Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 ..
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 \(\Rightarrow x=0\)
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...
Khi \(x=1,44\): \(A=\frac{1,44+7}{\sqrt{1,44}}=\frac{8,44}{1,2}=\frac{211}{30}\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-1}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}\)(ĐK: \(x\ge0,x\ne9\))
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2x-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+5\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(S=\frac{1}{B}+A=\frac{\sqrt{x}-3}{\sqrt{x}}+\frac{x+7}{\sqrt{x}}=\frac{x+\sqrt{x}+4}{\sqrt{x}}=\sqrt{x}+\frac{4}{\sqrt{x}}+1\)
\(\ge2\sqrt{\sqrt{x}.\frac{4}{\sqrt{x}}}+1=5\)
Dấu \(=\)khi \(\sqrt{x}=\frac{4}{\sqrt{x}}\Leftrightarrow x=4\)(thỏa mãn)
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)
\(A=\sqrt{x}+1-\frac{17}{1-\sqrt{x}}\)
\(\Leftrightarrow A=\frac{x-1+17}{\sqrt{x}-1}\)
\(\Leftrightarrow A=\frac{x+16}{\sqrt{x}-1}\)
\(B=\frac{x-7}{x-4\sqrt{x}+3}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}-3}\)
\(\Leftrightarrow B=\frac{x-7+\sqrt{x}-3-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow B=\frac{x-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow B=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow B=\frac{\sqrt{x}+3}{\sqrt{x}-1}\)
Vậy \(P=\frac{x-16}{\sqrt{x}-1}:\frac{\sqrt{x}+3}{\sqrt{x}-1}\)
\(\Leftrightarrow P=\frac{\left(x-16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow P=\frac{x-16}{\sqrt{x}+3}\)
b) Ta có : \(\sqrt{x}+3\ge3>0\)
Để P min \(\Leftrightarrow x-16\) min
Mà \(x-16\ge-16\)
Dấu " = " xảy ra \(\Leftrightarrow x=0\)
Vậy \(Min_P=\frac{-16}{3}\Leftrightarrow x=0\)
a) P= √x+1 √x−1 + x+2 x√x−1 - √x+1 x+√x+1 \(\Leftrightarrow0\)
b)\(\sqrt{x}\left(2x+2\right)+2x+abp^2-2\)
\(ĐKXĐ:x\ne1\)
a) \(Q=\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}+\frac{3-\sqrt{x}}{x-1}\)
\(=\frac{\sqrt{x}\left(1+\sqrt{x}\right)+\sqrt{x}\left(1-\sqrt{x}\right)+\sqrt{x}-3}{1-x}\)
\(=\frac{\sqrt{x}+x+\sqrt{x}-x+\sqrt{x}-3}{1-x}\)
\(=\frac{3\sqrt{x}-3}{1-x}=\frac{-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-3}{\sqrt{x}+1}\)
b) Ta có \(\sqrt{x}\ge0\)
\(\Rightarrow\sqrt{x}+1\ge1\)
\(\Rightarrow\frac{-3}{\sqrt{x}+1}\ge-3\)
Dấu "=" khi x = 0