Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\sqrt{2x+1}-\sqrt{5-x}+x-6=0\)
\(\Leftrightarrow\left(\sqrt{2x+1}-3\right)+\left(1-\sqrt{5-x}\right)+x-4=0\)
\(\Leftrightarrow\frac{2\left(x-4\right)}{\sqrt{2x+1}+3}+\frac{x-4}{\sqrt{5-x}+1}+x-4=0\)
\(\Leftrightarrow\left(x-4\right)\left(\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{5-x}+1}+1\right)=0\)
\(\Leftrightarrow x=4\)
\(1,\sqrt{\left(-0,3\right)^2}=\sqrt{0,09}=0,3\)
\(2,-\frac{1}{2}\sqrt{\left(0,3\right)^2}=-\frac{1}{2}.0,3=-0,15\)
\(3,\sqrt{a^{10}}=\sqrt{\left(a^5\right)^2}=a^5\left(a\ge0\right)\)
\(4,\sqrt{\left(2-x\right)^2}=\left|2-x\right|=2-x\left(x\le2\right)\)
\(5,\sqrt{x^2+2x+1}=\sqrt{\left(x+1\right)^2}=\left|x+1\right|\)
\(6,\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)(Vì \(1< \sqrt{2}\))
\(7,\sqrt{11+6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
\(8,\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)\)
\(=-2\)
\(9,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)
a/ \(A=\frac{1}{5+2\sqrt{6-x^2}}\)
Có: \(-x^2\le0\)với mọi x
=> \(6-x^2\le6\)
=> \(0\le\sqrt{6-x^2}\le\sqrt{6}\)
=> \(5\le5+2\sqrt{6-x^2}\le5+2\sqrt{6}\)
=> \(\frac{1}{5+2\sqrt{6}}\le\frac{1}{5+2\sqrt{6-x^2}}\le\frac{1}{5}\); với mọi x
=> \(\hept{\begin{cases}maxA=\frac{1}{5}\Leftrightarrow\sqrt{6-x^2}=0\Leftrightarrow x=\pm\sqrt{6}\\minA=\frac{1}{5+2\sqrt{6}}\Leftrightarrow\sqrt{6-x^2}=\sqrt{6}\Leftrightarrow x=0\end{cases}}\)
Vậy:...
b/ \(B=\sqrt{-x^2+2x+4}=\sqrt{-\left(x-1\right)^2+5}\)
Có: \(-\left(x-1\right)^2\le0\)với mọi x
=> \(-\left(x-1\right)^2+5\le5\)
=> \(0\le\sqrt{-\left(x-1\right)^2+5}\le\sqrt{5}\)
=> \(0\le B\le\sqrt{5}\)với mọi x
=> \(\hept{\begin{cases}maxB=\sqrt{5}\Leftrightarrow-\left(x-1\right)^2=0\Leftrightarrow x=1\\minB=0\Leftrightarrow\left(x-1\right)^2=5\Leftrightarrow x=\pm\sqrt{5}+1\end{cases}}\)
Vậy:...
a)Ta có:
\(0\le2\sqrt{6-x^2}\le2\sqrt{6}\)
\(\Leftrightarrow\frac{1}{5}\ge\frac{1}{5+2\sqrt{6-x^2}}\ge\frac{1}{5+2\sqrt{6}}=5-2\sqrt{6}\)
\(\Rightarrow\hept{\begin{cases}MAX\left(A\right)=\frac{1}{5}\\MIN\left(A\right)=5-2\sqrt{6}\end{cases}}\)Dấu "=" xảy ra khi \(\hept{\begin{cases}x=0\left(MIN\right)\\x=\sqrt{6}\left(MAX\right)\end{cases}}\)
các biểu thức trong căn pt hết về HĐT rồi phá ra là done
e/ \(\left(x-4\right)\sqrt{16-8x+x^2}=\left(x-4\right)\sqrt{\left(x-4\right)^2}=\left(x-4\right)\left(x-4\right)=\left(x-4\right)^2\)
f/ \(\left(2x-5\right)\sqrt{\dfrac{2}{\left(2x-5\right)^2}}=\left(2x-5\right)\cdot\dfrac{1}{\left|2x-5\right|}\cdot\sqrt{2}\)
+) với \(x>\dfrac{5}{2}\) có: \(\left(2x-5\right)\cdot\dfrac{1}{\left|2x-5\right|}\cdot\sqrt{2}=\dfrac{2x-5}{2x-5}\cdot\sqrt{2}=\sqrt{2}\)
+) với \(x< \dfrac{5}{2}\) có:
\(\left(2x-5\right)\cdot\dfrac{1}{\left|2x-5\right|}\cdot\sqrt{2}=\dfrac{2x-5}{-\left(2x-5\right)}\cdot\sqrt{2}=-1\cdot\sqrt{2}=-\sqrt{2}\)
g/ \(\sqrt{x-4\sqrt{x-4}}=\sqrt{x-4-2\cdot2\cdot\sqrt{2-4}+4}=\sqrt{\left(\sqrt{x-4}+2\right)^2}=\sqrt{x-4}+2\)
Với \(x\ge\dfrac{5}{2}\)có: \(A=x+\sqrt{2x-5}\ge\dfrac{5}{2}+0=\dfrac{5}{2}\)
Dấu '=' xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)
\(\Rightarrow A_{min}=\dfrac{5}{2}\)
đúng như mk dự đoán chớ mk thủ hết cách rk mà có dc à