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a) đkxđ x≥0 , x ≠1
\(K=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{x-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)b)
\(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2-1}{\sqrt{x}-2}=1-\dfrac{1}{\sqrt{x}-2}\)
để K ∈ z thì \(\dfrac{-1}{\sqrt{x}-2}\) nguyên
=> √x -2 ∈ Ư(-1)={-1;1}
=> x ∈ {1; 9}
vậy ...
a: \(=\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}\cdot\dfrac{x-1}{x-2\sqrt{x}}\)
\(=\dfrac{x-3\sqrt{x}}{x-2\sqrt{x}}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)
b: Để K là số nguyên thì \(\sqrt{x}-2-1⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\in\left\{1;-1\right\}\)
hay x=9
c: Để K là số âm thì \(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}< 0\)
=>4<x<9
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
\(a)A=\dfrac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\dfrac{2+\sqrt{8}}{1+\sqrt{2}}\\ A=\dfrac{\left(\sqrt{3}-\sqrt{6}\right)\left(1+\sqrt{2}\right)}{1^2-\left(\sqrt{2}\right)^2}-\dfrac{\left(2+\sqrt{8}\right)\left(1-\sqrt{2}\right)}{1^2-\left(\sqrt{2}\right)^2}\\ A=-\left(\sqrt{3}+\sqrt{6}-\sqrt{6}-2\sqrt{3}\right)+2-2\sqrt{2}+2\sqrt{2}-4\\ A=\sqrt{3}-2\)
\(b)B=\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right).\dfrac{x+2\sqrt{x}}{\sqrt{x}}\\ B=\left[\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\\ B=\dfrac{\sqrt{x}+2-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\left(\sqrt{x}+2\right)\\ B=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}+2\right)\\ B=\dfrac{4}{x-4}\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
Mọi ngươi giúp em với ạ chứ em làm câu a Bài 1 và 2 ra kết quả dài quá :(
Bài 1:
a: \(P=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b: Để P<1 thì P-1<0
\(\Leftrightarrow\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}< 0\)
=>căn a-2>0
=>a>4
\(a,\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\sqrt{3^2}-2\sqrt{3}+1}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}=-1\)
\(b,\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\left(dk:x\ne\pm\sqrt{2}\right)\\ =\dfrac{x^2+2\sqrt{2}x+\sqrt{2^2}}{x^2-\sqrt{2^2}}\\ =\dfrac{\left(x+\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}\\ =\dfrac{x+\sqrt{2}}{x-\sqrt{2}}\)
\(c,\sqrt{9x^2}-2x\left(dk:x< 0\right)\\ =\sqrt{3^2}.\sqrt{x^2}-2x\\ =3\left|x\right|-2x\\ =-3x-2x\\ =-5x\)
\(d,\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\\ =\sqrt{\sqrt{2^2}+2.3\sqrt{2}+3^2}-3+\sqrt{2}\\ =\sqrt{\left(\sqrt{2}+3\right)^2}-3+\sqrt{2}\\ =\sqrt{2}+3-3+\sqrt{2}\\ =2\sqrt{2}\)
\(e,\dfrac{x^2-5}{x+\sqrt{5}}\left(dk:x\ne-\sqrt{5}\right)\\ =\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}\\ =x-\sqrt{5}\)
b) (4√x + 4)/(x + 2√x + 5) ≥ 1
⇔ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0
Do x ≥ 0 ⇒ x + 2√x + 5 > 0
⇒ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0
⇔ (4√x + 4) - (x + 2√x + 5) ≤ 0
⇔ 4√x + 4 - x - 2√x - 5 ≤ 0
⇔ -x + 2√x - 1 ≤ 0
⇔ -(x - 2√x + 1) ≤ 0
⇔ -(√x - 1)² ≤ 0 (luôn đúng)
Vậy (4√x + 4)/(x + 2√x + 5) ≤ 1 với mọi x ≥ 0
a: \(P=\dfrac{x+8\sqrt{x}+8-x-4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}+2\right)}:\dfrac{x+\sqrt{x}+3+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4\left(\sqrt{x}+1\right)}{x+2\sqrt{x}+5}\)
b: 4(căn x+1)>=4
x+2căn x+5>=5
=>P<=4/5<1