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1: =>3x^2+5x-7=3x+14
=>2x=21
=>x=21/2
2;=>x+4=4
=>x=0
3: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{2}\\4x^2-20x+25-4x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{2}\\4x^2-24x+32=0\end{matrix}\right.\)
=>x>=5/2 và x^2-6x+8=0
=>x=4
4: \(\Leftrightarrow\left\{{}\begin{matrix}x>=1\\x^2+2x-1=x^2-2x+1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
5: \(\Leftrightarrow\sqrt{2x+16}=x-4\)
=>x>=4 và x^2-8x+16=2x+16
=>x>=4 và x^2-10x=0
=>x=10
a/ ĐKXĐ: \(x\ge1\)
Khi \(x\ge1\) ta thấy \(\left\{{}\begin{matrix}VT>0\\VP=1-x\le0\end{matrix}\right.\) nên pt vô nghiệm
b/ \(x\ge1\)
\(\sqrt{\sqrt{x-1}\left(x-2\sqrt{x-1}\right)}+\sqrt{\sqrt{x-1}\left(x+3-4\sqrt{x-1}\right)}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-1\right)^2}+\sqrt{\sqrt{x-1}\left(\sqrt{x-1}-2\right)^2}=\sqrt{x-1}\)
Đặt \(\sqrt{x-1}=a\ge0\) ta được:
\(\sqrt{a\left(a-1\right)^2}+\sqrt{a\left(a-2\right)^2}=a\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\Rightarrow x=1\\\sqrt{\left(a-1\right)^2}+\sqrt{\left(a-2\right)^2}=\sqrt{a}\left(1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left|a-1\right|+\left|a-2\right|=\sqrt{a}\)
- Với \(a\ge2\) ta được: \(2a-3=\sqrt{a}\Leftrightarrow2a-\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow a=\frac{9}{4}\Rightarrow\sqrt{x-1}=\frac{9}{4}\Rightarrow...\)
- Với \(0\le a\le1\) ta được:
\(1-a+2-a=\sqrt{a}\Leftrightarrow2a+\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-1}=1\Rightarrow...\)
- Với \(1< a< 2\Rightarrow a-1+2-a=\sqrt{a}\Leftrightarrow a=1\left(l\right)\)
c/ ĐKXĐ: \(x\ge\frac{49}{14}\)
\(\Leftrightarrow\sqrt{14x-49+14\sqrt{14x-49}+49}+\sqrt{14x-49-14\sqrt{14x-49}+49}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{14x-49}+7\right)^2}+\sqrt{\left(\sqrt{14x-49}-7\right)^2}=14\)
\(\Leftrightarrow\left|\sqrt{14x-49}+7\right|+\left|7-\sqrt{14x-49}\right|=14\)
Mà \(VT\ge\left|\sqrt{14x-49}+7+7-\sqrt{14x-49}\right|=14\)
Nên dấu "=" xảy ra khi và chỉ khi:
\(7-\sqrt{14x-49}\ge0\)
\(\Leftrightarrow14x-49\le49\Leftrightarrow x\le7\)
Vậy nghiệm của pt là \(\frac{49}{14}\le x\le7\)
5. \(y=\dfrac{-3x}{x+2}\)
xác định khi: \(x+2\ne0\Leftrightarrow x\ne-2\)
vậy D= (\(-\infty;+\infty\))\{-2}
6. \(y=\sqrt{-2x-3}\)
xác định khi: \(-2x-3\ge0\Leftrightarrow x\le\dfrac{-3}{2}\)
vậy D= (\(-\infty;\dfrac{-3}{2}\)]
7. \(y=\dfrac{3-x}{\sqrt{x-4}}\)
xác định khi: x-4 >0 <=> x>4
vậy D= (\(4;+\infty\))
8. \(y=\dfrac{2x-5}{\left(3-x\right)\sqrt{5-x}}\)
xác định khi: \(\left\{{}\begin{matrix}3-x\ne0\\5-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x< 5\end{matrix}\right.\)
vậy D= (\(-\infty;5\))\ {3}
9.\(y=\sqrt{2x+1}+\sqrt{4-3x}\)
xác định khi: \(\left\{{}\begin{matrix}2x+1\ge0\\4-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-1}{2}\\x\le\dfrac{4}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{-1}{2}\le x\le\dfrac{4}{3}\)
vậy D= [\(\dfrac{-1}{2};\dfrac{4}{3}\)]
1. \(y=\dfrac{3x-2}{x^2-4x+3}\)
xác định khi : \(x^2-4x+3\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne1\end{matrix}\right.\)
vậy tập xác định là: D = \(\left(-\infty;+\infty\right)\backslash\left\{3;1\right\}\)
2.\(y=2\sqrt{5-4x}\)
xác định khi \(5-4x\ge0\Leftrightarrow x\le\dfrac{5}{4}\)
vậy D= (\(-\infty;\dfrac{5}{4}\)]
3. \(y=\dfrac{2}{\sqrt{x+3}}+\sqrt{5-2x}\)
xác định khi: \(\left\{{}\begin{matrix}x+3>0\\5-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\le\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow-3< x\le\dfrac{5}{2}\)
vậy D= (\(-3;\dfrac{5}{2}\)]
4.\(\sqrt{9-x}+\dfrac{1}{\sqrt{x+2}-2}\)
xác định khi: \(\left\{{}\begin{matrix}9-x\ge0\\x+2\ge0\\x\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le9\\x\ge-2\\x\ne2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2\le x\le9\\x\ne2\end{matrix}\right.\)
Vậy D= [\(-2;9\)]\{2}
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
8) ĐKXĐ: $-2\leq x\leq 1$
PT $\Leftrightarrow (2x+4)-4\sqrt{2x+4}+4+[(1-x)-2\sqrt{1-x}+1]=0$
$\Leftrightarrow (\sqrt{2x+4}-2)^2+(\sqrt{1-x}-1)^2=0$
Dễ thấy: $(\sqrt{2x+4}-2)^2; (\sqrt{1-x}-1)^2\geq 0$ với mọi $x\in [-2;1]$ nên để tổng của chúng bằng $0$ thì:
$(\sqrt{2x+4}-2)^2=(\sqrt{1-x}-1)^2=0$
$\Leftrightarrow \sqrt{2x+4}=2; \sqrt{1-x}-1=0$
$\Leftrightarrow x=0$ (thỏa mãn)
Vậy.....
7)
ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow x^2+[(x+1)-2\sqrt{x+1}+1]=0$
$\Leftrightarrow x^2+(\sqrt{x+1}-1)^2=0$
Ta thấy:
$x^2\geq 0; (\sqrt{x+1}-1)^2\geq 0$ với mọi $x\geq -1$
Do đó để tổng của chúng bằng $0$ thì $x^2=(\sqrt{x+1}-1)^2=0$
$\Leftrightarrow x=0$ (thỏa mãn)
Vậy.......
ĐKXĐ: \(x\ge\frac{5}{2}\)
\(pt\Leftrightarrow\sqrt{2x-5+2\sqrt{2x-5}+1}+\sqrt{2x-5+6\sqrt{2x-5}+9}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)
\(\Leftrightarrow\sqrt{2x-5}+1+\sqrt{2x-5}+3=14\)
\(\Leftrightarrow\sqrt{2x-5}=5\)
\(\Leftrightarrow2x-5=25\)
\(\Leftrightarrow x=15\left(TM\right)\)
Vậy phương trình đã cho có nghiệm duy nhất \(x=15\)
@Nguyễn Việt Lâm