\(P=\left(\frac{x+2}{x\sqrt{x+1}}-\frac{1}{\sqrt{x}+1}\right)\frac{4\sqrt{x}}{3}\)
với \(x\ge0\)
a) Rút gọn
b) Tìm x để \(P=\frac{8}{9}\)
c) Tìm GTLN, GTNN của P
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a, \(P=\left(\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\cdot\frac{4\sqrt{x}}{3}=\frac{1+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
1/ \(C=\frac{x+9}{10\sqrt{x}}=\frac{\sqrt{x}}{10}+\frac{9}{10\sqrt{x}}\ge2.\frac{3}{10}=0,6\)
Đạt được khi x = 9
2/ \(E=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=x-3\sqrt{x}+2\)
\(=\left(x-\frac{2.\sqrt{x}.3}{2}+\frac{9}{4}\right)-\frac{1}{4}\)
\(=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Vậy GTNN là \(-\frac{1}{4}\)đạt được khi \(x=\frac{9}{4}\)
Không có GTLN nhé
a/ \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{x-4}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)
=> \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\right)\)
=> \(B=\frac{\sqrt{x}+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{1}{\sqrt{x}-2}\)
=> \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}\)
b/ B>2 <=> \(\frac{\sqrt{x}+5}{\sqrt{x}+2}>2\) <=> \(\sqrt{x}+5>2\sqrt{x}+4\)
<=> \(1>\sqrt{x}\)=> \(-1\le x\le1\)
c/ \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}=\frac{\sqrt{x}+2+3}{\sqrt{x}+2}=1+\frac{3}{\sqrt{x}+2}\)
Để Bmax thì \(\sqrt{x}+2\) đạt giá trị nhỏ nhất . Do \(\sqrt{x}+2\ge2\)=> Đạt nhỏ nhất khi x=0
Khí đó giá trị lớn nhất của B là: \(1+\frac{3}{2}=\frac{5}{2}\)Đạt được khi x=0
mk làm luôn
a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right).\)
=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)
=\(\frac{\left(3x+3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right).3}\)
=\(\frac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)
=
a/ \(A=\frac{\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)-\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{8\sqrt{x}}{9x-1}}{1-\frac{3\sqrt{x}+1-3}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{3x-4\sqrt{x}+1-3\sqrt{x}-1}{\left(3\sqrt{x}\right)^2-1}-\frac{8\sqrt{x}}{9x-1}}{1-1-\frac{3}{3\sqrt{x}+1}}\)
\(A=\frac{\frac{3x-7\sqrt{x}}{9x-1}-\frac{8\sqrt{x}}{9x-1}}{-\frac{3}{3\sqrt{x}+1}}\)
\(A=\frac{3x-7\sqrt{x}-8\sqrt{x}}{9x-1}\left(\frac{-\left(3\sqrt{x}+1\right)}{3}\right)\)
\(A=\frac{3x-15\sqrt{x}}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)
\(A=\frac{3\left(x-3\sqrt{x}\right)}{9x-1}\left(\frac{-3\sqrt{x}-1}{3}\right)\)
\(A=\frac{\left(x-3\sqrt{x}\right)\left(-3\sqrt{x}-1\right)}{9x-1}\)
\(A=\frac{3x\sqrt{x}+8x+3\sqrt{x}}{9x-1}\)
\(A=\frac{3x\sqrt{x}}{9x-1}+\frac{8x}{9x-1}+\frac{3\sqrt{x}}{9x-1}\)
\(A=\frac{x\sqrt{x}}{x-\frac{1}{3}}+\frac{8x}{9x-1}+\frac{\sqrt{x}}{x-\frac{1}{3}}\)
\(A=\frac{\sqrt{x}\left(x-1\right)}{x-\frac{1}{3}}+\frac{\frac{8}{3}x}{x-\frac{1}{3}}\)
\(A=\frac{\sqrt{x}\left(x-1\right)+\frac{8}{3}x}{x-\frac{1}{3}}\)