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a: Khi x=16 thì \(A=\dfrac{4+1}{4-1}=\dfrac{5}{3}\)
b: \(P=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{x-4}=\dfrac{x+\sqrt{x}-2}{x-4}=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=1+\dfrac{3}{\sqrt{x}-2}\)
Để P lớn nhất thì căn x-2=1
=>căn x=3
=>x=9
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x\ne4\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) Với \(x=3\)( thỏa mãn ĐKXĐ ) ta có \(P=\frac{3\sqrt{3}}{\sqrt{3}+2}=-9+6\sqrt{3}\)
c) A ở đâu ???? '-'
a, Để A nhận giá trị dương thì \(A>0\)hay \(x-1>0\Leftrightarrow x>1\)
b, \(B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}\)
\(=4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=\left(4-9+16\right)\sqrt{5}=11\sqrt{5}\)
( theo công thức \(A\sqrt{B}=\sqrt{A^2B}\))
c, Với \(a\ge0;a\ne1\)
\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)
1: Khi x=36 thì \(A=\dfrac{6}{2\cdot6-4}=\dfrac{6}{12-4}=\dfrac{6}{8}=\dfrac{3}{4}\)
2:
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >4\end{matrix}\right.\)
\(C=B:A\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{3\sqrt{x}-x}{x-4}\right):\dfrac{\sqrt{x}}{2\sqrt{x}-4}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+3\sqrt{x}-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+3\sqrt{x}-x}{\sqrt{x}+2}\cdot\dfrac{2}{\sqrt{x}}=\dfrac{2}{\sqrt{x}+2}\)
3: \(C\cdot\sqrt{x}< \dfrac{4}{3}\)
=>\(\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{3}< 0\)
=>\(\dfrac{2\sqrt{x}\cdot3-4\left(\sqrt{x}+2\right)}{3\left(\sqrt{x}+2\right)}< 0\)
=>\(6\sqrt{x}-4\sqrt{x}-8< 0\)
=>\(2\sqrt{x}-8< 0\)
=>\(\sqrt{x}< 4\)
=>\(0< =x< 16\)
Kết hợp ĐKXĐ của C, ta được: \(\left\{{}\begin{matrix}0< x< 16\\x< >4\end{matrix}\right.\)
a: Sửa đề: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
Khi x=9 thì \(B=\dfrac{\sqrt{9}+1}{\sqrt{9}+2}\)
\(=\dfrac{3+1}{3+2}=\dfrac{4}{5}\)
b: \(A=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{6+\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+6}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-5\sqrt{x}+6+x+2\sqrt{x}-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}+2}\)
c: P=A/B
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+2}:\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
\(P-2=\dfrac{2\sqrt{x}}{\sqrt{x}+1}-2=\dfrac{2\sqrt{x}-2\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\dfrac{-2}{\sqrt{x}+1}< 0\)
=>P<2
ques này nhiều ng` hỏi r` thay ab+bc+ca=1 vào rồi phân tích rút gọn
Do ab + bc + ca = 1 nên ta có :
\(a\sqrt{\frac{\left(b^2+1\right)\left(c^2+1\right)}{a^2+1}}=a\sqrt{\frac{\left(b^2+ab+ac+bc\right)\left(c^2+ab+ac+bc\right)}{a^2+ab+ac+bc}}\)
\(=a\sqrt{\frac{\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}{\left(a+b\right)\left(a+c\right)}}=a\sqrt{\left(b+c\right)^2}=a\left(b+c\right)=ab+ac\text{ }\left(1\right)\)
Tương tự : \(b\sqrt{\frac{\left(a^2+1\right)\left(c^2+1\right)}{b^2+1}}=ab+bc\) (2)và \(c\sqrt{\frac{\left(b^2+1\right)\left(a^2+1\right)}{c^2+1}}=bc+ac\) (3)
Cộng vế với vế của (1) ; (2) ; (3) lại ta được :
\(a\sqrt{\frac{\left(b^2+1\right)\left(c^2+1\right)}{a^2+1}}+b\sqrt{\frac{\left(a^2+1\right)\left(c^2+1\right)}{b^2+1}}+c\sqrt{\frac{\left(b^2+1\right)\left(a^2+1\right)}{c^2+1}}=2\left(ab+bc+ac\right)=2\)
\(\frac{3}{2}=a\sqrt{1-b^2}+b\sqrt{1-c^2}+c\sqrt{1-a^2}\)
\(\le\frac{a^2+1-b^2}{2}+\frac{b^2+1-c^2}{2}+\frac{c^2+1-a^2}{2}=\frac{3}{2}\)
=> \(\frac{3}{2}\le\frac{3}{2}\)( chỉ xảy ra dấu "=" )
Dấu "=" xảy ra <=> \(\hept{\begin{cases}a^2=1-b^2\\b^2=1-c^2\\c^2=1-a^2\end{cases}}\)=> \(a^2+b^2+c^2=3-\left(a^2+b^2+c^2\right)\)
=> \(B=a^2+b^2+c^2=\frac{3}{2}\)