\(\frac{x+1}{15}+\frac{x+2}{14}=\frac{x+3}{12}+\frac{x+4}{13}\)

\(\Leftrightarrow\frac{x+1}{15}+\frac{x+2}{14}-\frac{x+3}{12}-\frac{x+4}{13}=0\)

\(\Leftrightarrow\frac{x+1}{15}+1+\frac{x+2}{14}+1-\frac{x+3}{12}+1-\frac{x+4}{13}+1=0\)

\(\Leftrightarrow\frac{x+16}{15}+\frac{x+16}{14}-\frac{x+13}{12}-\frac{x+16}{13}=0\)

\(\Leftrightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{13}\right)=0\)

\(\Leftrightarrow x=-16\) (vì \(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{13}>0\))

Vậy: \(S=\left\{-16\right\}\)

a, MD là tia phân giác \(\Delta ABM\)

=> \(\frac{AD}{BD}=\frac{AM}{BM}\) (1)

ME là tia phân giác \(\Delta ACM\)

=> \(\frac{AE}{CE}=\frac{AM}{MC}\) (2)

AM là đường trung tuyến

=> MB = MC

=> \(\frac{AM}{BM}=\frac{AM}{MC}\)

Ta lét đảo => \(DE//BC\)

\(a,PT\Leftrightarrow8x^3-6x^2+4x-3=3x^3-36x^2+x-12\)

\(\Leftrightarrow5x^3+30x^2+3x+9=0\)

\(\Leftrightarrow x=-5,95...\)

\(b,PT\Leftrightarrow2x+22-3x^2-33x=6x-15x^2-4+10x\)

\(\Leftrightarrow12x^2-47x+26=0\)

<=> (3x - 2)(4x - 13) = 0

<=> x = 2/3 hoặc x = 13/4

c, Tách ra <=> (2x - 1)(2x - 5) = 0 <=> ...

4x⁴ - 21 x²y² + y⁴ 

= (4x⁴ + 4x²y² + y⁴) - 25x²y² 

=  [(2x²)² + 2x² . 2 . y² + (y²)²] - 25x²y²

= (2x² + y²) - 25x²y² 

= (2x² + y² - 5xy) (2x² + y² + 5xy)

4x⁴ - 21 x²y² + y⁴ 

= (4x⁴ + 4x²y² + y⁴) - 25x²y² 

=  [(2x²)² + 2x² . 2 . y² + (y²)²] - 25x²y²

= (2x² + y²) - 25x²y² 

= (2x² + y² - 5xy) (2x² + y² + 5xy)

Bạn tự vẽ hình.

a, Áp dụng định lí pitago vào \(\Delta ABC\left(\widehat{A}=90^o\right)\), từ đó tính được \(AC=8cm\)

Dễ dàng chứng minh được \(\Delta ABC~\Delta HBA\left(g.g\right)\)

=> \(\frac{BC}{BA}=\frac{AC}{HA}\)

Từ đó tính được \(AH=4,8cm\)

b, Chứng minh được \(\Delta EAD~\Delta EDB\left(g.g\right)\)

=> \(\frac{EA}{EC}=\frac{ED}{EB}\)

=> \(EA.EB=ED.EC\)

\(\Delta EAD,\Delta ECB:\)

\(\hept{\begin{cases}\widehat{E}:chung\\\frac{EA}{EC}=\frac{ED}{EB}\end{cases}}\)

=> \(\Delta EAD~\Delta ECB\left(c.g.c\right)\)

c, Chứng minh được \(\hept{\begin{cases}\Delta BDF~\Delta BED\left(g.g\right)\\\Delta EDF~\Delta EBD\left(g.g\right)\end{cases}}\)

=> \(\hept{\begin{cases}\frac{BD}{BF}=\frac{BE}{BD}\\\frac{DE}{EF}=\frac{BE}{DE}\end{cases}}\)

=> \(\hept{\begin{cases}BD^2=BE.BF\\DE^2=BE.EF\end{cases}}\)

=> \(\left(\frac{BD}{DE}\right)^2=\frac{BF.BE}{EF.BE}=\frac{BF}{FE}\)

a, 7 - x = -2x + 3

<=> 7 - x + 2x - 3 = 0

<=> x + 4 = 0 

<=> x = -4

Vậy....

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

<=> 6x + 2 + 2x - 5 = 0

<=> 8x - 3 = 0

<=> 8x = 3

<=> x = 3/8

Vậy....

c, 5x + 2(x-1) = 4x + 7

<=> 5x + 2x - 2 - 4x - 7 = 0

<=> 3x - 9 = 0

<=> 3x = 9

<=> x = 3

Vậy....

d, 10x² - 5x(2x + 3 ) = 15

<=> 10x² - 10x² - 15x = 15

<=> -15x = 15

<=> x = -1

Vậy....

e, \(\frac{2}{5}x-\frac{1}{10}=x+2\)

\(\Leftrightarrow\)\(\frac{2}{5}x-\frac{1}{10}-x-2=0\)

\(\Leftrightarrow\)\(\frac{-3}{5}x-\frac{21}{10}=0\)

\(\Leftrightarrow\)\(\frac{-3}{5}x=\frac{21}{10}\)

\(\Leftrightarrow\)\(x=\frac{-7}{2}\)

Vậy....

f, \(\frac{1}{2}\left(2x+3\right)+\frac{1}{3}x=\frac{1}{12}\)

\(\Leftrightarrow\)\(x+\frac{3}{2}+\frac{1}{3}x-\frac{1}{12}=0\)

\(\Leftrightarrow\)\(\frac{4}{3}x+\frac{17}{12}=0\)

\(\Leftrightarrow\)\(\frac{4}{3}x=\frac{-17}{12}\)

\(\Leftrightarrow\)\(x=\frac{-17}{16}\)

Vậy.....