A=(2+x)/x ; B=(x-1)/x + (2x+1)/(x^2+x)
1) tìm x để A/B > 3/2
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1 tháng 7 2023
`+)axx2+bxx1=cxx2+axx1<=>2a+b=2c+a<=>2c-a=b`
`+)cxx3+axx1=bxx2+axx1<=>3c+a=2b+a<=>3c=2b<=>c=2/3b`
mà `2c-a=b` nên `a=2c-b=4/3b-b=1/3b`
Khi đó: `cxx2+axx2=2(a+c)=2(1/3b+2/3b)=2b`
Vậy dấu hỏi chấm cần điền là `2`
19 tháng 1 2022
Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
19 tháng 5 2022
a: \(E=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x-3|=2
=>x-3=2 hoặc x-3=-2
=>x=5(nhận) hoặc x=1(loại)
Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)
c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow2x^2-x+1=0\)
hay \(x\in\varnothing\)
19 tháng 5 2022
f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)
-Vậy \(A_{min}=4\)
28 tháng 7 2021
a, ĐK : \(x\ne0;1\)
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x\left(x-1\right)}\right)\)
\(=\dfrac{x^2\left(x+1\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x^2}{x-1}\)
b, Thay x = 3 vào A ta được : \(\dfrac{9}{2}\)
c, \(A=4\Rightarrow\dfrac{x^2}{x-1}=4\Rightarrow x^2=4x-4\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
d, \(A< 2\Rightarrow\dfrac{x^2}{x-1}-2< 0\Leftrightarrow\dfrac{x^2-2x+1}{x-1}< 0\Rightarrow x-1< 0\Leftrightarrow x>1\)
28 tháng 7 2021
a,\(\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\dfrac{x}{x\left(x-1\right)}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x+1}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2}{x-1}\)
H
6 tháng 7 2023
\(a,A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{3+x}\right)\\ =\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{x^2-9}\right):\left(\dfrac{2\left(3+x\right)}{3+x}-\dfrac{x+5}{3+x}\right)\\ =\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{2\left(3+x\right)-\left(x+5\right)}{3+x}\\ =\left(\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{6+2x-x-5}{3+x}\)
\(=\dfrac{x^2-3x-\left(2x+6\right)-\left(x^2-1\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{3+x}\\ =\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3+x}{x+1}\\ =\dfrac{-5x-5}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3+x}{x+1}\\ =\dfrac{-5\left(x+1\right).\left(3+x\right)}{\left(x-3\right)\left(x+3\right).\left(x+1\right)}\\ =\dfrac{-5}{x-3}\)
\(b,A=x^2-x-2=0\\ \Leftrightarrow x^2+x-2x-2=0\\ \Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
\(c,\dfrac{-5}{x-3}=\dfrac{1}{2}\\ \Leftrightarrow-10=x-3\\ \Leftrightarrow-x+3=10\\ \Leftrightarrow-x=7\\ \Leftrightarrow x=7\)
Để `A=1/2` thì `x=7`
AH
Akai Haruma
Giáo viên
4 tháng 5 2023
Lời giải:
$A(x)+B(x)=(x^3-3x^2+3x-1)+(2x^3+x^2-x+5)$
$=3x^3-2x^2+2x+4$
b.
$A(x)C(x)=(x^3-3x^2+3x-1)(x-2)=x(x^3-3x^2+3x-1)-2(x^3-3x^2+3x-1)$
$=(x^4-3x^3+3x^2-x)-(2x^3-6x^2+6x-2)$
$=x^4-5x^3+9x^2-7x+2$