Cho biểu thức \(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=2\) với \(x\ge0,x\ne1\).Chứng minh A<1
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WH
21 tháng 4 2020
@Mai.T.Loan câu a pha cuối hơi tắt đó nhìn khó hiểu lắm
còn câu b kl sai r nha
CH
Cô Hoàng Huyền
Admin
VIP
30 tháng 10 2018
a) ĐK: \(x\ge0,x\ne1,x\ne\frac{1}{4}\)
\(A=1+\left(\frac{2x+\sqrt{x}-1}{1-x}-\frac{2x\sqrt{x}-\sqrt{x}+x}{1-x\sqrt{x}}\right)\frac{x-\sqrt{x}}{2\sqrt{x}-1}\)
\(A=1+\left[\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(1-\sqrt{x}\right)}-\frac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}\right]\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(A=1+\left[\frac{2\sqrt{x}-1}{1-\sqrt{x}}-\frac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}\right]\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(A=1-\sqrt{x}+\frac{x\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)
\(A=\frac{x+1}{x+\sqrt{x}+1}\)
Để \(A=\frac{6-\sqrt{6}}{5}\Rightarrow\frac{x+1}{x+\sqrt{x}+1}=\frac{6-\sqrt{6}}{5}\)
\(\Rightarrow5x+5=\left(6-\sqrt{6}\right)x+\left(6-\sqrt{6}\right)\sqrt{x}+6-\sqrt{6}\)
\(\Rightarrow\left(1-\sqrt{6}\right)x+\left(6-\sqrt{6}\right)\sqrt{x}+1-\sqrt{6}=0\)
\(\Rightarrow x-\sqrt{6}.\sqrt{x}+1=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{\sqrt{2}+\sqrt{6}}{2}\\\sqrt{x}=\frac{-\sqrt{2}+\sqrt{6}}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{cases}}\left(tmđk\right)\)
b) Xét \(A-\frac{2}{3}=\frac{x+1}{x+\sqrt{x}+1}-\frac{2}{3}=\frac{3x+3-2x-2\sqrt{x}-2}{3\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{3\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}\)
Do \(x\ge0,x\ne1,x\ne\frac{1}{4}\Rightarrow\left(\sqrt{x}-1\right)^2>0\)
Lại có \(x+\sqrt{x}+1=\left(\sqrt{x}+\frac{1}{2}\right)+\frac{3}{4}>0\)
Nên \(A-\frac{2}{3}>0\Rightarrow A>\frac{2}{3}\).
VH
12 tháng 10 2019
1.
\(A=\frac{1}{2}.2\sqrt{2}-\frac{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{22}.\sqrt{4}}{\sqrt{22}}-\frac{2}{\sqrt{2}}\)
\(=\sqrt{2}-\sqrt{2}+2-\sqrt{2}=2-\sqrt{2}\)
2.
a. \(P=\left[\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right].\left(\sqrt{x}-1\right)\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\sqrt{x}+1\)
b. \(P=2\Leftrightarrow\sqrt{x}+1=2\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\left(\text{KTM ĐKXĐ}\right)\)
\(\text{Vậy không tồn tại giá trị }x\text{ thỏa mãn }P=2\)
9 tháng 3 2020
\(1,A=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
2, Với x>1 ta có \(\frac{1}{A}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}\)
\(=\sqrt{x}-1+\frac{3}{\sqrt{x}-1}+3\)
Áp dụng bđt AM-GM ta có
\(\frac{1}{A}\ge2\sqrt{\left(\sqrt{x}-1\right).\frac{3}{\sqrt{x}-1}}+3=2\sqrt{3}+3\)
Dấu "=" xảy ra khi \(\left(\sqrt{x}-1\right)^2=3\Rightarrow\sqrt{x}=\pm\sqrt{3}+1\)
\(\Rightarrow x=\left(\pm\sqrt{3}+1\right)^2=4\pm2\sqrt{3}\)
NV
30 tháng 6 2016
a/ \(Q=\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right].\frac{2}{\sqrt{x}-1}\)
\(=\frac{x+2-x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(x-2\sqrt{x}+1\right).2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2}{x+\sqrt{x}+1}\)
b/ Ta có: \(x+\sqrt{x}+1=x+2.\frac{1}{2}.\sqrt{x}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow Q=\frac{2}{x+\sqrt{x}+1}>0\).
Vậy Q > 0
17 tháng 10 2019
\(1,P=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\frac{x-\sqrt{x}}{x\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
N
26 tháng 3 2019
\(P=\dfrac{x\sqrt{x}-x-\sqrt{x}-2}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)^2}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)\left(x-1\right)}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)\left(1-x^2\right)}{2\left(x+\sqrt{x}+1\right)}\)
bạn xem lại đề coi ?
đây là đề bài lấy từ đề thi huyện năm 2015-2016 của trường minh nha