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Câu 3 :
\(ĐKXĐ:x>0\)
\(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)
b) Để P = 3
\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)
\(\Leftrightarrow x-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(tm)
Vậy để \(P=3\Leftrightarrow x=4\)
Câu 1 : Hình như sai đề !! Mik sửa :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=-\frac{1}{\sqrt{x}-2}\)
b) Để A < 2
\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)
\(\Leftrightarrow-1< 2\sqrt{x}-4\)
\(\Leftrightarrow2\sqrt{x}>3\)
\(\Leftrightarrow\sqrt{x}>1,5\)
\(\Leftrightarrow x>2,25\)
Vậy để \(A< 2\Leftrightarrow x>2,25\)
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...
\(P=\left(\frac{3x+3}{x-9}-\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{3-\sqrt{x}}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right).ĐKXĐ:x\ge0,x\ne9\)
\(=\left(\frac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{3x+3-2x+6\sqrt{x}-x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(=\frac{3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3}{\sqrt{x}+3}\)
\(b,x=20-6\sqrt{11}=11-2.3\sqrt{11}+9\)
\(=\left(\sqrt{11}-3\right)^2\)
\(P=\frac{3}{\sqrt{x}+3}=\frac{3}{\sqrt{\left(\sqrt{11}-3\right)^2}+3}=\frac{3}{\sqrt{11}-3+3}=\frac{3\sqrt{11}}{11}\)
\(c,P>\frac{1}{2}\Rightarrow\frac{3}{\sqrt{x}+3}>\frac{1}{2}\)
\(\Leftrightarrow\frac{3}{\sqrt{x}+3}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow\frac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)\(\Leftrightarrow\frac{3-\sqrt{x}}{2\left(\sqrt{x}+3\right)}>0\)
vì \(2\left(\sqrt{x}+3\right)>0\) (nếu x=0 =>pt vô nghiệm)
\(\Rightarrow3-\sqrt{x}>0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\)
Kết hợp ĐKXĐ: \(0< x< 9\)
1. Áp dụng BĐT Bunhiakovski
a) \(\sqrt{x-2}+\sqrt{4-x}=\sqrt{\left(\sqrt{x-2}.1+\sqrt{4-x}.1\right)^2}\le\sqrt{\left(1^2+1^2\right)\left(x-2+4-x\right)}=2\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\sqrt{x-2}=\sqrt{4-x}\) \(\Leftrightarrow\) \(x=3\)
b) \(\sqrt{6-x}+\sqrt{x+2}=\sqrt{\left(\sqrt{6-x}.1+\sqrt{x+2}.1\right)^2}\le\sqrt{\left(1^2+1^2\right)\left(6-x+x+2\right)}=4\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\sqrt{6-x}=\sqrt{x+2}\) \(\Leftrightarrow\) \(x=2\)
c) \(\sqrt{x}+\sqrt{2-x}=\sqrt{\left(\sqrt{x}.1+\sqrt{2-x}.1\right)^2}\le\sqrt{\left(1^2+1^2\right)\left(x+2-x\right)}=2\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\sqrt{x}=\sqrt{2-x}\) \(\Leftrightarrow\) \(x=1\)
1.Điều kiện xđ \(x\ge2,x\le4\)
Từ ĐKXĐ ta có
\(x\ge2\Leftrightarrow x-2\ge0\Leftrightarrow\sqrt{x-2}\ge0\left(1\right)\)
\(x\le4\Leftrightarrow4-x\ge0\Leftrightarrow\sqrt{4-x}\ge0\left(2\right)\)
Từ (1),(2) cộng vế theo vế ta có:
\(\sqrt{x-2}+\sqrt{4-x}\ge0+0=0\)
ĐK \(\hept{\begin{cases}x\ge0\\x\ne4;x\ne9\end{cases}}\)
a. P=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right):\frac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+2+\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
b. Với \(x=4-2\sqrt{3}\Rightarrow P=\frac{\sqrt{4-2\sqrt{3}}+1}{4-2\sqrt{3}-4}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}+1}{-2\sqrt{3}}\)
\(=\frac{\sqrt{3}-1+1}{-2\sqrt{3}}=-\frac{1}{2}\)
c. Để \(\frac{1}{P}\le\frac{-5}{2}\Leftrightarrow\frac{x-4}{\sqrt{x}+1}+\frac{5}{2}\le0\Leftrightarrow\frac{2x-8+5\sqrt{x}+5}{2\left(\sqrt{x}+1\right)}\le0\)
\(\Leftrightarrow\frac{2x+5\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}\le0\Leftrightarrow2x+5\sqrt{x}-3\le0\)vì \(2\left(\sqrt{x}+1\right)\ge0\forall x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\le0\Leftrightarrow2\sqrt{x}-1\le0\Leftrightarrow0\le x\le\frac{1}{4}\left(tm\right)\)
Vậy với \(0\le x\le\frac{1}{4}\)thì \(\frac{1}{P}\le-\frac{5}{2}\)
d. Ta có \(B=P\left(\sqrt{x}-2\right)=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+2}=1-\frac{1}{\sqrt{x}+2}\)
Gỉa sử \(B\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(1\right)\Leftrightarrow\sqrt{x}+2\in\left\{-1;1\right\}\Leftrightarrow x\in\left\{\phi\right\}\)
Vậy B không nhận giá trị nguyên với mọi x để P có nghĩa