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@Mai.T.Loan câu a pha cuối hơi tắt đó nhìn khó hiểu lắm
còn câu b kl sai r nha
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. \(P=2\Leftrightarrow\sqrt{x}+1=2\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\left(\text{KTM ĐKXĐ}\right)\)
\(\text{Vậy không tồn tại giá trị }x\text{ thỏa mãn }P=2\)
\(1,A=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
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 \(\left(\sqrt{x}-1\right)^2=3\Rightarrow\sqrt{x}=\pm\sqrt{3}+1\)
\(\Rightarrow x=\left(\pm\sqrt{3}+1\right)^2=4\pm2\sqrt{3}\)
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
\(1,P=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\frac{x-\sqrt{x}}{x\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(P=\dfrac{x\sqrt{x}-x-\sqrt{x}-2}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)^2}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)\left(x-1\right)}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)\left(1-x^2\right)}{2\left(x+\sqrt{x}+1\right)}\)
Câu 1 bạn ghi lại đề, tử số cuối cùng bị thiếu
2/ \(a^2-4a+1=0\Rightarrow a^2+1=4a\Rightarrow a+\frac{1}{a}=4\)
\(\Rightarrow a^2+\frac{1}{a^2}+2=16\Rightarrow a^2+\frac{1}{a^2}=14\)
\(P=\frac{a^4+a^2+1}{a^2}=a^2+\frac{1}{a^2}+1=14+1=15\)
bạn xem lại đề coi ?
đây là đề bài lấy từ đề thi huyện năm 2015-2016 của trường minh nha