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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: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
\(\left(\dfrac{8-x\sqrt{x}}{2-\sqrt{x}}+2\sqrt{x}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2=\left(x+2\sqrt{x}+4+2\sqrt{x}\right).\dfrac{\left(2-\sqrt{x}\right)^2}{\left(\sqrt{x}+2\right)^2}=\left(\sqrt{x}+2\right)^2.\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+2\right)^2}=\left(\sqrt{x}-2\right)^2\)
a) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)
\(=\dfrac{6\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\left(\sqrt{5}-1\right)+\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\sqrt{5}-\dfrac{3}{2}-2\sqrt{5}+\dfrac{\sqrt{5}+1}{2}\)
\(=-\dfrac{1}{2}\sqrt{5}-\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}+\dfrac{1}{2}\)
=-1
Bài 1:
a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:
\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)
b) Ta có: \(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{x-9+\sqrt{x}+3-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-6-x+2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow3-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 3\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)
a: \(A=\left(\dfrac{\left(x-4\right)\left(\sqrt{x}+2\right)-x\sqrt{x}+8}{x-4}\right):\dfrac{x-2\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8-x\sqrt{x}+8}{x-4}\cdot\dfrac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)
\(=\dfrac{2x-4\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{1}{x-2\sqrt{x}+4}=\dfrac{2\sqrt{x}}{x-2\sqrt{x}+4}\)
b: \(A-1=\dfrac{2\sqrt{x}-x+2\sqrt{x}-4}{x-2\sqrt{x}+4}\)
\(=\dfrac{-x+4\sqrt{x}-4}{x-2\sqrt{x}+4}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-1\right)^2+3}< 0\)
=>A<1
c: \(2\sqrt{x}>=0;x-2\sqrt{x}+4=\left(\sqrt{x}-1\right)^2+3>0\)
=>A>=0 với mọi x thỏa mãn ĐKXĐ
mà A<1
nên 0<=A<1
=>Để A nguyên thì A=0
=>x=0
`A=((3sqrtx+6)/(x-4)+sqrtx/(sqrtx-2)):(x-9)/(sqrtx-3)(x>=0,x ne 4,x ne 9)`
`=((3(sqrtx+2))/((sqrtx-2)(sqrtx+2))+sqrtx/(sqrtx-2)):((sqrtx-3)(sqrtx+3))/(sqrtx-3)`
`=(3/(sqrtx-2)+sqrtx/(sqrtx-2)):(sqrtx+3)`
`=(sqrtx+3)/(sqrtx-2)*1/(sqrtx+3)`
`=1/(sqrtx-2)`
\(A=\left(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{x-9}{\sqrt{x}-3}\)
\(=\left(\dfrac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}-3}\)
\(=\left(\dfrac{3}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\left(\sqrt{x}+3\right)=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}.\dfrac{1}{\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
+) ta có : \(N=\dfrac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\dfrac{\sqrt{16-2\sqrt{15}}}{\sqrt{2}\left(\sqrt{30}-\sqrt{2}\right)}=\dfrac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\left(\sqrt{15}-1\right)}\)
\(=\dfrac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}=\dfrac{1}{2}\)
+) ta có : \(P=\left(\dfrac{8-x\sqrt{x}}{2-\sqrt{x}}+2\sqrt{x}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2\)
\(\Leftrightarrow P=\left(\dfrac{\left(2-\sqrt{x}\right)\left(4+2\sqrt{x}+x\right)}{2-\sqrt{x}}+2\sqrt{x}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2\) \(\Leftrightarrow P=\left(4+2\sqrt{x}+x+2\sqrt{x}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2\) \(\Leftrightarrow P=\left(2+\sqrt{x}\right)^2\dfrac{\left(2-\sqrt{x}\right)^2}{\left(2+\sqrt{x}\right)^2}=\left(2-\sqrt{x}\right)^2\)
<=>N=\(\dfrac{\sqrt{16-2\sqrt{15}}}{\sqrt{60}-2}\)
<=>N=\(\dfrac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\sqrt{15}-2}\)
<=>N=\(\dfrac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}\)
<=>N=\(\dfrac{1}{2}\)
P=\(\left(\dfrac{8-x\sqrt{x}}{2-\sqrt{x}}+2\sqrt{x}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2\)
P=\(\left(\dfrac{8-x\sqrt{x}+4\sqrt{x}-2x}{2-\sqrt{x}}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2\)
P=\(\dfrac{8+3\sqrt{x}+x}{2-\sqrt{x}}.\dfrac{\left(2-\sqrt{x}\right)^2}{\left(2+\sqrt{x}\right)^2}\)
P=\(\dfrac{\left(8+3\sqrt{x}+x\right)\left(2-\sqrt{x}\right)}{4+4\sqrt{x}+x}\)
a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\)
b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\sqrt{x}=a,\sqrt{y}=b\)
Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)
\(\Rightarrow B=x+\sqrt{xy}+y\)
Vậy...
c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\)
d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\)
a:b(a−4)2.√(a−4)4b2(b>0;a≠4)b(a−4)2.(a−4)4b2(b>0;a≠4)
= \(\dfrac{b}{\left(a-4\right)}.\dfrac{\sqrt{\left[\left(a-4\right)^2\right]^2}}{\sqrt{b^2}}\)
=\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\)
= 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn)
b:x√x−y√y√x−√y(x≥0;y≥0;x≠0)xx−yyx−y(x≥0;y≥0;x≠0)
=\(\dfrac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức )
= (x+\(\sqrt{xy}\)+y)
c:a(b−2)2.√(b−2)4a2(a>0;b≠2)a(b−2)2.(b−2)4a2(a>0;b≠2)
Tương tự câu a
d:x(y−3)2.√(y−3)2x2(x>0;y≠3)x(y−3)2.(y−3)2x2(x>0;y≠3)
tương tự câu a
e:2x +√1−6x+9x23x−1
= \(2x+\dfrac{\sqrt{\left(3x\right)^2-6x+1}}{3x-1}\)
= 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức)
=2x+\(\dfrac{3x-1}{3x-1}\)
=2x+1
a: \(M=\left(\dfrac{-\left(\sqrt{x}+2\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{4x}{x-4}\right)\cdot\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+3}\)
\(=\dfrac{-x-4\sqrt{x}-4+x-4\sqrt{x}+4-4x}{x-4}\cdot\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+3}\)
\(=\dfrac{-4x-8\sqrt{x}}{x-4}\cdot\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+3}\)
\(=\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(x-4\right)\left(\sqrt{x}+3\right)}=\dfrac{4\sqrt{x}}{\sqrt{x}+3}\)
b: \(x=\sqrt{5}-1-\left(\sqrt{5}-2\right)=\sqrt{5}-1-\sqrt{5}+2=1\)
Thay x=1 vào M, ta được:
\(M=\dfrac{4}{1+3}=\dfrac{4}{4}=1\)
c: Để M là số nguyên thì \(4\sqrt{x}-12+12⋮\sqrt{x}+3\)
\(\Leftrightarrow\sqrt{x}+3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;3;9\right\}\)
hay \(x\in\left\{0;9;81\right\}\)
\(a,A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\left(dk:x\ge0,x\ne4\right)\\ =\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\right)\\ =\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x-4+10-x}\)
\(=\dfrac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\sqrt{x}-2}.\dfrac{1}{6}\\ =\dfrac{-6}{\left(\sqrt{x}-2\right).6}\\
=-\dfrac{1}{\sqrt{x}-2}\)
\(b,A>0\Leftrightarrow-\dfrac{1}{\sqrt{x}-2}>0\Leftrightarrow\sqrt{x}-2< 0\\
\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với \(dk:x\ge0,x\ne4\), ta kết luận \(0\le x< 4\)
Ta có: \(\dfrac{\sqrt{x}\left(16-\sqrt{x}\right)}{x-4}+\dfrac{3+2\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-3\sqrt{x}}{\sqrt{x}+2}\)
\(=\dfrac{16\sqrt{x}-x-\left(3+2\sqrt{x}\right)\left(\sqrt{x}+2\right)+\left(3\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{16\sqrt{x}-x-3\sqrt{x}-6-2x-4\sqrt{x}+3x-6\sqrt{x}-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}\)
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