1 visit

2 hotel

3 near

4 travel

5 convenient

6 map

7 clothes

8 camera

9 take

10 attractions

Lời giải:

a. ĐKXĐ: $x\leq \frac{1}{5}$

PT $\Leftrightarrow 1-5x=3^2=9$

$\Leftrightarrow 5x=-8\Leftrightarrow x=\frac{-8}{5}$ (tm)

b. ĐKXĐ: $x\geq \frac{3}{5}$

PT $\Leftrightarrow 25x^2-9=4(5x-3)$

$\Leftrightarrow (5x-3)(5x+3)-4(5x-3)=0$

$\Leftrightarrow (5x-3)(5x-1)=0$

$\Leftrightarrow x=\frac{3}{5}$ (tm) hoặc $x=\frac{1}{5}$ (loại)

c. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow x-4\sqrt{x}+3=0$

$\Leftrightarrow (\sqrt{x}-1)(\sqrt{x}-3)=0$

$\Leftrightarrow \sqrt{x}=1$ hoặc $\sqrt{x}=3$

$\Leftrightarrow x=1$ hoặc $x=9$

d. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow (\sqrt{x}-2)^2-5=0$

$\Leftrightarrow (\sqrt{x}-2)^2=5$
$\Leftrightarrow \sqrt{x}-2=\pm \sqrt{5}$

$\Leftrightarrow \sqrt{x}=2+\sqrt{5}$ (chọn) hoặc $\sqrt{x}=2-\sqrt{5}$ (loại do âm)

$\Leftrightarrow x=(2+\sqrt{5})^2=9+4\sqrt{5}$

e.ĐKXĐ: $x\geq 3$
PT $\Leftrightarrow 2\sqrt{9}.\sqrt{x-3}-\frac{1}{5}.\sqrt{25}.\sqrt{x-3}-\frac{1}{7}\sqrt{49}.\sqrt{x-3}=20$

$\Leftrightarrow 6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20$

$\Leftrightarrow 4\sqrt{x-3}=20$

$\Leftrightarrow \sqrt{x-3}=5$

$\Leftrightarrow x-3=25$

$\Leftrightarrow x=28$

\(a,ĐK:x\le\dfrac{1}{5}\\ PT\Leftrightarrow1-5x=9\Leftrightarrow x=-\dfrac{8}{5}\\ b,ĐK:x\ge\dfrac{3}{5}\\ PT\Leftrightarrow\sqrt{5x-3}\left(\sqrt{5x+3}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x-3=0\\\sqrt{5x+3}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\5x+3=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=\dfrac{1}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{3}{5}\)

\(c,ĐK:x\ge0\\ PT\Leftrightarrow2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ge0\\ PT\Leftrightarrow x-4\sqrt{x}+4-3=0\\ \Leftrightarrow\left(\sqrt{x}-2-\sqrt{3}\right)\left(\sqrt{x}-2+\sqrt{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2+\sqrt{3}\\\sqrt{x}=2-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7+4\sqrt{3}\left(tm\right)\\x=7-4\sqrt{3}\left(tm\right)\end{matrix}\right.\\ e,ĐK:x\ge3\\ PT\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\Leftrightarrow x=28\left(tm\right)\)

a) \(\sqrt{1-5x}=3\left(đk:x\le\dfrac{1}{5}\right)\)

\(\Leftrightarrow1-5x=9\Leftrightarrow5x=-8\Leftrightarrow x=-\dfrac{8}{5}\left(tm\right)\)

b) \(\sqrt{25x^2-9}=2\sqrt{5x-3}\left(đk:x\ge\dfrac{3}{5}\right)\)

\(\Leftrightarrow25x^2-9=20x-12\)

\(\Leftrightarrow25x^2-20x+3=0\)

\(\Leftrightarrow\left(5x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=\dfrac{1}{5}\left(ktm\right)\end{matrix}\right.\)

c) \(ĐK:x\ge0\)

\(\Leftrightarrow2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

d) \(ĐK:x\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-2-\sqrt{5}\right)\left(\sqrt{x}-2+\sqrt{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2+\sqrt{5}\\\sqrt{x}=2-\sqrt{5}\left(VLý\right)\end{matrix}\right.\)

\(\Leftrightarrow x=9+4\sqrt{5}\left(tm\right)\)

e) \(ĐK:x\ge3\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\Leftrightarrow\sqrt{x-3}=5\)

\(\Leftrightarrow x-3=25\Leftrightarrow x=28\left(tm\right)\)

\(ĐKXĐ:x\ne-1;x\ne\frac{2}{3}\)

\(pt\Leftrightarrow\frac{7x-2\left(x+1\right)+\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=1\)

\(\Leftrightarrow7x-2\left(x+1\right)+\left(3x-2\right)=\left(3x-2\right)\left(x+1\right)\)

\(\Leftrightarrow8x-4=3x^2-2x+3x-2\)

\(\Leftrightarrow3x^2-7x+2=0\)

\(\Delta=7^2-4.3.2=25,\sqrt{\Delta}=5\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7+5}{6}=2\\x=\frac{7-5}{6}=\frac{1}{3}\end{cases}}\)

Tự cho đkxđ nha!!!

<=> \(\frac{x+1-x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2}{3x-2}\)

<=> \(\frac{3x-2}{\left(3x-2\right)\left(x+1\right)}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(3x-2\right)\left(x+1\right)}\)

<=> \(\frac{7x-2x-2-3x+2}{\left(3x-2\right)\left(x+1\right)}=0\)

<=> \(\frac{2x}{\left(3x-2\right)\left(x+1\right)}=0\)

=> 2x = 0

<=> x = 0 (TM)

Vậy ...

\(1-\frac{x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}+\frac{2}{2-3x}\)

\(\left(x+1\right)\left(3x-2\right)\left(2-3x\right)-x\left(3x-2\right)\left(2-3x\right)=7x\left(2-3x\right)+2\left(x+1\right)\left(3x-2\right)\)

\(-9x^2+12x-4=16x-15x^2-4\)

\(-9x^2+12x=16x-15x^2\)

\(9x^2-12x+16x-15x^2=0\)

\(-6x^2+4x=0\)

\(-2x\left(3x-2\right)=0\)

\(Th1:-2x=0\Leftrightarrow x=0\)

\(Th2:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)

kết quả x=2

\(x=2\)

Bài làm :

\(12x+3\left(x-1\right)=27\)

\(12x+3x-3=27\)

\(15x=30\)

\(x=30:15\)

\(x=2\)

Học tốt nhé

\(x^3-2x=-x^2+2\)

\(\Leftrightarrow x^3+x^2-2x-2=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-2\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Ta có: \(x^3-2x=-x^2+2\)

    \(\Leftrightarrow\left(x^3+x^2\right)-\left(2x+2\right)=0\)

    \(\Leftrightarrow x^2.\left(x+1\right)-2.\left(x+1\right)=0\)

    5\(\Leftrightarrow\left(x+1\right).\left(x^2-2\right)=0\)

    \(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x^2=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm\sqrt{2}\end{cases}}\)

Vậy \(S=\left\{-\sqrt{2};-1;\sqrt{2}\right\}\) 

\(x^3-2x=-x^2+2\)

\(x^3+x^2-2x-2=0\)

\(x^2\left(x+1\right)-2\left(x+1\right)=0\)

\(\left(x+1\right)\left(x^2-2\right)=0\)

\(Th1:x+1=0\Leftrightarrow x=-1\)

\(Th2:x^2-2=0\Leftrightarrow x^2=2\Leftrightarrow x=\pm\sqrt{2}\)

x²+4x-5=0

<=> x²-5x+x-5=0

<=> x(x-5)+(x-5)=0

<=> (x-5)(x+1)=0

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

lollolol là j

\(x^2-4x-5=0\)

\(\Rightarrow x^2-x+5x-5=0\)

\(\Rightarrow x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+5\right)=0\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}=>\orbr{\begin{cases}x=1\\x=-5\end{cases}}}\)

Học tốt

\(\left(4;7\right)\cup\left(-7;-4\right)=\left(-7;7\right)\)

,2x^2-12x+18+2xy-6y
= 2(x^2-6x+9) + 2y(x-3) 
= 2(x-3)^2 + 2y(x-3) 
= (x-3)(2x-6+2y)