Cho P = \(\dfrac{4\sqrt{x}}{\sqrt{x}+2}\left(ĐKXĐ:x\ge0;x\ne4\right)\). Tìm giá trị lớn nhất của P khi \(x\in N;x< 101\)
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Rút gọn B
B= \(\dfrac{x-2\sqrt{x}+4}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}-2}\left(đkxđ:x>0;x\ne4\right)\)
\(B=\dfrac{x-2\sqrt{x}+4}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}-2}\) (với \(x>0,x\ne4\))
\(B=\dfrac{x-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}-2}\)
\(B=\dfrac{x-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}}{\sqrt{x}\left(x-2\right)}\)
\(B=\dfrac{x-2\sqrt{x}+4-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
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\(B=\frac{2\left(x+4\right)}{x-3\sqrt{x}-4}+\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{8}{\sqrt{x}-4}\) \(ĐKXĐ:x\ge0;x\ne16\)
a) rút gọn B
\(B=\frac{2\left(x+4\right)}{x-3\sqrt{x}-4}+\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{8}{\sqrt{x}-4}\)
\(B=\frac{2\left(x+4\right)+\sqrt{x}\left(\sqrt{x}-4\right)-8\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(B=\frac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(B=\frac{3x-12\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(B=\frac{3\sqrt{x}}{\sqrt{x}+1}\)
vậy \(B=\frac{3\sqrt{x}}{\sqrt{x}+1}\)
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Cho biểu thức: \(A=\left(\frac{x+5\sqrt{x}-27}{x-16}+\frac{3-\sqrt{x}}{\sqrt{x}-4}\right):\frac{1}{\sqrt{x}+4}\)và \(B=\sqrt{x}-4\left(ĐKXĐ:x\ge0;x\ne16\right)\)
a) Rút gọn A
b) Tìm x sao cho B = -2A
a, Với x >= 0 ; x khác 16
\(A=\left(\frac{x+5\sqrt{x}-27+\left(3-\sqrt{x}\right)\left(\sqrt{x}+4\right)}{x-16}\right):\frac{1}{\sqrt{x}+4}\)
\(=\left(\frac{x+5\sqrt{x}-27+3\sqrt{x}+12-x-4\sqrt{x}}{x-16}\right):\frac{1}{\sqrt{x}+4}\)
\(=\left(\frac{4\sqrt{x}-15}{x-16}\right):\frac{1}{\sqrt{x}+4}=\frac{4\sqrt{x}-15}{\sqrt{x}-4}\)
b, Ta có \(B=-2A\Rightarrow\sqrt{x}-4=-\frac{8\sqrt{x}-30}{\sqrt{x}-4}\)
\(\Leftrightarrow x-8\sqrt{x}+16=-8\sqrt{x}+30\Leftrightarrow x-14=0\Leftrightarrow x=14\left(tm\right)\)
Bài 1: Cho \(A=\left(\dfrac{x-4}{\sqrt{x}-2}+\dfrac{x\sqrt{x}-8}{4-x}\right):\left[\dfrac{\left(\sqrt{x}-2\right)^2+2\sqrt{x}}{\sqrt{x}+2}\right]\)với \(x\ge0\); \(x\ne4\)
a, Rút gọn A
b, CMR: \(A< 1\) với \(x\ge0\); \(x\ne4\)
c, Tìm x để A nguyên
a: \(A=\left(\dfrac{\left(x-4\right)\left(\sqrt{x}+2\right)-x\sqrt{x}+8}{x-4}\right):\dfrac{x-2\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8-x\sqrt{x}+8}{x-4}\cdot\dfrac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)
\(=\dfrac{2x-4\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{1}{x-2\sqrt{x}+4}=\dfrac{2\sqrt{x}}{x-2\sqrt{x}+4}\)
b: \(A-1=\dfrac{2\sqrt{x}-x+2\sqrt{x}-4}{x-2\sqrt{x}+4}\)
\(=\dfrac{-x+4\sqrt{x}-4}{x-2\sqrt{x}+4}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-1\right)^2+3}< 0\)
=>A<1
c: \(2\sqrt{x}>=0;x-2\sqrt{x}+4=\left(\sqrt{x}-1\right)^2+3>0\)
=>A>=0 với mọi x thỏa mãn ĐKXĐ
mà A<1
nên 0<=A<1
=>Để A nguyên thì A=0
=>x=0
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Cho biểu thức \(Q=\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right):\left(1-\frac{2\sqrt{x}}{x+1}\right)ĐKXĐ:x\ge0;x\ne1\)
a, Rút gọn Q
b, Tìm x sao cho Q < 0.
\(P=\left(\frac{3x-3\sqrt{x}+3}{x-\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{x-1}\) \(ĐKXĐ:x\ge0;x\ne1\)
a) rút gọn
b) Tính P khi \(x=4-2\sqrt{3}\)
giúp mình giải bpt vs
\(\dfrac{\left|2x-1\right|-x}{2x}>1;\dfrac{2-\left|x-2\right|}{x^2-1}\ge0;\dfrac{\sqrt{x+4}-2}{4-9x^2}\le0;\dfrac{x^2-2x-3}{\sqrt[3]{3x-1}+\sqrt[3]{4-5x}}\ge0;\)\(3x^2-10x+3\ge0;\left(\sqrt{2}-x\right)\left(x^2-2\right)\left(2x-4\right)< 0;\dfrac{1}{x+9}-\dfrac{1}{x}>\dfrac{1}{2};\dfrac{2}{1-2x}\le\dfrac{3}{x+1}\)
Cho biểu thức: \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\) (với \(x\ge0;x\ne4\)). Tìm x để: \(\left|P-2\right|>P-2\)
\(P=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{x-4}\)
\(=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Để |P-2|>P-2 thì P-2>0
\(\Leftrightarrow3\sqrt{x}-2\sqrt{x}-4>0\)
hay x>16
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\(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
\(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
\(\left|P-2\right|>P-2\)
\(\Leftrightarrow2-P>P-2\) ;\(P< 2\) ( vì \(P-2>P-2\left(vô.lý\right)\) )
\(\Leftrightarrow4>2P\)
\(\Leftrightarrow P< 2\)
\(\rightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}< 2\)
\(\Leftrightarrow3\sqrt{x}< 2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}< 4\)
\(\Leftrightarrow x< 16\) ( t/m )
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24 tháng 10 2023 lúc 20:13
Ta có: A = \(\dfrac{4\sqrt{x}}{\sqrt{x}-2}\) và B = \(\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\) với \(x\ge0;x\ne4\)
Cho \(M=\dfrac{A}{B}\). So sánh \(M\) và \(\sqrt{M}\)
