\)
\(\Leftrightarrow [(x+2)(x+6)][(x-3)(x-4)]=16x^2\)
\(\Leftrightarrow (x^2+8x+12)(x^2-7x+12)=16x^2\)
\(\Leftrightarrow (a+8x)(a-7x)=16x^2\) (đặt \(x^2+12=a\) )
\(\Leftrightarrow a^2+ax-72x^2=0\)
\(\Leftrightarrow (a-8x)(a+9x)=0\Rightarrow \left[\begin{matrix} a-8x=0\\ a+9x=0\end{matrix}\right.\)
Nếu \(a-8x=0\Leftrightarrow x^2+12-8x=0\Leftrightarrow (x-2)(x-6)=0\Rightarrow \left[\begin{matrix} x=2\\ x=6\end{matrix}\right.\)
Nếu \(a+9x=0\Leftrightarrow x^2+12+9x=0\Leftrightarrow x=\frac{-9\pm \sqrt{33}}{2}\)
Vậy...........
2.
PT \(\Leftrightarrow [(4x+7)(2x+1)][(4x+5)(x+1)]=9\)
\(\Leftrightarrow (8x^2+18x+7)(4x^2+9x+5)=9\)
\(\Leftrightarrow (2a+7)(a+5)=9\) (đặt \(a=4x^2+9x\) )
\(\Leftrightarrow 2a^2+17a+26=0\)
\(\Leftrightarrow (a+2)(2a+13)=0 \)\(\Rightarrow \left[\begin{matrix} a+2=0\\ 2a+13=0\end{matrix}\right.\)
Nếu \(a+2=0\Leftrightarrow 4x^2+9x+2=0\Leftrightarrow (4x+1)(x+2)=0\)
\(\Rightarrow \left[\begin{matrix} x=\frac{-1}{4}\\ x=-2\end{matrix}\right.\)
Nếu \(2a+13=0\Leftrightarrow 8x^2+18x+13=0\) (pt này dễ thấy vô nghiệm)
Vậy.........