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\(t=\sqrt{2x-3}=>\frac{t^2+3}{2}=x\)
\(=>P=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}=\frac{\left(t-2\right)^2-1}{2}=\frac{\left(t-2\right)^2}{2}-\frac{1}{2}\)
ta có \(\frac{\left(t-2\right)^2}{2}\ge0\left(\forall t\right)\)
\(=>\frac{\left(t-2\right)^2}{2}-\frac{1}{2}\ge-\frac{1}{2}\left(\forall t\right)\)
minP=-1/2
dấu = xảy ra khi x=7/2
a) \(t=\sqrt{2x-3}\ge0\)
<=> \(t^2=2x-3\)
<=> \(x=\frac{t^2+3}{2}\)
=> \(P=\frac{t^2+3}{2}-2t\)
b) khi đó: \(P=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}=\frac{\left(t-2\right)^2-1}{2}\ge-\frac{1}{2}\)
Dấu "=" xảy ra <=> t = 2 khi đó: x = 7/2
ĐKXĐ:...
\(A=\frac{2\sqrt{x}\left(x+1\right)-3\left(x+1\right)}{2\sqrt{x}-3}=\frac{\left(2\sqrt{x}-3\right)\left(x+1\right)}{2\sqrt{x}-3}=x+1\)
\(B=\frac{2x\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\frac{2x}{\sqrt{x}}=2\sqrt{x}\)
\(A=x+1=\sqrt{4+\sqrt{7}}+1=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}+1=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}+1=\frac{1+\sqrt{14}+\sqrt{2}}{2}\)
\(B< -x+3\Leftrightarrow2\sqrt{x}< -x+3\Leftrightarrow x+2\sqrt{x}-3< 0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)< 0\Leftrightarrow\sqrt{x}-1< 0\Rightarrow x< 1\Rightarrow0< x< 1\)
Ta có:
\(A-B=x+1-2\sqrt{x}=\left(\sqrt{x}-1\right)^2\ge0\) \(\forall x\in TXĐ\)
Mà \(x\ne1\Rightarrow\) dấu "=" ko xảy ra
\(\Rightarrow A-B>0\Rightarrow A>B\)
a) \(P=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\frac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=2\sqrt{x}+1\)
\(Q=\frac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{x\sqrt{x}-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{x\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=\frac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}=x-1\)
b) \(P=Q\Leftrightarrow2\sqrt{x}+1=x-1\)
\(\Leftrightarrow x-2\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}+1-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=3\)
Vì \(\sqrt{x}-1\ge-1\) \(\Rightarrow\sqrt{x}-1=\sqrt{3}\)
\(\Rightarrow x=\left(\sqrt{3}+1\right)^2=4+2\sqrt{3}\)
Vậy...
1) \(B=\sqrt{x-1+2\sqrt[3]{x\sqrt{x}+3x+3\sqrt{x}+1}}\)
\(B=\sqrt{x-1+2\sqrt[3]{\sqrt{x^3}+3x+3\sqrt{x}+1}}\)
\(B=\sqrt{x-1+2\sqrt[3]{\left(\sqrt{x}+1\right)^3}}\)
\(B=\sqrt{x-1+2\left(\sqrt{x}+1\right)}\)
\(B=\sqrt{x-1+2\sqrt{x}+2}\)
\(B=\sqrt{\left(\sqrt{x}+1\right)^2}\)
\(B=\sqrt{x}+1\)
\(B=\sqrt{5}+1\)
2) Sửa đề :
\(C=\sqrt{2x-1+2\sqrt{x^2-x}}+\sqrt{2x-1-2\sqrt{x^2-x}}\)
\(C=\sqrt{x+2\sqrt{x\left(x-1\right)}+x-1}+\sqrt{x-2\sqrt{x\left(x-1\right)}+x-1}\)
\(C=\sqrt{\left(\sqrt{x}+\sqrt{x-1}\right)^2}+\sqrt{\left(\sqrt{x}-\sqrt{x-1}\right)^2}\)
\(C=\sqrt{x}+\sqrt{x-1}+\sqrt{x}-\sqrt{x-1}\)
\(C=2\sqrt{x}\)
\(C=2\cdot\sqrt{4}=4\)
a/ đkxđ: x>0; x khác 1
\(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}=\dfrac{x}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
b/ \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)
thay x vào A ta có:
\(A=\sqrt{\left(\sqrt{2}+1\right)^2}-1=\sqrt{2}+1-1=\sqrt{2}\)
1/ \(x-1=\sqrt[3]{2}\Rightarrow\left(x-1\right)^3=2\Rightarrow x^3-3x^2+3x-3=0\)
\(B=x^2\left(x^3-3x^2+3x-3\right)+x\left(x^3-3x^3+3x-3\right)+x^3-3x^2+3x-3+1945\)
\(B=1945\)
b/ Tương tự:
\(x-1=\sqrt[3]{2}+\sqrt[3]{4}\Rightarrow x^3-3x^2+3x-1=6+3\sqrt[3]{8}\left(\sqrt[3]{2}+\sqrt[3]{4}\right)\)
\(\Rightarrow x^3-3x^2+3x-1=6+6\left(x-1\right)\)
\(\Rightarrow x^3-3x^2-3x-1=0\)
\(P=x^2\left(x^3-3x^2-3x-1\right)-x\left(x^3-3x^2-3x-1\right)+x^3-3x^2-3x-1+2016\)
\(P=2016\)
Lời giải:
\(t=\sqrt{2x-3}\Rightarrow t^2=2x-3\Rightarrow x=\frac{t^2+3}{2}\)
Khi đó:
\(P=x-2\sqrt{2x-3}=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}\)