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Câu 11:
a: Thay x=-1 và y=1/2 vào y=ax+1, ta được:
\(a\cdot\left(-1\right)+1=\dfrac{1}{2}\)
=>\(1-a=\dfrac{1}{2}\)
=>\(a=1-\dfrac{1}{2}=\dfrac{1}{2}\)
b: Thay a=1/2 vào y=ax+1, ta được:
\(y=x\cdot\dfrac{1}{2}+1=\dfrac{1}{2}x+1\)
Bài 1:
a: (x-5)(x+2)+3(x-2)(x+2)-4x2+3x
\(=x^2-3x-10+3\left(x^2-4\right)-4x^2+3x\)
\(=-3x^2-10+3x^2-12\)
=-10-12
=-22
b: \(x^3+3x^2+5x+a⋮x+3\)
=>\(x^3+3x^2+5x+15+a-15⋮x+3\)
=>\(\left(x+3\right)\left(x^2+5\right)+a-15⋮x+3\)
=>a-15=0
=>a=15
Bài 2:
1: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
2: \(A=\left(\dfrac{x+1}{x-1}+\dfrac{8}{x^2-1}-\dfrac{x-1}{x+1}\right)\cdot\dfrac{x-1}{x+1}\)
\(=\left(\dfrac{x+1}{x-1}+\dfrac{8}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{x+1}\right)\cdot\dfrac{x-1}{x+1}\)
\(=\dfrac{\left(x+1\right)^2+8-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{x+1}\)
\(=\dfrac{x^2+2x+1+8-x^2+2x-1}{x+1}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{4x+8}{\left(x+1\right)^2}\)
Bài 7
Ta có:
VT = a³ + b³ = (a + b)(a² - ab + b²)
= (a + b)(a² - 2ab + b² + ab)
= (a + b)[(a - b)² + ab]
= VP
Vậy a³ + b³ = (a + b)[(a - b)² + ab]
Bài 8
(x + 2)³ + (x - 2)³ + x³ - 3x(x + 2)(x - 2)
= x³ + 6x² + 12x + 8 + x³ - 6x² + 12x - 8 + x³ - 3x(x² - 4)
= 3x³ + 24x - 3x³ + 12x
= 36x
22:
A=9x^2+12x+4+4x^2-28x+49-2(6x^2+15x+4x+10)
=13x^2-16x+53-12x^2-38x-20
=x^2-54x+33
=(-19)^2-54*(-19)+33=1420
Bài 4 :
\(A=5\left(x+3\right)\left(x-3\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)
\(A=5\left(x^2-9\right)+\left(4x^2+12x+9\right)+\left(x^2-12x+36\right)\)
\(A=5x^2-45+4x^2+12x+9+x^2-12x+36\)
\(A=10x^2\)
Với x = -1/5 => A = \(10.\left(\dfrac{-1}{5}\right)^2=\dfrac{2}{5}\)
Bài 3:
1) \(x^2-25=x^2-5^2=\left(x+5\right)\left(x-5\right)\)
2) \(9x^2-\dfrac{1}{16}y^2=\left(3x\right)^2-\left(\dfrac{1}{4}y\right)^2=\left(3x+\dfrac{1}{4}y\right)\left(3x-\dfrac{1}{4}y\right)\)
3) \(x^6-y^4=\left(x^3\right)^2-\left(y^2\right)^2=\left(x^3+y^2\right)\left(x^3-y^2\right)\)
4) \(\left(2x-5\right)^2-64=\left(2x-5\right)^2-8^2=\left[\left(2x-5\right)+8\right]\left[\left(2x-5\right)-8\right]=\left(2x+3\right)\left(2x-13\right)\)
5) \(81-\left(3x+2\right)^2=9^2-\left(3x+2\right)=\left[9-\left(3x+2\right)\right]\left[9+\left(3x+2\right)\right]=\left(7-3x\right)\left(11+3x\right)\)
17:x^2-9=(x-3)(x+3)
18: 4x^2-25=(2x-5)(2x+5)
19: =(x^2-y^2)(x^2+y^2)
=(x-y)(x+y)(x^2+y^2)
20: 9x^2+6xy+y^2=(3x+y)^2
21 6x-9-x^2
=-(x^2-6x+9)
=-(x-3)^2
22: x^2+4xy+4y^2
=x^2+2*x*2y+(2y)^2
=(x+2y)^2
23: =(x+y+x-y)(x+y-x+y)
=2x*2y=4xy
25: =(3x+1+x+1)(3x+1-x-1)
=(4x+2)*2x
=4x(2x+1)
27: =(2x-y)(4x^2+2xy+y^2)-6xy(2x-y)
=(2x-y)(2x-y)^2
=(2x-y)^3
Bài 2:
a)
ĐK: \(x\ne\pm2\)
\(A=\left(\dfrac{x+2}{x^2-4}-\dfrac{x-2}{x^2-4}\right):\dfrac{4}{\left(x+2\right)^2}\\ =\dfrac{x+2-x+2}{x^2-4}:\dfrac{4}{\left(x+2\right)\left(x+2\right)}\\ =\dfrac{4\left(x+2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right).4}\\ =\dfrac{x+2}{x-2}\)
b)
Thế \(x=-\dfrac{1}{2}\) vào A được:
\(A=\dfrac{x+2}{x-2}=\dfrac{-\dfrac{1}{2}+2}{-\dfrac{1}{2}-2}\\ =\dfrac{\dfrac{3}{2}}{-\dfrac{5}{2}}=\dfrac{3}{5}.\dfrac{-2}{2}\\ =\dfrac{3}{5}.-1=-\dfrac{3}{5}\)
c)
Để A nhận giá trị nguyên thì \(\left(x+2\right)⋮\left(x-2\right)\)
\(\Leftrightarrow\left(x-2+4\right)⋮\left(x-2\right)\\ \Rightarrow4⋮\left(x-2\right)\\ \Rightarrow\left(x-2\right)\in\left\{\pm1;\pm2;\pm4\right\}\\ \Rightarrow x\in\left\{3;1;4;0;6;-2\right\}\)
Bài thứ 2 ạ