Bài 2. (3 điểm) Cho biểu thức: $A=\dfrac{x^2-2x+1}{x^2-1}$ với $x \ne \pm 1$.
a) Rút gọn biểu thức $A$.
b) Tính giá trị của $A$ khi $x=3$ và $x=-\dfrac{3}{2}$.
c) Tìm $x \in \mathbb{Z}$ để biểu thức $A$ nhận giá trị nguyên.
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a: Ta có: |x+4|=1
=>x+4=1 hoặc x+4=-1
=>x=-3(loại) hoặc x=-5
Khi x=-5 thì \(A=\dfrac{\left(-5\right)^2-5}{3\left(-5+3\right)}=\dfrac{20}{3\cdot\left(-2\right)}=\dfrac{-10}{3}\)
b: \(B=\dfrac{x-1+x+1-3+x}{\left(x-1\right)\left(x+1\right)}=\dfrac{3x-3}{\left(x-1\right)\left(x+1\right)}=\dfrac{3}{x+1}\)
a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}
Thay x = 2, ta có B không tồn tại
Thay x = -1, ta có B = \(\dfrac{1}{3}\)
b)ĐKXĐ:x ≠ 2,-2
Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x
Do đó không tồn tại x thỏa mãn đề bài
a: TXĐ: D=[0;+\(\infty\))\{1}
\(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}-\dfrac{\sqrt{x}}{x-1}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\cdot2}\)
\(=\dfrac{-1}{\sqrt{x}+1}\)
\(a,ĐK:x\ge0\\ x\ne1\\ B=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{2\left(1-\sqrt{x}\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-1}{\sqrt{x}+1}\\ b,x=3\Leftrightarrow B=\dfrac{-1}{\sqrt{3}+1}=\dfrac{1-\sqrt{3}}{2}\\ c,\left|B\right|=\dfrac{1}{2}\Leftrightarrow\left|\dfrac{-1}{\sqrt{x}+1}\right|=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{2}\left(\sqrt{x}+1\ge1>0\right)\\ \Leftrightarrow\sqrt{x}+1=2\Leftrightarrow x=1\left(tm\right)\)
a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)
a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)
Có vài bước mình làm tắc á nha :>
ĐKXĐ : \(x\ne0;x\ne\pm1\)
a) Bạn ghi lại rõ đề.
b) \(B=\dfrac{x-1}{x+1}+\dfrac{3x-x^2}{x^2-1}=\dfrac{x-1}{x+1}+\dfrac{3x-x^2}{\left(x-1\right).\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2+3x-x^2}{\left(x-1\right).\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right).\left(x+1\right)}=\dfrac{1}{x-1}\)
c) \(P=A.B=\dfrac{x^2+x-2}{x.\left(x-1\right)}=\dfrac{\left(x-1\right).\left(x+2\right)}{x\left(x-1\right)}=\dfrac{x+2}{x}=1+\dfrac{2}{x}\)
Không tồn tại Min P \(\forall x\inℝ\)
1: Khi x=64 thì \(A=\dfrac{8+2}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)
2: \(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
3: A/B>3/2
=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{3}{2}>0\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{\sqrt{x}\cdot2}>0\)
=>\(-\sqrt{x}+2>0\)
=>-căn x>-2
=>căn x<2
=>0<x<4
1) Thay x=64 vào A ta có:
\(A=\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)
2) \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
3) Ta có:
\(\dfrac{A}{B}>\dfrac{3}{2}\) khi
\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)
\(\Leftrightarrow\dfrac{2-\sqrt{x}}{2\sqrt{x}}>0\)
Mà: \(2\sqrt{x}\ge0\forall x\)
\(\Leftrightarrow2-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 2\)
\(\Leftrightarrow x< 4\)
Kết hợp với đk:
\(0< x< 4\)
1) Sửa đề: x=0,09
Thay x=0,09 vào A, ta được:
\(A=\dfrac{\sqrt{0.09}}{\sqrt{0.09}-1}=\dfrac{0.3}{0.3-1}=\dfrac{0.3}{-0.7}=\dfrac{-3}{7}\)
a: \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\sqrt{x}-1\)
a) \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)
Đk: \(x>0\) và \(x\ne1\)
\(\Rightarrow A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)
\(=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-2x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
b) Thay \(x=3+2\sqrt{2}\) vào A ta được:
\(A=\sqrt{3+2\sqrt{2}}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1\)
\(=\sqrt{2}+1-1=\sqrt{2}\)
(Vì \(\sqrt{2}+1>0\Rightarrow\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\))
\(a,A=\dfrac{x^2-2x+1}{x^2-1}\\ =\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x-1}{x+1}\)
`b,` Khi `x=3` thì :
\(\dfrac{x-1}{x+1}=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)
Khi `x=-3/2` thì :
\(\dfrac{-\dfrac{3}{2}-1}{-\dfrac{3}{2}+1}\\ =\dfrac{-\dfrac{3}{2}-\dfrac{2}{2}}{-\dfrac{3}{2}+\dfrac{2}{2}}\\ =\dfrac{-\dfrac{5}{2}}{-\dfrac{1}{2}}\\ =-\dfrac{5}{2}\cdot\left(-2\right)=\dfrac{10}{2}=5\)
`c,` Để `A` nhận giá trị nguyên ta có :
\(\dfrac{x-1}{x+1}=\dfrac{x+1-2}{x+1}=\dfrac{x+1}{x+1}-\dfrac{2}{x+1}\)
Vậy \(x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
`-> x+1=1=>x=0`
`->x+1=-1=>x=-2`
`->x+1=2=>x=1`
`->x+1=-2=>x=-3`
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