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Bài 1 : Rút gọn biểu thức :
\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)
\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)
\(=-10\sqrt{2}+10-7+30\sqrt{2}\)
\(=20\sqrt{2}+3\)
Bài 2:
a) ĐKXĐ : x # 4 ; x # - 4
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
P =\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy, để P = 2 thì x = 16.
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
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) Rut gon H
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\)
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}-\dfrac{1}{\sqrt{a}-2}\)
DKXD : \(\left\{{}\begin{matrix}\sqrt{a}+3\ne0\\\sqrt{a}-2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a\ne9\\a\ne4\end{matrix}\right.\)
Ta co : \(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{a-\sqrt{a}-6}{a+\sqrt{a}-6}\)
Bài 1:
a: \(A=\dfrac{\sqrt{x}+2}{2\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4+x-4\sqrt{x}+4}{2\left(x-4\right)}\)
\(=\dfrac{2x+8}{2\left(x-4\right)}=\dfrac{x+4}{x-4}\)
b: Để A=8 thì x+4=8(x-4)
=>x+4=8x-32
=>-7x=-36
hay x=36/7(nhận)
Bài 2:
a: \(P=\dfrac{a-1}{2\sqrt{a}}\cdot\left(\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{a-1}\right)\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}=-2\sqrt{a}\)
b: Để P>=-2 thì P+2>=0
\(\Leftrightarrow-2\sqrt{a}+2>=0\)
=>0<=a<1
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
Bài 1:
a: \(A=\left(\dfrac{x-1}{2\sqrt{x}}\right)^2\cdot\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\)
\(=\dfrac{\left(x-1\right)^2}{4x}\cdot\dfrac{-4\sqrt{x}}{x-1}=\dfrac{-\left(x-1\right)}{\sqrt{x}}\)
b: Để B<0 thì -x+1<0
=>-x<-1
hay x>1
c: Để B=2 thì \(-\left(x-1\right)=2\sqrt{x}\)
\(\Leftrightarrow-x+1-2\sqrt{x}=0\)
\(\Leftrightarrow x+\sqrt{x}-1=0\)
\(\Leftrightarrow\sqrt{x}=\dfrac{\sqrt{5}-1}{2}\)
hay \(x=\dfrac{6-2\sqrt{5}}{4}\)
Bài 1 : ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Câu a :
\(B=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)^2\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)
\(=\left(\dfrac{\sqrt{x}.\sqrt{x}-1}{2\sqrt{x}}\right)^2\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\dfrac{x-1}{2\sqrt{x}}\right)^2\left(\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\left(x-1\right)^2}{\left(2\sqrt{x}\right)^2}\times\dfrac{-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(x-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{4x}\times\dfrac{-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=-\dfrac{x-1}{\sqrt{x}}\)
Câu b :
Để \(B< 0\Leftrightarrow-\dfrac{x-1}{\sqrt{x}}< 0\Leftrightarrow\dfrac{x-1}{\sqrt{x}}>0\Leftrightarrow x-1>0\Leftrightarrow x>1\)
Vậy \(x>1\) thì \(B< 0\)
Câu c :
Để \(B=-2\Leftrightarrow-\dfrac{x-1}{\sqrt{x}}=-2\)
\(\Leftrightarrow\left(\dfrac{-\left(x-1\right)}{\sqrt{x}}\right)^2=\left(-2\right)^2\)
\(\Leftrightarrow\dfrac{x^2-2x+1}{x}=4\)
\(\Leftrightarrow\dfrac{x^2-2x+1}{x}=\dfrac{4x}{x}\)
\(\Leftrightarrow x^2-2x+1=4x\)
\(\Leftrightarrow x^2-6x+1=0\)
\(\Delta=\left(-6\right)^2-4=32>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{6+\sqrt{32}}{2}=3+2\sqrt{2}\\x_1=\dfrac{6-\sqrt{32}}{2}=3-2\sqrt{2}\end{matrix}\right.\)
Vậy \(x=3+2\sqrt{2}\) hoặ \(x=3-2\sqrt{2}\) thì \(B=-2\)
a) điều kiện : \(x\ge0;x\ne m^2\)
ta có : \(P=\dfrac{2\sqrt{x}}{\sqrt{x}+m}+\dfrac{\sqrt{x}}{\sqrt{x}-m}-\dfrac{m^2}{4x-4m^2}\)
\(\Leftrightarrow P=\dfrac{2\sqrt{x}}{\sqrt{x}+m}+\dfrac{\sqrt{x}}{\sqrt{x}-m}-\dfrac{m^2}{4\left(\sqrt{x}+m\right)\left(\sqrt{x}-m\right)}\) \(\Leftrightarrow P=\dfrac{2\sqrt{x}\left(\sqrt{x}-m\right)+\sqrt{x}\left(\sqrt{x}+m\right)-m^2}{4\left(\sqrt{x}+m\right)\left(\sqrt{x}-m\right)}\) \(\Leftrightarrow P=\dfrac{3x-\sqrt{x}m-m^2}{4\left(\sqrt{x}+m\right)\left(\sqrt{x}-m\right)}\)b) ta có : \(P=0\Leftrightarrow3x-m\sqrt{x}-m^2\)
ta có : \(\Delta=\left(m^2\right)-3.4\left(-m^2\right)=13m^2\ge0\)
th1: \(m=0\) \(\Rightarrow\) phương trình có nghiệm duy nhất : \(\sqrt{x}=\dfrac{m}{6}\Leftrightarrow x=\dfrac{m^2}{36}\)
th2: \(m\ne0\) \(\Rightarrow\) phương trình có 2 nghiệm phân biệt : \(\left\{{}\begin{matrix}\sqrt{x}=\dfrac{m+\sqrt{13m^2}}{6}\\\sqrt{x}=\dfrac{m-\sqrt{13m^2}}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14m^2+2\sqrt{13}m^2}{36}\\x=\dfrac{14m^2-2\sqrt{13}m^2}{36}\end{matrix}\right.\)
c) để \(x>1\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2>36\\14m^2+2\sqrt{13}m^2>36\\14m^2-2\sqrt{13}m^2>36\end{matrix}\right.\)
\(\Rightarrow m^2>36\Leftrightarrow\left[{}\begin{matrix}m>6\\m< -6\end{matrix}\right.\)
hình như rút gọn sai thì phải bn ơi