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4.a)\(x-2\sqrt{x}+3\)
\(=x-2\sqrt{x}+1+2\)
\(=\left(\sqrt{x}-1\right)^2+2\)
Vì \(\left(\sqrt{x}-1\right)^2\ge0,\forall x\)
\(\left(\sqrt{x}-1\right)^2+2\ge2\)
\(\Rightarrow Min_{bt}=2\) khi \(\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
b)Ta có:
\(x-4\sqrt{y}+13\ge0\)
\(\Leftrightarrow x-4\sqrt{y}\ge-13\)
Dấu "=" xảy ra khi \(x-4\sqrt{y}=0\Leftrightarrow x=4\sqrt{y}\)
Vậy \(min_{bt}=0\) khi \(x=4\sqrt{y}\)
c)Ta có:
\(2x-4\sqrt{y}+6\ge0\)
\(\Leftrightarrow x-2\sqrt{y}+3\ge0\)
\(\Leftrightarrow x-2\sqrt{y}\ge-3\)
Dấu "=" xảy ra khi \(x-2\sqrt{y}=0\Leftrightarrow x=2\sqrt{y}\)
Vậy \(Min_{bt}=0\) khi \(x=2\sqrt{y}\)
d)Ta có:
\(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\)
Vì \(\left(x+1\right)^2\ge0,\forall x\)
\(\Leftrightarrow\left(x+1\right)^2+4\ge4\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)^2+4}\le\frac{1}{4}\)
\(\Leftrightarrow-\frac{1}{\left(x+1\right)^2+4}\ge-\frac{1}{4}\)
\(\Leftrightarrow-\frac{4}{\left(x+1\right)^2+4}\ge-1\)
Vậy \(Min_{bt}=-1\) khi \(x+1=0\Leftrightarrow x=-1\)
e/ \(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)
\(\Leftrightarrow4+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\)
\(\Leftrightarrow2\sqrt{-x^2+8x-12}=x^2-8x+20\)
Đặt \(\sqrt{-x^2+8x-12}=a\left(a\ge0\right)\)thì pt thành
\(2a=-a^2+8\)
\(\Leftrightarrow a^2+2a-8=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-4\left(l\right)\\a=2\end{cases}}\)
\(\Leftrightarrow\sqrt{-x^2+8x-12}=2\)
\(\Leftrightarrow-x^2+8x-12=4\)
\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)
a/ \(4x^2+3x+3-4x\sqrt{x+3}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1-2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2+\left(1-\sqrt{2x-1}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=\sqrt{x+3}\\1=\sqrt{2x-1}\end{cases}\Leftrightarrow}x=1\)
\(1.x^2-4x-2\sqrt{2x-5}+5=0\left(x>=\dfrac{5}{2}\right)\)
\(\text{⇔}2x-5-2\sqrt{2x-5}+1+x^2-6x+9=0\)
\(\text{⇔}\left(\sqrt{2x-5}-1\right)^2+\left(x-3\right)^2=0\)
\(\text{⇔}\sqrt{2x-5}-1=0\) hoặc \(x-3=0\)
\(\text{⇔}x=3\left(TM\right)\)
KL...........
\(2.x+y+4=2\sqrt{x}+4\sqrt{y-1}\)
\(\text{⇔}x-2\sqrt{x}+1+y-1-4\sqrt{y-1}+4=0\)
\(\text{⇔}\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-2\right)^2=0\)
\(\text{⇔}x=1;y=5\)
KL..........
\(3.\sqrt{x-2}+\sqrt{y-3}+\sqrt{z-5}=\dfrac{1}{2}\left(x+y+z-7\right)\)
\(\text{⇔}2\sqrt{x-2}+2\sqrt{y-3}+2\sqrt{z-5}=x+y+z-7\)
\(\text{⇔}x-2-2\sqrt{x-2}+1+y-3-2\sqrt{y-3}+1+z-5-2\sqrt{z-5}+1=0\)
\(\text{⇔}\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-5}-1\right)^2=0\)
\(\text{⇔}x=1;y=4;z=6\)
KL...........
\(d.Tuong-tự-nhé-bn\)
7/
ĐKXĐ: \(-3\le x\le\frac{2}{3}\)
\(\Leftrightarrow2x+8\sqrt{x+3}+4\sqrt{3-2x}=2\)
\(\Leftrightarrow8\sqrt{x+3}+4\sqrt{3-2x}-\left(3-2x\right)+1=0\)
\(\Leftrightarrow8\sqrt{x+3}+\sqrt{3-2x}\left(4-\sqrt{3-2x}\right)+1=0\)
Do \(x\ge-3\Rightarrow3-2x\le9\Rightarrow\sqrt{3-2x}\le3\)
\(\Rightarrow4-\sqrt{3-2x}>0\)
\(\Rightarrow VT>0\)
Phương trình vô nghiệm (bạn coi lại đề)
5/
\(\Leftrightarrow8x^2-3x+6-4x\sqrt{3x^2+x+2}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{3x^2+x+2}+3x^2+x+2\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x^2+x+2}\right)^2+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-\sqrt{3x^2+x+2}=0\\x-2=0\end{matrix}\right.\) \(\Rightarrow x=2\)
6/
ĐKXĐ: ....
\(\Leftrightarrow\left(x-2000-2\sqrt{x-2000}+1\right)+\left(y-2001-2\sqrt{y-2001}+1\right)+\left(z-2002-2\sqrt{z-2002}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2001\\y=2002\\z=2003\end{matrix}\right.\)