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ĐKXĐ: \(x-\dfrac{1}{x}\ge0;x\ne0\)
Chia 2 vế của pt cho x:
\(x+2\sqrt{x-\dfrac{1}{x}}=3+\dfrac{1}{x}\)
\(\Leftrightarrow x-\dfrac{1}{x}+2\sqrt{x-\dfrac{1}{x}}-3=0\)
Đặt \(\sqrt{x-\dfrac{1}{x}}=t\ge0\)
\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-\dfrac{1}{x}}=1\Rightarrow x^2-x-1=0\)
\(\Rightarrow...\)
Lời giải:
PT $\Leftrightarrow (4x^2-4x+1)-3|2x-1|+2=0$
$\Leftrightarrow (2x-1)^2-3|2x-1|+2=0$
$\Leftrightarrow |2x-1|^2-3|2x-1|+2=0$
$\Leftrightarrow (|2x-1|-1)(|2x-1|-2)=0$
$\Rightarrow |2x-1|=1$ hoặc $|2x-1|=2$
$\Leftrightarrow 2x-1=\pm 1$ hoặc $2x-1=\pm 2$
$\Rightarrow x\in \left\{0; 1; \frac{3}{2}; \frac{-1}{2}\right\}$
a) \(\Leftrightarrow x^2+10x+25-x^2+8x-15=-8\\ \Leftrightarrow18x=-18\\ \Leftrightarrow x=-1\)
b) \(\Leftrightarrow\left(2x+1\right)^2-3\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(2x+1-3\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
\(\Leftrightarrow2x^2+10x-x^2+6x-9=x^2+6\)
=>16x-9=6
=>16x=15
hay x=15/16
\(PT\Leftrightarrow2x^2+10x-x^2+6x-9-x^2-6=0.\)
\(\Leftrightarrow16x-15=0.\\ \Leftrightarrow x=\dfrac{15}{16}.\)
ĐK: \(x\ge0\)
Dễ thấy \(1-\sqrt{2\left(x^2-x+1\right)}\le1-\sqrt{2}< 0\)
Khi đó bất phương trình tương đương:
\(x-\sqrt{x}\le1-\sqrt{2\left(x^2-x+1\right)}\)
\(\Leftrightarrow\sqrt{x}-\dfrac{1}{\sqrt{x}}-1+\sqrt{2\left(x+\dfrac{1}{x}-1\right)}\le0\)
\(\Leftrightarrow\sqrt{x}-\dfrac{1}{\sqrt{x}}-1+\sqrt{2\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)^2+2}\le0\)
\(\Leftrightarrow t-1+\sqrt{2t^2+2}\le0\)
Ta có:
(2 - 3x)(x + 8) = (3x - 2)(3 - 5x)
⇔ (2 - 3x)(x + 8) - (3x - 2)(3 - 5x) = 0
⇔ (2 - 3x)(x + 8) + (2 - 3x)(3 - 5x) = 0
⇔ (2 - 3x)(x + 8 + 3 - 5x) = 0
⇔ (2 - 3x)(11 - 4x) = 0
⇔ 2 - 3x = 0 hay 11 - 4x = 0
⇔ 2 = 3x hay 11 = 4x
⇔ x = \(\dfrac{2}{3}\) hay x = \(\dfrac{11}{4}\)
Vậy tập nghiệm của pt S = \(\left\{\dfrac{2}{3};\dfrac{11}{4}\right\}\)
<=> (2-3x ) (x+8) + (2-3x ) (3-5x)=0
<=> (2-3x ) ( x+8 + 3-5x ) =0
<=> (2-3x ) ( 11 - 4x ) = 0
=> 2-3x =0 hoặc 11-4x =0
3x = 2 4x =11
x = 2/3 x = 11/4
ĐKXĐ:\(x-1\ge0\Rightarrow x\ge1\)
\(x-\sqrt{x-1}-7=0\\ \Leftrightarrow\sqrt{x-1}=x-7\left(x\ge7\right)\\ \Leftrightarrow x-1=x^2-14x+49\\ \Leftrightarrow x^2-15x+50=0\\ \Leftrightarrow\left(x^2-10x\right)-\left(5x-50\right)=0\\ \Leftrightarrow x\left(x-10\right)-5\left(x-10\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=5\left(ktm\right)\end{matrix}\right.\)
f) ĐKXĐ : \(x\ge1\)
\(x-\sqrt{x-1}-7=0\)
<=>\(x-1-3\sqrt{x-1}+2\sqrt{x-1}-6=0\)
<=> \(\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+2\right)=0\)
<=> \(\left[{}\begin{matrix}\sqrt{x-1}=3\\\sqrt{x-1}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x\in\varnothing\end{matrix}\right.\Leftrightarrow x=10\)
Vậy x = 10 là nghiệm phương trình
d) ĐKXĐ: \(x\ne5,x\ne-2\)
\(pt\Leftrightarrow\dfrac{x^2-x-1}{\left(x-5\right)\left(x+2\right)}-\dfrac{1}{x-5}=0\Leftrightarrow\dfrac{x^2-x-1-\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x^2-x-1-x-2=0\Leftrightarrow x^2-2x-3=0\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
e) ĐKXĐ: \(x\ne1,x\ne2\)
\(pt\Leftrightarrow\dfrac{x+1}{x-1}+\dfrac{1}{x-2}-\dfrac{3x+4}{\left(x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x-2\right)+\left(x-1\right)-3x-4}{\left(x-1\right)\left(x-2\right)}=0\Leftrightarrow x^2-x-2+x-1-3x-4=0\Leftrightarrow x^2-3x-7=0\)
\(\Delta=\left(-3\right)^2-4.\left(-7\right)=37>0\)
PT có 2 nghiệm: \(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{37}}{2}\left(tm\right)\\x_2=\dfrac{3+\sqrt{37}}{2}\left(tm\right)\end{matrix}\right.\)
18) ĐKXĐ: \(x\ne-1,x\ne4\)
\(pt\Leftrightarrow\dfrac{x^2-x+8-2x\left(x-4\right)}{\left(x+1\right)\left(x-4\right)}=0\Leftrightarrow-x^2+7x+8=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=8\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
19) ĐKXĐ: \(x\ne\pm3\)
\(pt\Leftrightarrow\dfrac{\left(x+3\right)^2-48-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=0\Leftrightarrow12x-48=0\Leftrightarrow x=4\left(tm\right)\)
20) ĐKXĐ: \(x\ne\pm1\)
\(pt\Leftrightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2-16}{\left(x-1\right)\left(x+1\right)}=0\Leftrightarrow4x-16=0\Leftrightarrow x=4\left(tm\right)\)
đó là kt cơ bản nhất luôn , mấy bài này k bt lm thì lm b j nx :<