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\(\sqrt{2x+1}-\sqrt{5-x}+x-6=0\)
\(\Leftrightarrow\left(\sqrt{2x+1}-3\right)+\left(1-\sqrt{5-x}\right)+x-4=0\)
\(\Leftrightarrow\frac{2\left(x-4\right)}{\sqrt{2x+1}+3}+\frac{x-4}{\sqrt{5-x}+1}+x-4=0\)
\(\Leftrightarrow\left(x-4\right)\left(\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{5-x}+1}+1\right)=0\)
\(\Leftrightarrow x=4\)
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 \(\hept{\begin{cases}x=0\left(MIN\right)\\x=\sqrt{6}\left(MAX\right)\end{cases}}\)
\(B=\dfrac{xy}{xy}+\dfrac{\left(x-y\right)x}{x\left(x-y\right)}-\dfrac{y\left(x-y\right)}{y\left(x-y\right)}=1\)
Ta có: \(\sqrt{2x^2-4x+5}=\sqrt{2x^2-4x+2+3}=\sqrt{\left(\sqrt{2}x-\sqrt{2}\right)^2+3}\)
Lại có: \(\left(\sqrt{2}x-\sqrt{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(\sqrt{2}x-\sqrt{2}\right)^2+3\ge3\)
\(\Rightarrow\sqrt{\left(\sqrt{2}x-\sqrt{2}\right)^2+3}\ge\sqrt{3}\)
Vậy Min y là \(2+\sqrt{3}\)
\(y=2+\sqrt{2x^2-4x+5}=2+\sqrt{2x^2-4x+2+3}\)
\(=2+\sqrt{2\left(x^2-2x+1\right)+3}=2+\sqrt{2\left(x-1\right)^2+3}\)
Vì \(\left(x-1\right)^2\ge0\)\(\forall x\)
\(\Rightarrow2\left(x-1\right)^2\ge0\)\(\forall x\)\(\Rightarrow2\left(x-1\right)^2+3\ge3\)\(\forall x\)
\(\Rightarrow\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\)\(\forall x\)
\(\Rightarrow y=2+\sqrt{2\left(x-1\right)^2+3}\ge2+\sqrt{3}\)
Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)
Vậy \(miny=2+\sqrt{3}\)\(\Leftrightarrow x=1\)
Làm sai kìa !
Cái chỗ \(\left|\sqrt{x-2}-5+3-\sqrt{x-2}\right|\ge2\) chứ ? Trị tuyệt đối luôn dương mà
Cái trên là vừa phát hiện trong khi giải cái dưới
Vấn đề là giá trị của x cơ
\(1,\sqrt{\left(-0,3\right)^2}=\sqrt{0,09}=0,3\)
\(2,-\frac{1}{2}\sqrt{\left(0,3\right)^2}=-\frac{1}{2}.0,3=-0,15\)
\(3,\sqrt{a^{10}}=\sqrt{\left(a^5\right)^2}=a^5\left(a\ge0\right)\)
\(4,\sqrt{\left(2-x\right)^2}=\left|2-x\right|=2-x\left(x\le2\right)\)
\(5,\sqrt{x^2+2x+1}=\sqrt{\left(x+1\right)^2}=\left|x+1\right|\)
\(6,\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)(Vì \(1< \sqrt{2}\))
\(7,\sqrt{11+6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
\(8,\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)\)
\(=-2\)
\(9,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)
các biểu thức trong căn pt hết về HĐT rồi phá ra là done
\(\sqrt{x-3}+\sqrt{5-x}=x^2-8x+18.\)
ĐK: \(3\le x\le5\)
\(PT\Leftrightarrow\sqrt{x-3}-1+\sqrt{5-x}-1=x^2-8x+18-2\)
\(\Leftrightarrow\frac{x-3-1}{\sqrt{x-3}-1}+\frac{5-x-1}{\sqrt{5-x}+1}=\left(x-4\right)^2\)
\(\Leftrightarrow\frac{x-4}{\sqrt{x-3}+1}+\frac{4-x}{\sqrt{5-x}+1}=\left(x-4\right)^2\)
\(\Leftrightarrow\left(x-4\right)^2-\frac{x-4}{\sqrt{x-3}+1}+\frac{x-4}{\sqrt{5-x}+1}=0\)
\(\Leftrightarrow\left(x-4\right).\left(x-4-\frac{1}{\sqrt{x-3}-1}+\frac{1}{\sqrt{5-x}+1}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-4-\frac{1}{\sqrt{x-3}-1}+\frac{1}{\sqrt{5-x}+1}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4\left(TM\right)\\x-4-\frac{1}{\sqrt{x-3}-1}+\frac{1}{\sqrt{5-x}+1}=0\end{cases}}\) (Vô nghiệm)
Vậy pt có nghiệm x-4
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