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Bài 1 :
a) \(P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(P=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(P=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}+1}{x}\)
b) \(P>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-2x}{x}>0\)
\(\Leftrightarrow\sqrt{x}-2x+1>0\left(x>0\right)\)
\(\Leftrightarrow\sqrt{x}+x^2-2x+1-x^2>0\)
\(\Leftrightarrow\sqrt{x}+x^2+\left(x-1\right)^2>0\left(\forall x>0\right)\)
Vậy P > 1/2 với mọi x> 0 ; x khác 1
Bài 2 :
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+a}+\frac{2}{a-1}\right)\)
\(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{2}{a-1}\right)\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1+2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)\left(\sqrt{a}+1\right)}\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1+2a+2\sqrt{a}}\)
\(K=\frac{\left(a-1\right)^2}{3a+2\sqrt{a}-1}\)
b) \(a=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)( thỏa mãn ĐKXĐ )
Thay a vào biểu thức K , ta có :
\(K=\frac{\left(3+2\sqrt{2}-1\right)^2}{3\left(3+2\sqrt{2}\right)+2\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{9+6\sqrt{2}+2\left|\sqrt{2}+1\right|-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{8+6\sqrt{2}+2\sqrt{2}+2}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{10+8\sqrt{2}}\)
a)ĐKXĐ : tự làm nha
\(A=\left(\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+1}\right)\times\left(1-\frac{1}{\sqrt{x}}\right)\)
\(A=\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\times\left(1-\frac{1}{\sqrt{x}}\right)\)
\(A=\left(\frac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\times\left(1-\frac{1}{\sqrt{x}}\right)\)
\(A=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\times\left(1-\frac{1}{\sqrt{x}}\right)\)
\(A=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\times\left(\frac{\sqrt{x}-1}{\sqrt{x}}\right)\)
\(A=\frac{2}{\sqrt{x}+1}\)(1)
b) Thay \(x=3-2\sqrt{2}\)vào (1) , ta có:
\(A=\frac{2}{\sqrt{3-2\sqrt{2}}+1}=\frac{2}{\sqrt{2}-1+1}=\sqrt{2}\)
c) Ta có: \(x.A=\frac{8}{3}\Leftrightarrow x.\left(\frac{2}{\sqrt{x}+1}\right)=\frac{8}{3}\)
\(\Leftrightarrow\frac{2x}{\sqrt{x}+1}=\frac{8}{3}\Rightarrow6x=8\sqrt{x}+8\)
Đến đây bn tự giải x ra nhé .
P/s : mình sửa đề dấu chia thành dấu nhân nha
b, A = \(2-\sqrt{2}\) bn xem lại
c, mục đích của mik là tìm x , thế nên mik mới hỏi
\(a,ĐKXĐ:\hept{\begin{cases}a\ge0,\sqrt{a}\ne0\\\sqrt{a}-1\ne0\\\sqrt{a}-2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}a>0\\a\ne1\\a\ne4\end{cases}}}\)
\(b,\)Rút gọn : \(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(Q=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\)
\(Q=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a^2-1-a^2+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(Q=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\)
\(Q=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
c, bn thay vào rồi tính nha
a/ Với x = \(23-12\sqrt{3}\) ta có:
\(x-11=23-12\sqrt{3}-11=12-12\sqrt{3}=12\left(1-\sqrt{3}\right)\)
\(\sqrt{x-2}-3=\sqrt{23-12\sqrt{3}-2}-3=\sqrt{21-12\sqrt{3}}-3=\sqrt{3^2-2.3.2\sqrt{3}+\left(2\sqrt{3}\right)^2}-3=\sqrt{\left(3-2\sqrt{3}\right)^2}-3=2\sqrt{3}-6\) \(=2\sqrt{3}\left(1-\sqrt{3}\right)\)
=>\(\frac{x-11}{\sqrt{x-2}-3}=\frac{12\left(1-\sqrt{3}\right)}{2\sqrt{3}\left(1-\sqrt{3}\right)}=\frac{12}{2\sqrt{3}}=\frac{2\sqrt{3}.2\sqrt{3}}{2\sqrt{3}}=2\sqrt{3}\)
b/ \(\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}=\frac{1-\sqrt{a}}{2\left(1-a\right)}+\frac{1+\sqrt{a}}{2\left(1-a\right)}-\frac{a^2+2}{\left(1-a\right)\left(1-a+a^2\right)}\)
=\(\frac{2}{2\left(1-a\right)}-\frac{a^2+2}{\left(1-a\right)\left(1-a+a^2\right)}=\frac{1-a+a^2-a^2-2}{\left(1-a\right)\left(1-a+a^2\right)}=\frac{-a-1}{1-a^3}\)
Thay : \(a=\sqrt{2}tacó:\)
\(\frac{-\sqrt{2}-1}{1-\sqrt{2}^3}=\frac{-\left(1+\sqrt{2}\right)}{1-2\sqrt{2}}\)
a) \(ĐKXĐ:-1< a< 1\)
\(B=\left(\frac{1}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{3}{\sqrt{1-a^2}}+1\right)\)
\(=\left(\frac{3}{\sqrt{1+a}}+\frac{\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right):\left[\frac{3}{\sqrt{\left(1-a\right)\left(1+a\right)}}+\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{\left(1-a\right)\left(1+a\right)}}\right]\)
\(=\left[\frac{3}{\sqrt{1+a}}+\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\right]:\frac{3+\sqrt{\left(1+a\right)\left(1-a\right)}}{\sqrt{\left(1+a\right)\left(1-a\right)}}\)
\(=\frac{3+\sqrt{\left(1+a\right)\left(1-a\right)}}{\sqrt{1+a}}.\frac{\sqrt{\left(1+a\right)\left(1-a\right)}}{3+\sqrt{\left(1+a\right)\left(\sqrt{1-a}\right)}}\)
\(=\sqrt{1-a}\)
b) \(a=\frac{\sqrt{3}}{2+\sqrt{3}}\)\(\Rightarrow1-a=1-\frac{\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}-\sqrt{3}}{2+\sqrt{3}}=\frac{2}{2+\sqrt{3}}\)
\(=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{4-2\sqrt{3}}{4-3}=4-2\sqrt{3}\)
\(=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
Thay \(1-a=\left(\sqrt{3}-1\right)^2\)vào biểu thức ta được:
\(B=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
a/ Điều kiện xác định tự tìm nhé.
\(\sqrt{a}=\sqrt{2\sqrt{2}+3}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
Vậy \(A=\frac{2\sqrt{2}+3-1}{\sqrt{2}+1}=\frac{2\sqrt{2}+2}{\sqrt{2}+1}=\frac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\)
b/ \(\frac{a-1}{\sqrt{a}}=a-2\Leftrightarrow a-1=a\sqrt{a}-2\sqrt{a}\)
Đặt \(t=\sqrt{a},t>0\) thì : \(t^2-1=t^3-2t\Leftrightarrow t^3-t^2-2t+1=0\)
Giải pt trên để tìm a.
đk: \(a\ge0;a\ne1\)
Ta có:
\(B=\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}\)
\(B=\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(B=\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)+\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)-2\left(a^2+2\right)\left(1+\sqrt{a}\right)}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(B=\frac{2a+2\sqrt{a}+2-2a^2\sqrt{a}-2a^2-4-4\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(B=\frac{-2a^2\sqrt{a}-2a^2+2a-2\sqrt{a}-2}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(B=\frac{-a^2\sqrt{a}-a^2+a-\sqrt{a}-1}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
Tại \(a=\sqrt{2}\) thì giá trị của B là:
\(B=\frac{-\left(\sqrt{2}\right)^2.\left(\sqrt{\sqrt{2}}\right)-\left(\sqrt{2}\right)^2+\sqrt{2}-\sqrt{\sqrt{2}}-1}{\left(1+\sqrt{\sqrt{2}}\right)\left(1-\sqrt{\sqrt{2}}\right)\left(\sqrt{2}+\sqrt{\sqrt{2}}+1\right)}\)
\(B\approx3,45267\)
\(ĐKXĐ:x>1\)
\(B=\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}\)
\(=\frac{1-\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}+\frac{1+\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}+\frac{a^2+2}{a^3-1}\)
\(=\frac{\left(1-\sqrt{a}\right)+\left(1+\sqrt{a}\right)}{2\left(1-a\right)}+\frac{a^2+2}{a^3-1}\)
\(=\frac{2}{2\left(1-a\right)}+\frac{a^2+2}{a^3-1}=\frac{1}{1-a}+\frac{a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-a^2-a-1+a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{-a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{-1}{a^2+a+1}\)
Với \(a=\sqrt{2}\)( thỏa mãn ĐKXĐ ), ta có:
\(B=\frac{-1}{\left(\sqrt{2}\right)^2+\sqrt{2}+1}=\frac{-1}{2+\sqrt{2}+1}=\frac{-1}{3+\sqrt{2}}\)