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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.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
a: \(A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{x-9}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\sqrt{x}+3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
b: Khi \(x=4-2\sqrt{3}\) vào A, ta được:
\(A=\dfrac{-3\left(\sqrt{3}-1\right)+3}{\left(\sqrt{3}-1+3\right)\left(\sqrt{3}-1+1\right)}\)
\(=\dfrac{-3\sqrt{3}+6}{\sqrt{3}\cdot\left(\sqrt{3}+2\right)}=\dfrac{-3+2\sqrt{3}}{2+\sqrt{3}}\)
a) ĐKXĐ : \(a\ge0;a\ne4;a\ne25\) Rút gọn:\(M=\dfrac{5}{\sqrt{a}+2}\) (xin lỗi, mình đã làm rồi nhưng bài giải ko được gửi đi, đây là M sau khi rút gọn, bạn tìm cách rút nha, cũng dễ lắm ^_^). b) M<1 khi \(\dfrac{5}{\sqrt{a}+2}< 1\Leftrightarrow\dfrac{5}{\sqrt{a}+2}-1< 0\Leftrightarrow\dfrac{5-\sqrt{a}-2}{\sqrt{a}+2}< 0\Leftrightarrow\dfrac{3-\sqrt{a}}{\sqrt{a}+2}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3-\sqrt{a}< 0\\\sqrt{a}+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}3-\sqrt{a}>0\\\sqrt{a}+2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{a}>3\\\sqrt{a}>-2\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{a}< 3\\\sqrt{a}< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}>3\Leftrightarrow a>9\left(a\ne25\right)\\\sqrt{a}< -2\Leftrightarrow a\in\varnothing\end{matrix}\right.\) Vậy M<1 khi a>9 (a khác 25).
c) tìm GTLN của M=\(\dfrac{5}{\sqrt{a}+2}\), ta có:\(a\ge0\Leftrightarrow\sqrt{a}\ge0\Leftrightarrow\sqrt{a}+2\ge2\Leftrightarrow\dfrac{5}{\sqrt{a}+2}\le\dfrac{5}{2}\Rightarrow maxM=\dfrac{5}{2}khi\sqrt{a}=0\Leftrightarrow a=0\)
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
a: \(K=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{a-1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\dfrac{a-1}{\sqrt{a}+1}=\dfrac{a-1}{\sqrt{a}}\)
b: Thay \(a=3+2\sqrt{2}\) vào K, ta được:
\(K=\dfrac{3+2\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{2\sqrt{2}+2}{\sqrt{2}+1}=2\)
c: Để K<0 thì a-1<0
hay 0<a<1
a: \(P=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b: Để A>1/6 thì A-1/6>0
\(\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}>0\)
\(\Leftrightarrow2\sqrt{a}-4-\sqrt{a}>0\)
\(\Leftrightarrow\sqrt{a}>4\)
hay a>16
c: Để A=0 thì \(\sqrt{a}-2=0\)
hay a=4(loại)
1. \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
\(=\left(1+\sqrt{2}\right)^2-\sqrt{3}^2\)
\(=1+2\sqrt{2}+2-3\)
\(=2\sqrt{2}\)
3. \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\cdot\left(1+\dfrac{1}{\sqrt{x}}\right)\)(1)
ĐKXĐ \(x>0,x\ne1\)
pt (1) <=> \(\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+1+\sqrt{x}-1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\sqrt{x}\cdot2}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\)
b) Để \(\sqrt{A}>A\Leftrightarrow\sqrt{\dfrac{2}{\sqrt{x}-1}}>\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}>\dfrac{4}{x-2\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}-\dfrac{4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\cdot\left(\sqrt{x}-1\right)-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-2-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-6}{x-2\sqrt{x}+1}>0\)
Vì \(2\sqrt{2}-6< 0\Rightarrow x-2\sqrt{x}+1< 0\)
mà \(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\ge0\forall x\)
Vậy không có giá trị nào của x thỏa mãn \(\sqrt{A}>A\)
(P/s Đề câu b bị sai hay sao vậy, chả có số nào mà \(\sqrt{A}>A\) cả, check lại đề giùm với nhé)
\(a,\) ĐKXĐ\(:\) \(a\) ≥ \(0\) \(;\) \(a\) ≠ \(4\) \(;\) \(a\) ≠ \(9\)
\(M=\left(1-\dfrac{\sqrt{a}}{1+\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}+\dfrac{\sqrt{a}+2}{3-\sqrt{a}}+\dfrac{\sqrt{a}+2}{a-5\sqrt{a}+6}\right)\\ =\left(\dfrac{1+\sqrt{a}-\sqrt{a}}{\sqrt{a}+1}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}+\dfrac{\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\right)\\ =\dfrac{1}{\sqrt{a}+1}:\left(\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}+1}:\left(\dfrac{a-9-a+4+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\right)\\ =\dfrac{1}{\sqrt{a}+1}:\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\\ =\dfrac{1}{\sqrt{a}+1}:\dfrac{1}{\sqrt{a}-2}\\ =\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\)
\(b,\) Với \(a\) ≥ \(0\) \(;\) \(a\) ≠ \(4\) \(;\) \(a\) ≠ \(9\) ta có\(:\)
\(M< 0\) ⇔ \(\dfrac{\sqrt{a}-2}{\sqrt{a}+1}< 0\)
Mà\(:\) \(\sqrt{a}+1>0\) với ∀ \(a\) ≥ \(0\) \(;\) \(a\) ≠ \(4\) \(;\) \(a\) ≠ \(9\)
⇒\(M< 0\) ⇔ \(\sqrt{a}-2< 0\)
⇔ \(\sqrt{a}< 2\)
⇔ \(a< 4\)
Kết hợp với ĐKXĐ ta có\(:\) \(0\) ≤ \(a< 4\)
Vậy \(0\) ≤ \(a< 4\) thì \(M< 0\)