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a: \(P=\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2+x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{3x+\sqrt{x}+3-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
b: Thay \(x=3-2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{2\cdot\left(3-2\sqrt{2}\right)+2\left(\sqrt{2}-1\right)+2}{\sqrt{2}-1}\)
\(=\dfrac{6-4\sqrt{2}+2\sqrt{2}-2+2}{\sqrt{2}-1}=\dfrac{6-2\sqrt{2}}{\sqrt{2}-1}=4\sqrt{2}+2\)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
a) A=\(\dfrac{\sqrt{x}[\left(\sqrt{x}\right)^3-1]}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
A=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\) A=\(\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
A=\(x-\sqrt{x}+1\)
b) A=\(\dfrac{3}{4}\)
=> \(x-\sqrt{x}+1=\dfrac{3}{4}\)
\(x-\sqrt{x}+\dfrac{1}{4}=0\)
\(\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\)
=> \(\sqrt{x}=\dfrac{1}{2}\)
=> \(x=\dfrac{1}{4}\)
Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
a: \(Q=\dfrac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-\left(\sqrt{x}+2\right)}{\sqrt{x}+3}\)
\(P=\dfrac{x-9-x+3\sqrt{x}}{x-9}=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}=\dfrac{3}{\sqrt{x}+3}\)
\(M=\dfrac{3}{\sqrt{x}+3}:\dfrac{-\left(\sqrt{x}+2\right)}{\sqrt{x}+3}=\dfrac{-3}{\sqrt{x}+2}\)
Để M>0 thì căn x+2<0(vô lý)
b: \(M\left(x+1\right)=\dfrac{-3\left(x+1\right)}{\sqrt{x}+2}\)
\(=\dfrac{-3x-3}{\sqrt{x}+2}=\dfrac{-3x+12-15}{\sqrt{x}+2}=-\dfrac{3\left(x-4\right)}{\sqrt{x}+2}-\dfrac{15}{\sqrt{x}+2}\)
\(=-3\left(\sqrt{x}-2\right)-\dfrac{15}{\sqrt{x}+2}\)
\(=-3\sqrt{x}+6-\dfrac{15}{\sqrt{x}+2}\)
\(=-3\sqrt{x}-6-\dfrac{15}{\sqrt{x}+2}+12\)
\(=-\left(3\sqrt{x}+6+\dfrac{15}{\sqrt{x}+2}\right)+12\)
\(3\left(\sqrt{x}+2\right)+\dfrac{15}{\sqrt{x}+2}>=2\cdot\sqrt{3\left(\sqrt{x}+2\right)\cdot\dfrac{15}{\sqrt{x}+2}}=6\sqrt{5}\)
=>M(x+1)<=-6căn 5+12
Dấu = xảy ra khi \(3\left(\sqrt{x}+2\right)^2=15\)
=>(căn x+2)^2=5
=>căn x+2=căn 5
=>căn x=căn 5-2
=>x=9-4căn 5
a) Mạn phép ko chép lại đề , mk làm luôn.
ĐKXĐ : x > 0 ; x # 1
\(Q=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{\left(x+2\sqrt{x}\right)\left(\sqrt{x}-1\right)-\left(x-2\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{x\sqrt{x}-x+2x-2\sqrt{x}-\left(x\sqrt{x}+x-2x-2\sqrt{x}\right)}{x\left(x-1\right)}.\)
\(Q=\dfrac{2}{x-1}\)
b) Để Q ∈ Z ⇒ x ∈ Z
⇒ x - 1 ∈ Ư(2)
+) x - 1 = 1 ⇔ x = 2 ( TM )
+) x - 1 = - 1 ⇔ x = 0 ( KTM)
+) x - 1 = 2 ⇔ x = 3 ( TM)
+) x - 1 = - 2⇔ x = -1 ( KTM)
KL.....
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)