t =\(\frac{6}{\sqrt{\left(1+x^2\right)}}\)
tính t theo x
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TH
3 tháng 8 2019
a) ĐKXĐ \(\sqrt{x}\ne2\Leftrightarrow x\ne4\)
b) \(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{1}{\sqrt{x}-2}\)
\(A=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
c) x = 6 + \(4\sqrt{2}\) = \(\left(2+\sqrt{2}\right)^2\)
=> A = \(\frac{\sqrt{\left(2+\sqrt{2}\right)^2}-4}{\sqrt{\left(2+\sqrt{2}\right)^2}-2}=\frac{\sqrt{2}-2}{\sqrt{2}}\)
TH
3 tháng 8 2019
2.
a) đkxđ: \(x\ne4;x\ne9\)
A=\(\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
=\(\left(\frac{x-\sqrt{x}-6+x\sqrt{x}-9\sqrt{x}-2x+18-x\sqrt{x}+2x-4\sqrt{x}-8}{\left(x-4\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(\frac{x-14\sqrt{x}-5}{\left(x-4\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
b) A = -2/5
(k biết là do đề sai hay mình sai chứ đến đây nản quá! bạn làm nốt nhé!)
16 tháng 12 2019
a/ Rút gọn D
D = \(\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\)\(\left(\frac{\chi+\sqrt{x}}{\sqrt{x}+1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
= \(\left(\frac{x}{2\sqrt{x}}-\frac{1}{2\sqrt{x}}\right)\)\(\left[\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}\right)}{\left(\sqrt[]{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
= \(\frac{x-1}{2\sqrt{x}}\)\(\left[\frac{x\sqrt{x}+x-x-\sqrt{x}-\left(x\sqrt{x}+x+x+\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
= \(\frac{x-1}{2\sqrt{x}}\)\(\left[\frac{x\sqrt{x}+x-x-\sqrt{x}-x\sqrt{x}-x-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
=\(\frac{x-1}{2\sqrt{x}}\) \(\frac{-2x-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{x-1}{2\sqrt{x}}\) \(\frac{-2\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
= \(-\sqrt{x}-1\)
b/Tìm x để D > -6
Ta có D> -6
hay \(-\sqrt{x}-1\)> -6
⇔ \(-\sqrt{x}\)> -5
⇔ \(\sqrt{x}\) < 5
⇔ \(x\) <25
Chúc bạn học tốt ;>
8 tháng 2 2017
\(\sqrt{x}=y\\ \)
ĐK: \(x\ne0,1,4\Leftrightarrow\left\{\begin{matrix}y>0\\y\ne1\&4\end{matrix}\right.\) ko sửa được y khác 1 &2
\(P=\left(\frac{\left(1-y\right)}{\left(y-2\right)}+\frac{y}{\left(y-1\right)}+\frac{y+2}{\left(y-1\right)\left(y-2\right)}\right):\left(\frac{2}{y-2}-\frac{y-1}{y\left(y-2\right)}\right)\)
\(P=\left(\frac{2y-y^2-1}{\left(y-2\right)\left(y-1\right)}+\frac{y^2-2y}{\left(y-1\right)\left(y-2\right)}+\frac{y+2}{\left(y-1\right)\left(y-2\right)}\right):\left(\frac{2y-y+1}{y\left(y-2\right)}\right)\)
\(P=\left(\frac{y+1}{\left(y-1\right)\left(y-2\right)}\right).\left(\frac{y\left(y-2\right)}{\left(y+1\right)}\right)=\frac{y}{y-1}\)
a) \(P=\frac{\sqrt{x}}{\sqrt{x}-1}\)
b)\(x=6-2\sqrt{5}=5-2\sqrt{5}+1=\left(\sqrt{5}-1\right)^2\)
\(p=\frac{\left(\sqrt{5}-1\right)}{\sqrt{5}-2}=\left(\sqrt{5}-1\right)\left(\sqrt{5}+2\right)=3-\sqrt{5}\)
C)\(\frac{P}{\sqrt{x}}=\frac{1}{\sqrt{x}-1}\ge-1\) tuy nhiên đk: x khác 0=> dấu đẳng thức không xẩy ra (xem lại đề)
9 tháng 2 2017
Xem lại 1/(căn(x)-1) có cực trị duy nhất khi x=0 tuy nhiên nó cũng không phải GTLN : rất có thể rút gọn P bị sai nếu không đề sai.
S
15 tháng 7 2019
\(ĐK:x\ne4;x\ne9;x\ge0\)
\(D=\left(\frac{x-2\sqrt{x}}{x-4}-1\right):\left(\frac{4-x}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)=\left(\frac{\sqrt{x}}{\sqrt{x}+2}-1\right):\left(\frac{4-x}{x-\sqrt{x}-6}+\frac{\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)=\frac{-2}{\sqrt{x}+2}:\left(\frac{4-x}{x-\sqrt{x}-6}+\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)=\frac{-2}{\sqrt{x}+2}:\frac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\frac{-2}{\sqrt{x}+2}:\frac{\sqrt{x}-3}{\sqrt{x}+2}=\frac{-2}{\sqrt{x}-3}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 5 2019
ĐKXĐ: \(x>0;x\ne4;9\)
\(P=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{x+2\sqrt{x}+1-\sqrt{x}-1}{x+2\sqrt{x}+1}\right)\)
\(=\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2}\right)\)
\(=\left(\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right).\left(\frac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\left(\frac{\sqrt{x}+1}{\sqrt{x}}\right)=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P< 0\Rightarrow\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}< 0\Rightarrow\sqrt{x}-2< 0\Rightarrow x< 4\)
Vậy để \(P< 0\Rightarrow0< x< 4\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2019
c/
\(\left(x-4\right)P+y^2+2xy+1+\left|2x+3y+1\right|=0\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x^2-1\right)}{x-4}+y^2+2xy+1+\left|2x+3y+1\right|=0\)
\(\Leftrightarrow x^2+y^2+2xy+\left|2x+3y+1\right|=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left|2x+3y+1\right|=0\)
Do \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left|2x+3y+1\right|\ge0\end{matrix}\right.\) \(\forall x;y\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\2x+3y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2019
ĐKXĐ: \(x\ge0;x\ne4\)
\(P=\left(\frac{\sqrt{x}+2}{\sqrt{x}+3}+\frac{x^2-x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\right)\)
\(P=\left(\frac{x-4+x^2-x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{x+3\sqrt{x}+\sqrt{x}+4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\right)\)
\(P=\left(\frac{x^2-1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2}\right)\)
\(P=\frac{x^2-1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}.\left(\frac{\sqrt{x}+3}{\sqrt{x}+2}\right)\)
\(P=\frac{x^2-1}{x-4}\)
b/ Để \(P\ge0\Leftrightarrow\frac{x^2-1}{x-4}\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-1\ge0\\x-4>0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-1\le0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>4\\-1\le x\le1\end{matrix}\right.\)
Kết hợp với ĐKXĐ \(x\ge0\), \(\Leftrightarrow\left[{}\begin{matrix}x>4\\0\le x\le1\end{matrix}\right.\)