a: =>10|x-2|=50

=>|x-2|=5

=>x-2=5 hoặc x-2=-5

=>x=7 hoặc x=-3

b: =>|3x-12|+|5x-20|=56

=>8|x-4|=56

=>|x-4|=7

=>x-4=7 hoặc x-4=-7

=>x=11 hoặc x=-3

   ĐKXĐ: \(x\ne0\) 

\(\frac{7x}{10}=\frac{14}{x}\Rightarrow7x^2=140\Rightarrow x^2=20\Rightarrow\orbr{\begin{cases}x=2\sqrt{5}\\x=-2\sqrt{5}\end{cases}}\)

Vậy ............................... chúc bn hok tốt

\(\frac{7x}{5}+\frac{14}{x}\)

=>\(7x.x=5.14\)

=>\(7x^2=70\)

=>\(x^2=70:7=10\)

=>\(x=\sqrt{x}\)

7x/5=14/x

=>7x*x=14*5

7*x*x=70

x*x=70/7=10

x^2=10

khó + lười + nhiều = không làm

Hello

ko thích làm

a)  \(7x^2-16x=2x^3-56\)

\(\Leftrightarrow\)\(2x^3-7x^2+16x-56=0\)

\(\Leftrightarrow\)\(2x\left(x^2+8\right)-7\left(x^2+8\right)=0\)

\(\Leftrightarrow\)\(\left(2x-7\right)\left(x^2+8\right)=0\)

\(\Leftrightarrow\)\(2x-7=0\)

\(\Leftrightarrow\)\(x=3,5\)

Vậy...

b)  \(x^7+x^3+2x^5+2x=0\)

\(\Leftrightarrow\)\(x.\left(x^6+x^2+2x^4+2\right)=0\)

\(\Leftrightarrow\)\(x\left(x^2+2\right)\left(x^4+1\right)=0\)

\(\Leftrightarrow\)\(x=0\)

Vậy...

c)  \(\left(2x+1\right)x-5\left(x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\)\(2x\left(x+\frac{1}{2}\right)-5\left(x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\)\(\left(2x-5\right)\left(x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-5=0\\x+\frac{1}{2}=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2,5\\x=-0,5\end{cases}}\)

Vậy...

\(-5.\left(2-x\right)+4\left(x-3\right)=10x+15\)

       \(-10+5x+4x-12=10x+15\)

                                \(9x-22=10x+15\)

                              \(10x-9x=-22-15\)

                                              \(x=-37\)

\(7.\left(x-9\right)-5\left(6-x\right)=-6+11x\)

       \(7x-63-30+5x=-6+11x\)

                          \(12x-93=-6+11x\)

                        \(12x-11x=-6+93\)

                                          \(x=87\)

\(xy+14+2y+7x=-10\)

\(\left(xy+2y\right)+\left(14+7x\right)=-10\)

\(y\left(x+2\right)+7\left(2+x\right)=-10\)

\(\left(x+2\right)\left(y+7\right)=-10\)

Câu cuối mk chỉ biết làm đến đó thôi 

bn tự làm nha

Ta có : \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\) \(\frac{1}{x^2+15x+56}=\frac{1}{14}\)

<=>\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)+...+ \(\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)

<=> \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}\)= \(\frac{1}{14}\)

<=> \(\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)

<=> \(\frac{x+8-x-1}{\left(x+1\right)\left(x+8\right)}=\frac{1}{14}\)

<=>\(\frac{7.14}{14\left(x+1\right)\left(x+8\right)}=\frac{\left(x+1\right)\left(x+8\right)}{14\left(x+1\right)\left(x+8\right)}\)

<=> \(x^2+9x+8=98\)<=> \(x^2+9x-90=0\)

<=> (x-6)(x+15) =0 

<=> \(\orbr{\begin{cases}x=6\\x=-15\end{cases}}\)

Vậy phương trình có 2 nghiệm  x  \(\in\left(6,15\right)\)

==============

- Do ko biết viết dấu ngoặc nhọn nên thay = dấu ngoặc tròn

- Đề ko rõ ràng , lần sau nhớ ghi yêu cầu ?  

Tìm x:

\(a.\dfrac{35}{28}-x=\dfrac{5}{14}\\ x=\dfrac{35}{28}-\dfrac{5}{14}\\ x=\dfrac{25}{28}\\ b.\dfrac{6}{7}\times x=\dfrac{2}{3}\\ x=\dfrac{2}{3}:\dfrac{6}{7}\\ x=\dfrac{7}{9}\\ c.x\times7=\dfrac{3}{4}\\ x=\dfrac{3}{4}:7\\ x=\dfrac{3}{28}\\ d.2:x-\dfrac{1}{3}=\dfrac{2}{5}\\ 2:x=\dfrac{2}{5}+\dfrac{1}{3}\\ 2:x=\dfrac{11}{15}\\ x=2:\dfrac{11}{15}\\ x=\dfrac{30}{11}.\)

\(a,\dfrac{35}{28}-x=\dfrac{5}{14}\)

\(x=\dfrac{35}{28}-\dfrac{5}{14}\)

\(x=\dfrac{25}{28}\)

\(b,\dfrac{6}{7}\times x=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}:\dfrac{6}{7}\)

\(x=\dfrac{7}{9}\)

\(c,x\times7=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}:7\)

\(x=\dfrac{3}{28}\)

\(d,2:x-\dfrac{1}{3}=\dfrac{2}{5}\)

\(2:x=\dfrac{2}{5}+\dfrac{1}{3}\)

\(2:x=\dfrac{11}{15}\)

\(x=2:\dfrac{11}{15}\)

\(x=\dfrac{30}{11}\)

#YVA

2x^2 + x - 6

= 2x^2 + 4x - 3x - 6

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

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

7x^2 + 50x + 7 

= 7x^2 + x + 49x + 7

= 7x(x + 7) + x + 7

= (7x + 1)(x + 7)

12x^2 + 7x - 12

15x^2 +  7x - 2

= 15x^2 - 3x + 10x - 2

= 3x(5x - 1) + 2(5x - 1) 

= (3x + 2)(5x - 1)

a^2 - 5a - 14

= a^2 + 2a - 7a - 14

= a(a + 2) - 7(a + 2)

= (a - 7)(a + 2)

2x^2 + 5x + 2

= 2x^2 + x + 4x + 2

= 2x(x + 2) + x + 2

= (2x + 1)(x + 2)

\(2x^2+x-6=2x^2+4x-3x-6\)

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

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

\(7x^2+50x+7\)

\(=7x^2+x+49x+7\)

\(=x\left(7x+1\right)+7\left(7x+1\right)\)

\(=\left(7x+1\right)\left(x+7\right)\)

\(12x^2+7x-12\)

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

\(=4x\left(3x+4\right)-3\left(3x+4\right)\)

\(=\left(3x+4\right)\left(4x-3\right)\)

\(4p^2-36p+56\)

\(=4p^2+8p+28p+56\)

\(=4p\left(p+2\right)+28\left(p+2\right)\)

\(=\left(p+2\right)\left(4p+28\right)\)