cho \(P=(\frac{3}{1-x}+\frac{1}{\sqrt{x}+1}):\frac{1}{\sqrt{x}+1}\)
tìm giá trị nhỏ nhất của biểu thức \(M=\frac{x+12}{\sqrt{x}-1}\times\frac{1}{P}\)
K
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3 tháng 8 2019
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\)
\(=\frac{3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{1}{\sqrt{x}+1}\)\(+\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+1=\frac{3}{\sqrt{x}-1}+1\)
\(=\frac{\sqrt{x}-1+3}{\sqrt{x}-1}=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{\sqrt{x}-1+3}{\sqrt{x}-1}=1+\frac{3}{\sqrt{x}-1}\)
\(P\in Z\Leftrightarrow1+\frac{3}{\sqrt{x}-1}\in Z\Rightarrow\frac{3}{\sqrt{x}-1}\in Z\)
\(\Rightarrow\sqrt{x}-1\inƯ_3\)
Mà \(Ư_3=\left\{\pm1;\pm3\right\}\)
\(Th1:\sqrt{x}-1=1\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(Th2:\sqrt{x}-1=-1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
\(Th3:\sqrt{x}-1=3\Rightarrow\sqrt{x}=4\Rightarrow x=16\)
\(Th4:\sqrt{x}-1=-3\Rightarrow\sqrt{x}=-2\Rightarrow x\in\varnothing\)
\(\Rightarrow x\in\left\{0;4;16\right\}\)
\(M=\frac{x+12}{\sqrt{x}-1}.\left(1\div\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{x+12}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{x+12}{\sqrt{x}+2}\)
\(=\frac{x-4+16}{\sqrt{x}+2}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\frac{16}{\sqrt{x}+2}\)
\(=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}=\sqrt{x}+2+\frac{16}{\sqrt{x}+2}-4\)
Áp dụng Bất đẳng thức Cô - Si cho hai số nguyên dương \(\sqrt{x}+2;\frac{16}{\sqrt{x}+2}\)ta có :
\(\sqrt{x}+2+\frac{16}{\sqrt{x}+2}\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{16}{\sqrt{x}+2}}\)
\(\Rightarrow\sqrt{x}+2+\frac{16}{\sqrt{x}+2}\ge2.\sqrt{16}=2.4=8\)
\(\Rightarrow\sqrt{x}+2+\frac{16}{\sqrt{x}+2}-4\ge4\)
\(\Rightarrow M_{min}=4\Leftrightarrow\sqrt{x}+2=\frac{16}{\sqrt{x}+2}\)
\(\Rightarrow\left(\sqrt{x}+2\right)^2=16\)
\(\Rightarrow\sqrt{x}+2=4\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(KL:M_{min}=4\Leftrightarrow x=4\)
LM
10 tháng 9 2017
Kết quả rút gọn: \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(M=\frac{x+12}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{x+12}{\sqrt{x}+2}\)
\(M=\frac{x-4+16}{\sqrt{x}+2}=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}=\left(\sqrt{x}+2+\frac{16}{\sqrt{x}+2}\right)-4\)
Âp dụng BĐT AM-GM cho 2 số không âm ta có:
\(M\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{16}{\sqrt{x}+2}}-4=2.4-4=4\)
Vậy min M =4. Dấu bằng xảy ra \(\Leftrightarrow\left(\sqrt{x}+2\right)^2=16\Leftrightarrow\sqrt{x}+2=4\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
26 tháng 10 2017
\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\) \(ĐKXĐ:x\ne1\)
\(P=\left(\frac{3}{x-1}+\frac{\sqrt{x}-1}{x-1}\right):\frac{1}{\sqrt{x}+1}\)
\(P=\frac{\sqrt{x}+2}{x-1}.\left(\sqrt{x}+1\right)\)
\(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
b) theo câu a) \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}\) với \(ĐKXĐ:x\ne1\)
theo bài ra \(P=\frac{5}{4}\)thì \(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\)
\(\Leftrightarrow\left(\sqrt{x}+2\right).4=\left(\sqrt{x}-1\right).5\)
\(\Leftrightarrow4\sqrt{x}+8=5\sqrt{x}-5\)
\(\Leftrightarrow-\sqrt{x}+13=0\)
\(\Leftrightarrow-\sqrt{x}=-13\)
\(\Leftrightarrow\sqrt{x}=13\)
\(\Leftrightarrow x=169\)
vậy \(x=169\)khi \(P=\frac{5}{4}\)
23 tháng 3 2020
a)\(B=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\)
\(B=\left(\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{1}{\sqrt{x}+1}\right)\times\left(\sqrt{x}+1\right)\)
\(B=\frac{3+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\left(\sqrt{x}+1\right)\)
\(B=\frac{2+\sqrt{x}}{\sqrt{x}-1}\)
3 tháng 2 2022
a: \(P=\dfrac{x+\sqrt{x}+1+11\sqrt{x}-11+34}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+12\sqrt{x}+24}{\sqrt{x}+2}\)
b: Thay \(x=3-2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{3-2\sqrt{2}+12\left(\sqrt{2}-1\right)+24}{\sqrt{2}-1+2}\)
\(=\dfrac{27-2\sqrt{2}+12\sqrt{2}-12}{\sqrt{2}+1}=5+5\sqrt{2}\)
7 tháng 7 2017
a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
b. M =\(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2-5\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{1-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-1}{\sqrt{x}+1}\)
c. \(M=\frac{-1}{\sqrt{x}+1}\ge-1\)
Vậy Min M =-1 khi x=0