Cho biểu thức A=( x/x+2 + 5x-12/5x2-15x - 8/5x2 +10x ): x2 - 2x + 2/x2 - x - 6
Tìm đkcđ và rút gọn
Tính A khi x = 1 và x = 3
Tìm x để A min
Tìm x thuộc Z để A thuộc Z
K
Khách
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6 tháng 1 2022
a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)
b: \(A=\dfrac{x^2-4-5+x+3}{\left(x-2\right)\left(x+3\right)}=\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}=\dfrac{x+2}{x-2}\)
c: Để A=3/4 thì 4x-8=3x+6
=>x=14
d: Để A nguyên thì \(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{3;1;4;0;6;-2\right\}\)
27 tháng 7 2021
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
3 tháng 7 2021
a) ĐKXĐ: \(x\notin\left\{0;3;1\right\}\)
Sửa đề: \(A=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}\)
Ta có: \(A=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}\)
\(=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-6x+18}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-6\left(x-3\right)}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-3}{x-1}\)
b) Để A nguyên thì \(-3⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;-2;4\right\}\)
H
3 tháng 5 2023
\(a,N\left(x\right)=x^2+3x^4-2x-x^2+2x^3=3x^4+2x^3+\left(x^2-x^2\right)-2x\\ =3x^4+2x^3-2x\\ P\left(x\right)=-8+5x-6x^3-4x+6=-6x^3+\left(5x-4x\right)+\left(-8+6\right)\\ =-6x^3+x-2\)
Bậc của N(x) là 4
Bậc của P(x) là 3
\(b,P\left(x\right)+N\left(x\right)=3x^4+2x^3-2x-6x^3+x-2\\ =3x^4+\left(2x^3-6x^3\right)+\left(-2x+x\right)-2\\ =3x^4-4x^3-x-2\)
\(c,B\left(x\right)=-2x^2\left(x^3-2x+5x^2-1\right)\\ =\left(-2x^2\right).x^3+\left(-2x^2\right).\left(-2x\right)+\left(-2x^2\right).5x^2+\left(-2x^2\right).\left(-1\right)\\ =-2x^5+4x^3-10x^4+2x^2\\ =-2x^5-10x^4+4x^3+2x^2\)
20 tháng 1 2022
a: DKXĐ: \(x\notin\left\{3;-3\right\}\)
b: \(A=\left(\dfrac{x}{\left(x-3\right)\left(x+3\right)}+\dfrac{-1}{x-3}\right)\cdot\dfrac{x+3}{3}\)
\(=\dfrac{x-x-3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{3}=\dfrac{-1}{x-3}\)
c: Thay x=5 vào A, ta được:
\(A=\dfrac{-1}{5-3}=-\dfrac{1}{2}\)
d: Để A là số nguyên thì \(x-3\in\left\{1;-1\right\}\)
hay \(x\in\left\{4;2\right\}\)
20 tháng 1 2022
ab, đk x khác 3 ; -3
\(A=\left(\dfrac{x}{x^2-9}-\dfrac{1}{x-3}\right):\dfrac{3}{x+3}\Leftrightarrow=\left(\dfrac{x-x-3}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{3}{x+3}=-\dfrac{1}{x-3}\)
c, x^2 - 8x + 15 = 0 <=> (x-3)(x-5) = 0 <=> x = 3 (ktm) ; x= 5
Thay x = 5 vào A ta được : A =-1/2
d, \(\Rightarrow x-3\inƯ\left(-1\right)=\left\{\pm1\right\}\)
TH1 : x - 3 = 1 <=> x = 4
TH2 : x - 3 = -1 <=> x = 2
14 tháng 3 2022
a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)
\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)
b. -Để M thuộc Z thì:
\(\left(x^2+x-2\right)⋮\left(x+3\right)\)
\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)
\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)
\(\Rightarrow4⋮\left(x+3\right)\)
\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)
c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)
\(=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2-x+3x-3-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)
a: \(A=\left(\dfrac{x}{x+2}+\dfrac{4x-12}{5x^2-15x}-\dfrac{8}{5x^2+10x}\right):\dfrac{x^2-2x+2}{x^2-x-6}\)
\(=\left(\dfrac{x}{x+2}+\dfrac{4x-12}{5x\left(x-3\right)}-\dfrac{8}{5x\left(x+2\right)}\right)\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{x^2-2x+2}\)
\(=\left(\dfrac{x}{x+2}+\dfrac{4}{5x}-\dfrac{8}{5x\left(x+2\right)}\right)\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{x^2-2x+2}\)
\(=\dfrac{5x^2+4x+8-8}{5x\left(x+2\right)}\cdot\dfrac{\left(x-3\right)\left(x+2\right)}{x^2-2x+2}\)
\(=\dfrac{5x^2+4x}{5x}\cdot\dfrac{x-3}{x^2-2x+2}=\dfrac{\left(5x+4\right)\left(x-3\right)}{5\left(x^2-2x+2\right)}\)
b: Khi x=1 thì \(A=\dfrac{\left(5+4\right)\left(1-3\right)}{5\left(1-2+2\right)}=\dfrac{9\cdot\left(-2\right)}{5}=\dfrac{-18}{5}\)
Khi x=3 thì \(A=\dfrac{\left(5\cdot3+4\right)\left(3-3\right)}{A}=0\)