Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(7\sqrt{AB}+7B-\sqrt{A}-\sqrt{B}=7\sqrt{B}\left(\sqrt{A}+\sqrt{B}\right)-\left(\sqrt{A}+\sqrt{B}\right)\)\(=\left(\sqrt{A}+\sqrt{B}\right)\left(7\sqrt{B}-1\right)\)
b, \(a\sqrt{b}-b\sqrt{a}+\sqrt{a}-\sqrt{b}=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)+\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)\)
c,\(\sqrt{x^2-25y^2}-\sqrt{x-5y}=\sqrt{x-5y}.\sqrt{x+5y}-\sqrt{x-5y}\)
\(=\sqrt{x-5y}\left(\sqrt{x+5y}-1\right)\)
\(a,7\sqrt{AB}+7B-\sqrt{A}-\sqrt{B}\)( Với A>= 0, B>=0)
\(=\left(7\sqrt{AB}-\sqrt{A}\right)+\left(7B-\sqrt{B}\right)\)
\(=7\sqrt{A}\left(\sqrt{B}-1\right)+7\sqrt{B}\left(\sqrt{B}-1\right)\)
\(=\left(\sqrt{B}-1\right)\left(7\sqrt{A}+7\sqrt{B}\right)\)
\(=7\left(\sqrt{B}-1\right)\left(\sqrt{A}+\sqrt{B}\right)\)
\(b,a\sqrt{b}-b\sqrt{a}+\sqrt{a}-\sqrt{b}\)Với a>= 0, b>=0)
\(=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)+\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)\)
\(c,\sqrt{x^2-25y^2}-\sqrt{x-5y}\)
\(=\sqrt{\left(x-5y\right)\left(x+5y\right)}-\sqrt{x-5y}\)
\(=\sqrt{x-5y}.\sqrt{x+5y}-\sqrt{x-5y}\)
\(=\sqrt{x-5y}\left(\sqrt{x+5y}-1\right)\)
\(a)\) \(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=a-b\)
\(b)\) \(B=a-b=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)\(\Rightarrow\)\(B^2=\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2=2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\)
\(B^2=4-2\sqrt{4-3}=4-2=2\)\(\Rightarrow\)\(B=\sqrt{2}\) ( vì \(B>0\) )
...
2.\(\sqrt{9x}-5\sqrt{x}=6-4\sqrt{x}\)
\(\Leftrightarrow3\sqrt{x}-5\sqrt{x}=6-4\sqrt{x}\)
\(\Leftrightarrow3\sqrt{x}-5\sqrt{x}+4\sqrt{x}=6\)
\(\Leftrightarrow2\sqrt{x}=6\)
\(\Leftrightarrow\sqrt{x}=\frac{6}{2}\)
\(\Leftrightarrow\sqrt{x}=3\)
\(\Leftrightarrow\left(\sqrt{x}\right)^2=\left(3\right)^2\)
\(\Leftrightarrow x=9\)
vậy x=9
mình chỉ giúp bạn được vậy thui :)
chúc bạn học tốt nha:)))
ĐK: \(x\ge-7\)
PT \(\Leftrightarrow\left(\sqrt[3]{x-8}-\left(x-8\right)\right)+\left[\sqrt{x+7}-4\right]+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\frac{-\left(x-9\right)\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}+\frac{x-9}{\sqrt{x+7}+4}+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left[x^2+x+2+\frac{1}{\sqrt{x+7}+4}-\frac{\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}\right]=0\)
\(\Leftrightarrow x=9\)
P/s:em chả biết đánh giá cái ngoặc to thế nào nữa:((((
\(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a-1}}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}\right)^2-2\sqrt{a}+1}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}\)
Mà \(\sqrt{a}-1< \sqrt{a}\) => \(\frac{\sqrt{a}-1}{\sqrt{a}}< 1\)
Vậy M < 1.
1.
ĐK \(a\ge0;a\ne1\)
Ta có \(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right).\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\)
\(=\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(a-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{a-1}{\sqrt{a}}\)
\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
\(=\frac{4a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}=4a\)
2. Với \(a=\frac{\sqrt{6}}{2+\sqrt{6}}\Rightarrow A=\frac{4\sqrt{6}}{2+\sqrt{6}}\)
Để \(\sqrt{A}>A\Rightarrow\sqrt{4a}>4a\Rightarrow2\sqrt{a}-4a>0\Rightarrow2\sqrt{a}\left(1-2\sqrt{a}\right)>0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{a}>0\\1-2\sqrt{a}>0\end{cases}\Rightarrow\hept{\begin{cases}a>0\\a>\frac{1}{4}\end{cases}\Rightarrow}a>\frac{1}{4}}\)
Vậy để \(\sqrt{A}>A\)thì \(a>\frac{1}{4};a\ne1\)
\(a,\)\(7\sqrt{ab}+7b-\sqrt{a}-\sqrt{b}\)
\(=7\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}+\sqrt{b}\right)\left(7\sqrt{b}-1\right)\)
\(b,a\sqrt{b}-b\sqrt{a}+\sqrt{a}-\sqrt{b}\)
\(=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)+\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}-1\right)\)
\(c,\sqrt{x^2-25y^2}-\sqrt{x-5y}\)
\(=\sqrt{\left(x-5y\right)\left(x+5y\right)}-\sqrt{x-5y}\)
\(=\sqrt{x-5y}\left(\sqrt{x-5y}-1\right)\)