Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, Ta có : \(x=25\Rightarrow\sqrt{x}=\sqrt{25}=5\)
\(\Rightarrow Q=\frac{5-1}{5+1}=\frac{4}{6}=\frac{2}{3}\)
b, \(P=\frac{x\sqrt{x}-1}{x-\sqrt{x}}+\frac{x\sqrt{x}+1}{x+\sqrt{x}}-\frac{4}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{4}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1+x-\sqrt{x}+1-4}{\sqrt{x}}=\frac{2x-2}{\sqrt{x}}\)
c, Ta có : \(P.Q.\sqrt{x}< 8\)hay \(\frac{2x-2}{\sqrt{x}}.\sqrt{x}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)< 8\)
\(\Leftrightarrow\frac{2\left(x-1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< 8\Leftrightarrow2\left(\sqrt{x}-1\right)^2< 8\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2< 4\Leftrightarrow\sqrt{x}-1< 2\Leftrightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)
ĐK ; \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
a, \(Q=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x-8\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-7\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-7}{\sqrt{x}+1}\)
b. \(Q< \frac{1}{2}\Rightarrow\frac{\sqrt{x}-7}{\sqrt{x}+1}-\frac{1}{2}< 0\Rightarrow\frac{\sqrt{x}-15}{2\left(\sqrt{x}+1\right)}< 0\Rightarrow\sqrt{x}-15< 0\)
\(\Rightarrow0\le x< 225\)và \(x\ne4\)
c. \(Q=\frac{\sqrt{x}-7}{\sqrt{x}+1}=1-\frac{8}{\sqrt{x}+1}\)
Ta thấy \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\frac{-8}{\sqrt{x}+1}\ge-8\Rightarrow1-\frac{8}{\sqrt{x}+1}\ge-7\)
\(\Rightarrow Q\ge-7\)
Vậy \(MinQ=-7\). Dấu bằng xảy ra \(\Rightarrow x=0\)
a) Thay x=4 zô là đc . ra kết quả \(\frac{7}{6}\)là dúng
b) \(B=\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\)
\(=\frac{3x+3\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\)
\(=>P=A.B=\frac{3\sqrt{x}+1}{x+\sqrt{x}}.\frac{3\left(x+\sqrt{x}\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}=\frac{3}{3\sqrt{x}-1}\)
c) xét \(\frac{1}{P}=\frac{3\sqrt{x}-1}{3}\)
do \(\sqrt{x}\ge0=>3\sqrt{x}-1\ge-1\)\(=>\frac{3\sqrt{x}-1}{3}\ge-\frac{1}{3}\)
\(=>\frac{1}{P}\ge-\frac{1}{3}\)
dấu = xảy ra khi x=0
zậy ..
1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
\(a,x=7-4\sqrt{3}=4-2.2\sqrt{3}+3\) (Thỏa mãn ĐKXĐ)
\(=\left(2-\sqrt{3}\right)^2\)
\(B=\frac{2}{\sqrt{x}-2}=\frac{2}{\sqrt{\left(2-\sqrt{3}\right)^2}-2}\)
\(=\frac{2}{2-\sqrt{3}-2}=-\frac{2\sqrt{3}}{3}\)
\(b,P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\)
\(=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\frac{2}{\sqrt{x}-2}:\frac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2}{\sqrt{x}-2}:\frac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2}{\sqrt{x}-2}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(P=\frac{4}{3}\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}+1}=\frac{4}{3}\)
\(\Leftrightarrow3\left(\sqrt{x}+2\right)=4\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow3\sqrt{x}+6=4\sqrt{x}+4\)
\(\Leftrightarrow6-4=4\sqrt{x}-3\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)(ko thỏa mãn ĐKXĐ)
=>pt vo nghiệm
d,\(\left(\sqrt{x}+1\right)P-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\frac{\sqrt{x}+2}{\sqrt{x}+1}-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)
\(\Leftrightarrow\sqrt{x}+2-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)
\(\Leftrightarrow-4\sqrt{x-1}+28=-6x+10\sqrt{5x}\)
\(\Leftrightarrow x=5\)
bài này lp 8 cx làm dc , CTV mà ngu lonee :)
nhờ vào năng lực rinegan của chúa pain , ta có thể dễ dàng nhìn ra ......
\(1-x=\left(\sqrt{x}+1\right)\left(1-\sqrt{x}\right).\) dkxd , x dương và x khác 1
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}\left(1-\sqrt{x}\right)-\sqrt{x}+4}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)\)
\(P=\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\left(\frac{\sqrt{x}-x-\sqrt{x}+4}{1-x}\right)\)
\(p=\frac{\left(\sqrt{x}-2\right)}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(1-\sqrt{x}\right)}{-\left(x-4\right)}\)
\(P=\frac{\left(\sqrt{x}-2\right)}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(1-\sqrt{x}\right)}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{1-\sqrt{x}}{-\sqrt{x}-2}\)
B) dkxd có x luôn dương
vậy ta suy ra \(-\left(\sqrt{x}+2\right)< 0\) " âm"
vậy để \(\frac{1-\sqrt{x}}{-\left(\sqrt{x}+2\right)}< 0\)
thì \(1-\sqrt{x}>0\) " vì số dương chia cho số âm luôn bé hơn 0 "
\(-\sqrt{x}>-1\Leftrightarrow\sqrt{x}< 1\)
để p dương thì ................ 0<x<1
c)
\(\frac{1-\sqrt{x}}{-\sqrt{x}+2}=\frac{2-\sqrt{x}+1}{-\sqrt{x}+2}=1+\frac{1}{-\sqrt{x}+2}\)
vì x dương " dkxd "
suy ra \(\orbr{\begin{cases}\sqrt{x}+2\ge2\\-\sqrt{x}+2\le2\end{cases}}\)
vì " năm ở mẫu "
\(\frac{1}{-\sqrt{x}+2}\ge\frac{1}{2}\)
\(1+\frac{1}{-\sqrt{x}+2}\ge1+\frac{1}{2}=\frac{3}{2}\)
dấu = xảy ra khi x = 0
d!t , sửa lại câu C , thành 2-1 , ko phải 2 +1 :)