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Bài 1:
27x3 - 8 : (6x + 9x2 +4)
= (3x - 2) (9x2 + 6x + 4) : (9x2 + 6x + 4)
= 3x - 2
Bài 3:
a, 81x4 + 4 = (9x2)2 + 36x2 + 4 - 36x2
= (9x2 + 2)2 - (6x)2
= (9x2 + 6x + 2)(9x2 - 6x + 2)
b, x2 + 8x + 15 = x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c, x2 - x - 12 = x2 + 3x - 4x - 12
= x(x + 3) - 4(x + 3)
= (x + 3) (x - 4)
Câu 1:
(27x3 - 8) : (6x + 9x2 + 4)
= (3x - 2)(9x2 + 6x + 4) : (6x + 9x2 + 4)
= 3x - 2
Câu 2:
a) (3x - 5)(2x+ 11) - (2x + 3)(3x + 7)
= 6x2 + 33x - 10x - 55 - 6x2 - 14x - 9x - 21
= -76
⇒ đccm
b) (2x + 3)(4x2 - 6x + 9) - 2(4x3 - 1)
= 8x3 + 27 - 8x3 + 2
= 29
⇒ đccm
Câu 3:
a) 81x4 + 4
= (9x2)2 + 22
= (9x2 + 2)2 - (6x)2
= (9x2 - 6x + 2)(9x2 + 6x + 2)
b) x2 + 8x + 15
= x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c) x2 - x - 12
= x2 - 4x + 3x - 12
= x(x - 4) + 3(x - 4)
= (x - 4)(x + 3)
Bài 1:
\(A=\left(x-y\right)\left(x^2+xy+y^2\right)+2y^3\)
\(A=x^3-y^3+2y^3\)
\(A=x^3+y^3\)
Thay \(x=\dfrac{2}{3},y=\dfrac{1}{3}\) vào A, ta có:
\(A=\left(\dfrac{2}{3}\right)^3+\left(\dfrac{1}{3}\right)^3=\dfrac{8}{27}+\dfrac{1}{27}=\dfrac{9}{27}=\dfrac{1}{3}\)
\(B=x^2-x+\dfrac{1}{2}=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}>0\)
Câu a : Ta có :
\(B=x^2-x+\dfrac{1}{2}=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}>0\)
Câu b : Ta có :
\(C=\left(2n+1\right)^2-1=\left(2n+1-1\right)\left(2n+1+1\right)=2n\left(2n+2\right)=4n^2+4n=8n\left(\dfrac{1}{2}n+\dfrac{1}{2}\right)\)
Do có thừa số là 8 nên \(8n\left(\dfrac{1}{2}n+\dfrac{1}{2}\right)\) luôn chia hết cho 8
\(\Rightarrow C=\left(2n+1\right)^2-1\) chia hết cho 8 ( đpcm )
1, a, để A có giá trị xác định <=> 5x-5y \(\ne\) 0 => 5x\(\ne\)5y =>x\(\ne\)y b, A=\(\dfrac{x^2-y^2}{5x-5y}=\dfrac{\left(x+y\right)\left(x-y\right)}{5\left(x-y\right)}=\dfrac{\left(x+y\right)}{5}\) 2, a,
A=\(\dfrac{2x^3+4x}{x^3-4x}+\dfrac{x^2-4}{x^2+2x}+\dfrac{2}{2-x}\) =\(\dfrac{2x\left(x+2\right)}{x\left(x^2-4\right)}+\dfrac{\left(x+2\right)\left(x-2\right)}{x\left(x+2\right)}-\dfrac{2}{x-2}\) =\(\dfrac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x}-\dfrac{2}{x-2}\) =\(\dfrac{2x}{x\left(x-2\right)}+\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}-\dfrac{2x}{x\left(x-2\right)}\) =\(\dfrac{2x+\left(x-2\right)^2-2x}{x\left(x-2\right)}\) =\(\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}\) =\(\dfrac{\left(x-2\right)}{x}\)
b, thay x=4 vào A ta có : A=\(\dfrac{4-2}{4}\) =\(\dfrac{2}{4}=\dfrac{1}{2}\)
c, để A \(\in\) Z => (x-2)\(⋮\)x mà x\(⋮\)x =>-2\(⋮\)x => x\(\in\){ \(\pm1;\pm2\)} mà x\(\ne\)\(\pm2\) => x\(\in\left\{-1,+1\right\}\)
Bài 3 : a, Ta có B= 2.(-1)2+-(-1)+1 =2+1+1=4 b, Ta có A=2x3 +5x2 -2x +a =(2x3 -x2 +x )+(6x2-3x +3) +(a-3) \(⋮\) 2x2-x+1 => x(2x2-x+1)+3(2x2-x+1) +(a-3)\(⋮\) 2x2-x+1
=>a-3=0 (vì a-3 là số dư )=>a-3 Vậy a=3 thì A\(⋮\)B c,B=1 => 2x2 -x+1=1 =>x(2x-1)=0 => x=0 hoặc 2x-1 =0 => x=0 hoặc x=\(\dfrac{1}{2}\)
Bài 3:
a: \(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
=-5n chia hết cho 5
b: \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)\)
\(=n^2+4n-n-4-\left(n^2+n-4n-4\right)\)
\(=n^2+3n-4-\left(n^2-3n-4\right)\)
\(=6n⋮6\)
Bài 2:
a. \(x\left(x^2+5\right)=x^3+5x\)
b. \(\left(3x-5\right)\left(2x+1\right)-\left(6x^2-5\right)\)
\(=6x^2-7x-5-6x^2+5=-7x\)
c. \(\left(2x+3\right)\left(2x-3\right)-\left(2x+1\right)^2\)
\(=4x^2-9-4x^2-4x-1=-4x-10=\)
d. \(\left(2x^4+x^3-3x^2+5x-2\right):\left(x^2-x+1\right)=2x^2+3x-2\)
Bài 3:
a. \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
b. \(x^2-2x-y^2+1=\left(x-1\right)^2-y^2=\left(x+y-1\right)\left(x-y-1\right)\)
Câu 1:
a,
\(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right).\dfrac{3x}{1-2x+x^2}\)
= \(\left[\dfrac{1}{x\left(x+1\right)}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right].\dfrac{3x}{\left(x-1\right)^2}\)
= \(\dfrac{1-2x+x^2}{x\left(x+1\right)}.\dfrac{3x}{\left(x-1\right)^2}\)
= \(\dfrac{\left(x-1\right)^2.3x}{x\left(x+1\right)\left(x-1\right)^2}\)
= \(\dfrac{3}{x+1}\)
b, Để A đạt giá trị nguyên:
=> x + 1 thuộc Ư(3) = {-3;-1;1;3}
| x+1 | -3 | -1 | 1 | 3 |
| x | -4 | -2 | 0 | 2 |
Vậy x thuộc {-4;-2;0;2}.
Bị tự tin quá khả năng nhẩm mồm, sai em xin lỗi ...
a, Ta có \(P\left(x\right)=8x^3+2x^2-3x-3x^3+10-x-2x^2-3\)
\(=5x^3-4x-7\)
\(Q\left(x\right)=9x^3-4x^2+2x-3+2x+3x^2+4x^3-2\)
\(=13x^3-x^2+4x-5\)
b, Ta có : \(P\left(-\frac{1}{2}\right)=5.\left(-\frac{1}{2}\right)^3-4.\left(-\frac{1}{2}\right)-7=-\frac{45}{8}\)
c , \(M\left(x\right)=P\left(x\right)+Q\left(x\right)\)
\(5x^3-4x-7+13x^3-x^2+4x-5=18x^3-x^2-12\)
\(N\left(x\right)=P\left(x\right)-Q\left(x\right)\)
\(5x^3-4x-7-13x^3+x^2-4x+5=-8x^3-8x-2+x^2\)
d, Đặt \(5x^3-4x-7=0\)( vô nghiệm )