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a)ĐK:\(\begin{cases}25x^2-9 \ge 0\\5x+3 \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}(5x-3)(5x+3) \ge 0\\5x+3 \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}\left[ \begin{array}{l}x\ge \dfrac35\\x \le -\dfrac35\end{array} \right.\\\end{cases}\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x \ge \dfrac35\end{array} \right.\)
`pt<=>\sqrt{5x+3}(\sqrt{5x-3}-2)=0`
`<=>` \(\left[ \begin{array}{l}5x+3=0\\\sqrt{5x-3}=2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\5x-3=4\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x=7/5\end{array} \right.\)
`b)sqrt{x-3}/sqrt{2x+1}=2`
ĐK:\(\begin{cases}x-3 \ge 0\\2x+1>0\\\end{cases}\)
`<=>x>=3`
`pt<=>sqrt{x-3}=2sqrt{2x+1}`
`<=>x-3=8x+4`
`<=>7x=7`
`<=>x=1(l)`
`c)sqrt{x^2-2x+1}+sqrt{x^2-4x+4}=3`
`<=>sqrt{(x-1)^2}+sqrt{(x-2)^2}=3`
`<=>|x-1|+|x-2|=3`
`**x>=2`
`pt<=>x-1+x-2=3`
`<=>2x=6`
`<=>x=3(tm)`
`**x<=1`
`pt<=>1-x+2-x=3`
`<=>3-x=3`
`<=>x=0(tm)`
`**1<=x<=2`
`pt<=>x-1+2-x=3`
`<=>=-1=3` vô lý
Vậy `S={0,3}`
a) Áp dụng bđt AM-GM có:
\(\sqrt[3]{\left(9-x\right).8.8}\le\dfrac{9-x+8+8}{3}=\dfrac{25-x}{3}\)\(\Leftrightarrow\sqrt[3]{9-x}\le\dfrac{25-x}{12}\)
\(\sqrt[3]{\left(7+x\right).8.8}\le\dfrac{7+x+8+8}{3}=\dfrac{23+x}{3}\)\(\Leftrightarrow\sqrt[3]{7+x}\le\dfrac{23+x}{12}\)
Cộng vế với vế \(\Rightarrow\sqrt[3]{9-x}+\sqrt[3]{7+x}\le4\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}9-x=8\\7+x=8\end{matrix}\right.\)\(\Rightarrow x=1\)
Vậy...
b)Đk:\(x\ge2\)
Pt \(\Leftrightarrow\left(x-1\right)^2.\left(x^2-4\right)=\left(x-2\right)^2.\left(x^2-1\right)\)
\(\Leftrightarrow\left(x-1\right)^2\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\left(x-1\right)\)
Do \(x\ge2\Rightarrow x-1>0\)
Chia cả hai vế của pt cho x-1 ta được:
\(\left(x-1\right)\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x-1\right)\left(x+2\right)-\left(x-2\right)\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2+x-2-x^2+3x-2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
Vậy S={2}
c)Đk:\(\left\{{}\begin{matrix}9-x^2\ge0\\x^2-1\ge0\\x-3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le3\\\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Rightarrow x=3\)
Thay x=3 vào pt thấy thỏa mãn
Vậy S={3}
a) Quên mất, ko áp dụng đc AM-GM, xin lỗi
Pt \(\Leftrightarrow\sqrt[3]{9-x}-2=2-\sqrt[3]{7+x}\)
\(\Leftrightarrow\dfrac{9-x-8}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\dfrac{8-\left(7-x\right)}{4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}}\)
\(\Leftrightarrow\dfrac{1-x}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\dfrac{1-x}{4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{1}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\dfrac{1}{4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4=4+2\sqrt[3]{7+x}+\sqrt[3]{\left(7+x\right)^2}\left(1\right)\end{matrix}\right.\)
Từ (1) \(\Leftrightarrow\sqrt[3]{\left(9-x\right)^2}-\sqrt[3]{\left(7+x\right)^2}+2\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)=0\)
\(\Leftrightarrow\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)\left(\sqrt[3]{9-x}+\sqrt[3]{7+x}\right)+2\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)=0\)
\(\Leftrightarrow\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right).4+2\left(\sqrt[3]{9-x}-\sqrt[3]{7+x}\right)=0\)
\(\Leftrightarrow\sqrt[3]{9-x}-\sqrt[3]{7+x}=0\)
\(\Leftrightarrow\sqrt[3]{9-x}=\sqrt[3]{7+x}\)\(\Leftrightarrow9-x=7+x\)
\(\Leftrightarrow x=1\)
Vậy S={1}
Bạn xem lại đề câu b và c nhé !
a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)
\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)
\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ
\(\Rightarrow x\ge2\) thỏa mãn đề.
d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)
\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)
Pt tương đương :
\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )
e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)
\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)
Phương trình (1) tương đương :
\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )
\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)
\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)
\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)
a) \(3x-2\sqrt{x-1}=4\) (ĐK: x ≥ 1)
\(\Rightarrow3x-2\sqrt{x-1}-4=0\)
\(\Rightarrow3x-6-2\sqrt{x-1}+2=0\)
\(\Rightarrow3\left(x-2\right)-2\left(\sqrt{x-1}-1\right)=0\)
\(\Rightarrow3\left(x-2\right)-2.\dfrac{x-2}{\sqrt{x-1}+1}=0\)
\(\Rightarrow\left(x-2\right)\left[3-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)
*TH1: x = 2 (t/m)
*TH2: \(3-\dfrac{2}{\sqrt{x-1}+1}=0\)
\(\Rightarrow3=\dfrac{2}{\sqrt{x-1}+1}\)
\(\Rightarrow3\sqrt{x-1}+3=2\)
\(\Rightarrow3\sqrt{x-1}=-1\) (vô lí)
Vậy S = {2}
b) \(\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\) (ĐK: \(-\dfrac{1}{4}\le x\le3\) )
\(\Rightarrow\sqrt{4x+1}-3-\sqrt{x+2}+2-\sqrt{3-x}+1=0\)
\(\Rightarrow\dfrac{4x-8}{\sqrt{4x+1}+3}-\dfrac{x-2}{\sqrt{x+2}+2}+\dfrac{x-2}{\sqrt{3-x}+1}=0\)
\(\Rightarrow\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}-\dfrac{1}{\sqrt{x+2}+2}+\dfrac{1}{\sqrt{3-x}+1}\right)=0\)
=> x = 2
\(a,3x-2\sqrt{x-1}=4\left(x\ge1\right)\\ \Leftrightarrow-2\sqrt{x-1}=4-3x\\ \Leftrightarrow4\left(x-1\right)=16-24x+9x^2\\ \Leftrightarrow9x^2-28x+20=0\\ \Leftrightarrow\left(x-2\right)\left(9x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{10}{9}\left(tm\right)\end{matrix}\right.\)
\(b,\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\left(-\dfrac{1}{4}\le x\le3\right)\\ \Leftrightarrow4x+1+x+2-2\sqrt{\left(4x+1\right)\left(x+2\right)}=3-x\\ \Leftrightarrow-2\sqrt{\left(4x+1\right)\left(x+2\right)}=2-6x\\ \Leftrightarrow\sqrt{4x^2+9x+2}=3x-1\\ \Leftrightarrow4x^2+9x+2=9x^2-6x+1\\ \Leftrightarrow5x^2-15x-1=0\\ \Leftrightarrow\Delta=225+20=245\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15-\sqrt{245}}{10}=\dfrac{15-7\sqrt{5}}{10}\left(ktm\right)\\x=\dfrac{15+\sqrt{245}}{10}=\dfrac{15+7\sqrt{5}}{10}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{15+7\sqrt{5}}{10}\)
Lời giải:
a. Đề thiếu
b. PT $\Leftrightarrow \sqrt{(x-1)^2}+\sqrt{(x-2)^2}=3$
$\Leftrightarrow |x-1|+|x-2|=3$
Nếu $x\geq 2$ thì pt trở thành:
$x-1+x-2=3$
$\Leftrightarrow 2x-3=3$
$\Leftrightarrow x=3$ (tm)
Nếu $1\leq x< 2$ thì:
$x-1+2-x=3\Leftrightarrow 1=3$ (vô lý)
Nếu $x< 1$ thì:
$1-x+2-x=3$
$\Leftrightarrow x=0$ (tm)
a) \(\sqrt{-x^2+x+4}=x-3\left(đk:x\ge3\right)\)
\(-x^2+x+4=x^2-6x+9\)
\(2x^2-7x-5=0\)
\(\Delta=49-4.2.\left(-5\right)=89\)
\(\left[{}\begin{matrix}x=\dfrac{7+\sqrt{89}}{4}\left(TM\right)\\x=\dfrac{7-\sqrt{89}}{4}\left(L\right)\end{matrix}\right.\)
b) \(\sqrt{-2x^2+6}=x-1\left(đk:x\ge1\right)\)
\(-2x^2+6=x^2-2x+1\)
\(3x^2-2x-5=0\)
\(\Delta=4+4.3.5=64\)
\(\left[{}\begin{matrix}x=\dfrac{2-8}{6}=-1\left(L\right)\\x=\dfrac{2+8}{6}=\dfrac{5}{3}\left(TM\right)\end{matrix}\right.\)
c) \(\sqrt{x+2}=1+\sqrt{x-3}\left(Đk:x\ge3\right)\)
\(x+2=1+x-3+2\sqrt{x-3}\)
\(\sqrt{x-3}=2\)
\(x-3=4\)
\(x=7\)