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Những câu hỏi liên quan
16 tháng 12 2023
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: \(A=\left(\dfrac{x+1}{x-1}+\dfrac{8}{x^2-1}-\dfrac{x-1}{x+1}\right)\cdot\dfrac{x-1}{x+1}\)
\(=\left(\dfrac{x+1}{x-1}+\dfrac{8}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{x+1}\right)\cdot\dfrac{x-1}{x+1}\)
\(=\dfrac{\left(x+1\right)^2+8-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{x+1}\)
\(=\dfrac{x^2+2x+1+8-x^2+2x-1}{x+1}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{4x+8}{\left(x+1\right)^2}\)
28 tháng 12 2023
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:
\(y=x\cdot\dfrac{1}{2}+1=\dfrac{1}{2}x+1\)
H9
HT.Phong (9A5)
CTVHS
4 tháng 7 2023
Câu 44:
\(12x-9-4x^2=-4x^2+12x-9=-\left(4x^2-12x+9\right)\)
\(=\left[\left(2x\right)^2-2\cdot3\cdot2x+3^2\right]=-\left(2x-3\right)^2\)
⇒ Chọn C
Câu 45:
\(x^3-6x^2y+12xy^2-8y^3=x^3-3\cdot2y\cdot x^2+3\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(x-2y\right)^3\)
⇒ Chọn D
GH
27 tháng 6 2023
5
\(=4^2-x^2=\left(4-x\right)\left(4+x\right)\)
6
\(=4^2-\left(3x+1\right)^2=\left(4-3x-1\right)\left(4+3x+1\right)=\left(3-3x\right)\left(5+3x\right)\\ =3\left(1-x\right)\left(5+3x\right)\)
7
\(=\left(2x+5\right)^2-\left(3x\right)^2=\left(2x+5+3x\right)\left(2x+5-3x\right)\\ =\left(5x+5\right)\left(5-x\right)\\ =5\left(x+1\right)\left(5-x\right)\)
8
\(=\left(2x-1-3x+1\right)\left(2x-1+3x-1\right)=\left(-x\right)\left(5x-2\right)\)
9
\(=\left(2x\right)^2-2.2x.y+y^2=\left(2x-y\right)^2\)
10
\(=\left(x+1\right)^2-\left(3y\right)^2=\left(x+1-3y\right)\left(x+1+3y\right)\)
11
\(=\left(x^2y^2\right)^2+2.2x^2y^2+2^2=\left(x^2y^2+2\right)\)
12
\(=\left(y-2\right)^2-x^2=\left(y-2-x\right)\left(y-2+x\right)\)
13
\(=1-\left(3\sqrt{3}x\right)^3=\left(1-3\sqrt{3}x\right)\left[1^2+3\sqrt{3}.x+\left(3\sqrt{3}.x\right)^2\right]=\left(1-3\sqrt{3}x\right)\left(1+3\sqrt{3}x+27x^2\right)\)
H9
HT.Phong (9A5)
CTVHS
27 tháng 6 2023
Bài 5:
a) \(M=\left(2x-1\right)^2+2\left(2x-1\right)\left(3x+1\right)+\left(3x+1\right)^2\)
\(M=\left[\left(2x-1\right)+\left(3x+1\right)\right]^2\)
\(M=\left(5x\right)^2\)
Thay \(x=-\dfrac{1}{5}\) vào biểu thức M ta có:
\(\left(5\cdot-\dfrac{1}{5}\right)^2=\left(-1\right)^2=1\)
Vậy: ...
b) \(N=\left(3x-1\right)^2-2\left(9x^2-1\right)+\left(3x+1\right)^2\)
\(N=\left(3x-1\right)^2-2\left(3x+1\right)\left(3x-1\right)+\left(3x+1\right)^2\)
\(N=\left[\left(3x-1\right)-\left(3x+1\right)\right]^2\)
\(N=\left(-2\right)^2=4\)
Vậy: ...
28 tháng 6 2023
14: =x^3-3x^2+3x-1-1
=(x-1)^3-1
=(x-1-1)(x^2-2x+1+x-1+1)
=(x-2)(x^2-x+1)
15: =x^3+x^2-4x^2-4x+7x+7
=(x+1)(x^2-4x+7)
16: =2x^3-x^2-2x^2+x+2x-1
=(2x-1)(x^2-x+1)
28 tháng 6 2023
1: =x^2-2*x*2y+4y^2
=(x-2y)^2
2: 9x^2-y^2=(3x-y)(3x+y)
5: x^3-8=(x-2)(x^2+2x+4)
4: =x^2-(y-2)^2
=(x-y+2)(x+y-2)
7: 8x^3+12x^2+6x+1=(2x+1)^3
14: =x^3-3x^2+3x-1-1
=(x-1)^3-1
=(x-1-1)(x^2-2x+1+x-1+1)
=(x-2)(x^2-x+1)
15: =x^3+x^2-4x^2-4x+7x+7
=(x+1)(x^2-4x+7)
16: =2x^3-x^2-2x^2+x+2x-1
=(2x-1)(x^2-x+1)
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