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a)
ĐKXĐ: \(x\ne-4\)
Để A nguyên thì \(3x+21⋮x+4\)
\(\Leftrightarrow3x+12+9⋮x+4\)
mà \(3x+12⋮x+4\)
nên \(9⋮x+4\)
\(\Leftrightarrow x+4\inƯ\left(9\right)\)
\(\Leftrightarrow x+4\in\left\{1;-1;3;-3;9;-9\right\}\)
\(\Leftrightarrow x\in\left\{-3;-5;-1;-7;5;-13\right\}\)(nhận)
Vậy: Để A nguyên thì \(x\in\left\{-3;-5;-1;-7;5;-13\right\}\)
b) ĐKXĐ: \(x\ne\dfrac{1}{2}\)
Để B nguyên thì \(2x^3-7x^2+7x+5⋮2x-1\)
\(\Leftrightarrow2x^3-x^2-6x^2+3x+4x-2+7⋮2x-1\)
\(\Leftrightarrow x^2\left(2x-1\right)-3x\left(2x-1\right)+2\left(2x-1\right)+7⋮2x-1\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-3x+2\right)+7⋮2x-1\)
mà \(\left(2x-1\right)\left(x^2-3x+2\right)⋮2x-1\)
nên \(7⋮2x-1\)
\(\Leftrightarrow2x-1\inƯ\left(7\right)\)
\(\Leftrightarrow2x-1\in\left\{1;-1;7;-7\right\}\)
\(\Leftrightarrow2x\in\left\{2;0;8;-6\right\}\)
hay \(x\in\left\{1;0;4;-3\right\}\)(nhận)
Vậy: \(x\in\left\{1;0;4;-3\right\}\)
a, (x + 2)(x2 - 2x + 4 ) - x(x + 3)(x - 3) = 26
<=> x3 + 8 - x(x2 - 9) = 26
<=> x3 + 8 - x3 + 9x = 26
<=> 9x - 18 = 0
<=> 9x = 18
<=> x = 2
b, (x - 3)(x2 + 3x + 9) - x(x - 4)(x + 4) = 21
<=> x3 - 27 - x(x2 - 16) = 21
<=> x3 - 27 - x3 + 16x = 21
<=> 16x - 48 = 0
<=> 16x = 48
<=> x = 3
c, (2x - 1)(4x2 + 2x + 1) - 4x(2x2 - 3) = 23
<=> 8x3 - 1 - 8x3 + 12x = 23
<=> 12x - 24 = 0
<=> 12x = 24
<=> x = 2
a/\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x+3\right)\left(x-3\right)=26\)
↔ \(x^3+2^3\)\(-x\left(x^2-3^2\right)\)= 26
↔\(x^3+8-x^3+9x=26\)
↔\(9x=18\leftrightarrow x=2\)
Vậy x=2
b/\(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=21\)
\(\Leftrightarrow x^3-3^3-x\left(x^2-4^2\right)=21\)
\(\Leftrightarrow x^3-9-x^3+16x=21\)
\(\Leftrightarrow16x=30\)
\(\Leftrightarrow x=\frac{15}{8}\)
Vậy \(x=\frac{15}{8}\)
c/\(\left(2x-1\right)\left(4x^2+2x+1\right)-4x\left(2x^2-3\right)=23\)
↔\(\left(2x\right)^3-1^3-4x\left(2x^2-3\right)=23\)
↔\(8x^3-1-8x^3+12x=23\)
↔\(12x=24\leftrightarrow x=2\)
Vậy x=2
a. \(2x\left(x-5\right)+21=x\left(2x+1\right)-12\)
\(2x^2-10x+21=2x^2+x-12\)
\(\left(2x^2-2x^2\right)-\left(10x+x\right)=-12-21\)
\(-11x=-33\Rightarrow x=3\)
b. \(\left(x^2-4\right)\left(x-2\right)\left(3-2x\right)=0\)
\(\left(x-2\right)^2\left(x+2\right)\left(3-2x\right)=0\)
\(\left[{}\begin{matrix}\left(x-2\right)^2=0\\x+2=0\\3-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=\frac{3}{2}\end{matrix}\right.\)
a) \(x\left(x-1\right)-x^2+2x=5\)
\(\Leftrightarrow x^2-x-x^2+2x=5\)
\(\Leftrightarrow x=5\)
b) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)
\(\Leftrightarrow-x=21\)
\(\Leftrightarrow x=-21\)
c) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)
\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)
\(\Leftrightarrow x=-4\)
a) x ( x - 1 ) - x^2 + 2x = 5
=>x2-x-x2+2x=5
=>-x+2x=5
=>x=5
b) 4x ( 3x + 2 ) - 6x ( 2x + 5 ) + 21 ( x - 1 ) = 0
=>12x2+8x-12x2-30x-21+21x=0
=>-x-21=0
=>x=-21
c) 2x( x + 1) - x^2 ( x + 2 ) + x^3 - x + 4 = 0
=>2x2+2x-x3-2x2+x3-x+4=0
=>x+4=0
=>x=-4
\(21-\left(2x-4\right)\left(x+1\right)\)
\(=21-2x^2-2x+4x+4\)
\(=-2x^2+2x+25\)