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a)Đk:\(0\le x\le1\)
\(\sqrt{x}+\sqrt{1-x}+\sqrt{x+1}=2\)
\(pt\Leftrightarrow\sqrt{x}+\sqrt{1-x}-1+\sqrt{x+1}-1=0\)
\(\Leftrightarrow\sqrt{x}+\frac{1-x-1}{\sqrt{1-x}+1}+\frac{x+1-1}{\sqrt{x+1}-1}=0\)
\(\Leftrightarrow\frac{x}{\sqrt{x}}-\frac{x}{\sqrt{1-x}+1}+\frac{x}{\sqrt{x+1}-1}=0\)
\(\Leftrightarrow x\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{1-x}+1}+\frac{1}{\sqrt{x+1}-1}\right)=0\)
\(\Rightarrow x=0\)
b)\(\frac{3x+3}{\sqrt{x}}=4+\frac{x+1}{\sqrt{x^2-x+1}}\)
\(pt\Leftrightarrow\frac{3x+3}{\sqrt{x}}-6=\frac{x+1}{\sqrt{x^2-x+1}}-2\)
\(\Leftrightarrow\frac{3x+3-6\sqrt{x}}{\sqrt{x}}=\frac{x+1-2\sqrt{x^2-x+1}}{\sqrt{x^2-x+1}}\)
\(\Leftrightarrow\frac{\frac{\left(3x+3\right)^2-36x}{3x+3+6\sqrt{x}}}{\sqrt{x}}=\frac{\frac{\left(x+1\right)^2-4\left(x^2-x+1\right)}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}\)
\(\Leftrightarrow\frac{\frac{9x^2+18x+9-36x}{3x+3+6\sqrt{x}}}{\sqrt{x}}=\frac{\frac{x^2+2x+1-4x^2+4x-4}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}\)
\(\Leftrightarrow\frac{\frac{9x^2-18x+9}{3x+3+6\sqrt{x}}}{\sqrt{x}}-\frac{\frac{-3x^2+6x-3}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\frac{\frac{9\left(x-1\right)^2}{3x+3+6\sqrt{x}}}{\sqrt{x}}+\frac{\frac{3\left(x-1\right)^2}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow3\left(x-1\right)^2\left(\frac{\frac{3}{3x+3+6\sqrt{x}}}{\sqrt{x}}+\frac{\frac{1}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}\right)=0\)
Dêx thấy: \(\frac{\frac{3}{3x+3+6\sqrt{x}}}{\sqrt{x}}+\frac{\frac{1}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}>0\forall....\)
\(\Rightarrow3\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
b)\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
\(pt\Leftrightarrow\frac{4}{x}+\sqrt{x-\frac{1}{x}}-\sqrt{\frac{3}{2}}=x+\sqrt{2x-\frac{5}{x}}-\sqrt{\frac{3}{2}}\)
\(\Leftrightarrow\left(\frac{4}{x}-x\right)+\frac{x-\frac{1}{x}-\frac{3}{2}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}=\frac{2x-\frac{5}{x}-\frac{3}{2}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\)
\(\Leftrightarrow\frac{-\left(x-2\right)\left(x+2\right)}{x}+\frac{\frac{\left(x-2\right)\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(x-2\right)\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{-\left(x+2\right)}{x}+\frac{\frac{\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\right)=0\)
Pt trong ngoặc VN suy ra x=2
a)\(x^2+3\sqrt{x^2-1}=\sqrt{x^4-x^2+1}\)
\(\Leftrightarrow x^2+3\sqrt{x^2-1}-1=\sqrt{x^4-x^2+1}-1\)
\(\Leftrightarrow\frac{x^2\left(3\sqrt{x^2-1}+1\right)}{3\sqrt{x^2-1}+1}+\frac{9\left(x^2-1\right)-1}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2+1-1}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{9x^2-10+3x^2\sqrt{x^2-1}+x^2}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{\sqrt{x^2-1}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}=\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{\sqrt{\left(x-1\right)\left(x+1\right)}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(\frac{\frac{1}{\sqrt{x^2-1}}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2}{\sqrt{x^4-x^2+1}+1}\right)=0\)
pt trong căn vô nghiệm
suy ra x=1; x=-1
-1; -6
b) ĐK: \(x^2+7x+7\ge0\) (đk xấu quá em ko giải đc;v)
PT \(\Leftrightarrow3x^2+21x+18+2\left(\sqrt{x^2+7x+7}-1\right)=0\)
\(\Leftrightarrow3\left(x+1\right)\left(x+6\right)+2\left(\frac{x^2+7x+6}{\sqrt{x^2+7x+7}+1}\right)=0\)
\(\Leftrightarrow3\left(x+1\right)\left(x+6\right)+\frac{2\left(x+1\right)\left(x+6\right)}{\sqrt{x^2+7x+7}+1}=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left[3+\frac{1}{\sqrt{x^2+7x+7}+1}\right]=0\)
Hiển nhiên cái ngoặc vuông > 0 nên vô nghiệm suy ra x = -1 (TM) hoặc x = -6 (TM)
Vậy....
P/s: Cũng may nghiệm đẹp chứ chứ nghiệm xấu thì tiêu rồi:(
a, dk \(x\ge0\)
ap dung bdt cosi ta co
\(\sqrt{x+3}+\frac{4x}{\sqrt{x+3}}\ge2\sqrt{4x}=4\sqrt{x}\)
dau = xay ra \(\Leftrightarrow\sqrt{x+3}=\frac{4x}{\sqrt{x+3}}\Leftrightarrow x+3=4x\Rightarrow x=1\)(tm dk)
kl x=1 la no cua pt
a) \(\text{Đ}K\text{X}\text{Đ}:\frac{3}{2}\le x\le\frac{5}{2}\)
Áp dụng BĐT Bunhiacopxki ta có:
\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)
Dấu '=' xảy ra khi \(\sqrt{2x-3}=\sqrt{5-2x}\Leftrightarrow x=2\)
Lại có: \(VP=3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)
Dấu '=' xảy ra khi x=2
Do đó VT=VP khi x=2
b) ĐK: \(x\ge0\). Ta thấy x=0 k pk là nghiệm của pt, chia 2 vế cho x ta có:
\(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\Leftrightarrow x-2-\sqrt{x}-\frac{2}{\sqrt{x}}+\frac{4}{x}=0\)
\(\Leftrightarrow\left(x+\frac{4}{x}\right)-\left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)-2=0\)
Đặt \(\sqrt{x}+\frac{2}{\sqrt{x}}=t>0\Leftrightarrow t^2=x+4+\frac{4}{x}\Leftrightarrow x+\frac{4}{x}=t^2-4\), thay vào ta có:
\(\left(t^2-4\right)-t-2=0\Leftrightarrow t^2-t-6=0\Leftrightarrow\left(t-3\right)\left(t+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=-2\end{cases}}\)
Đối chiếu ĐK của t
\(\Rightarrow t=3\Leftrightarrow\sqrt{x}+\frac{2}{\sqrt{x}}=3\Leftrightarrow x-3\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)