\(\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{5x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

a ) ĐKXĐ : \(x\ne0,x\ne-5\)

b ) Rút gọn trước cái đã

\(\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{5x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+10x^2+50x-10x-50+50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+12x^2+35x}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x+5\right)\left(x+7\right)}{2x\left(x+5\right)}=\dfrac{x+7}{2x}\)

Khi \(A=1\), thì :

\(\dfrac{x+7}{2x}=1\Leftrightarrow x=7\)

Khi A = 3, thì :

\(\dfrac{x+7}{2x}=3\Leftrightarrow x=-1.\)

Bài 2 :

a ) ĐKXĐ : x\(\ne-3;2\)

b ) \(\dfrac{x-2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{2-x}\)

\(=\dfrac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)

c ) Khi \(A=-\dfrac{3}{4}\), thì :

\(\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)

\(\Leftrightarrow4x-16=-3x+6\)

\(\Leftrightarrow x=\dfrac{22}{7}\)

d ) Ta có :

\(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)

Để A nguyên thi \(x-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

Thay vào rồi tìm ra nếu x có trong đkxđ thì loại .

e ) \(x^2-9=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Thay từng x vào A là tìm ra

a, ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne-5\end{matrix}\right.\)

b, \(P=\dfrac{x^2}{5x+25}+\dfrac{2x-10}{x}+\dfrac{50+5x}{x^2+5x}\)

\(=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x-5\right)}{x}+\dfrac{5\left(x+10\right)}{x\left(x+5\right)}\)

\(=\dfrac{x^2\cdot x}{5x\left(x+5\right)}+\dfrac{2\left(x-5\right)\cdot5\left(x+5\right)}{5x\left(x+5\right)}+\dfrac{5\left(x+10\right)\cdot5}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+10\left(x^2-25\right)+25x+250}{5x\left(x+5\right)}\)\(=\dfrac{x^3+10x^2-250+25x+250}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+10x^2+25x}{5x\left(x+5\right)}=\dfrac{x\left(x^2+2\cdot x\cdot5+5^2\right)}{5x\left(x+5\right)}=\dfrac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\dfrac{x+5}{5}\)

c, \(P=-4\Leftrightarrow\dfrac{x+5}{5}=-4\)

\(\Leftrightarrow x+5=-20\Leftrightarrow x=-25\)

d, \(P=\dfrac{x+5}{5}\Rightarrow\dfrac{1}{P}=\dfrac{5}{x+5}\)

\(\Rightarrow\dfrac{1}{P}\in Z\Leftrightarrow5⋮x+5\)

\(\Leftrightarrow x+5\inƯ\left(5\right)\in\left\{\pm1;\pm5\right\}\)

\(\Leftrightarrow x\in\left\{-10;-6;-4;0\right\}\)

e, \(Q=P+\dfrac{x+25}{x+5}=\dfrac{x+5}{5}+\dfrac{x+25}{x+5}\)

=\(\dfrac{\left(x+5\right)^2+5\left(x+25\right)}{5\left(x+5\right)}=\dfrac{x^2+15x+150}{5\left(x+5\right)}=\dfrac{x\left(x+5\right)+10\left(x+5\right)+100}{5\left(x+5\right)}=x+2+\dfrac{50}{x+5}=\left[\left(x+5\right)+\dfrac{50}{x+5}\right]-3\ge2\sqrt{\left(x+5\right)\cdot\dfrac{50}{x+5}}-3=2\sqrt{50}-3=10\sqrt{2}-3\)

Dấu bằng khi \(\left(x+5\right)^2=50\Leftrightarrow x=5\sqrt{2}-5\left(x>0\right)\)

Bài 1:

Gọi bốn số liên tiếp cần tìm là a;a+1;a+2;a+3(Điều kiện: a∈N)

Theo đề bài, ta có:

\(a\cdot\left(a+1\right)+146=\left(a+2\right)\left(a+3\right)\)
\(\Leftrightarrow a^2+a+146=a^2+5a+6\)

\(\Leftrightarrow a^2+a+146-a^2-5a-6=0\)

\(\Leftrightarrow-4a+140=0\)

\(\Leftrightarrow-4a=-140\)

hay a=35(nhận)

Vậy: Bốn số liên tiếp cần tìm là 35;36;37;38

Bài 2:

Ta có: \(N=3\cdot\frac{1}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot5\frac{118}{119}-\frac{5}{117\cdot119}+\frac{8}{39}\)

\(=3\cdot\frac{1}{117\cdot119}-2852\cdot\frac{1}{117\cdot119}-5\cdot\frac{1}{117\cdot119}+\frac{8}{39}\)

\(=\frac{-2854}{117\cdot119}+\frac{8}{39}\)

\(=\frac{-2854}{39\cdot357}+\frac{2856}{39\cdot357}=\frac{2}{20943}\)

a,+b, \(A=\dfrac{4}{x+2}+\dfrac{3}{x-2}-\dfrac{5x-6}{x^2-4}\left(x\ne\pm2\right)\)

\(=\dfrac{4\left(x-2\right)+3\left(x+2\right)-5x+6}{x^2-4}\)

=\(\dfrac{4x-8+3x+6-5x+6}{x^2-4}\)

= \(\dfrac{2x+4}{x^2-4}\)

= \(\dfrac{2}{x-2}\)

c, Thay \(x=-4\) vào A ta được :

\(A=\dfrac{2}{-4-2}=\dfrac{-1}{3}\)

d, Để A có giá trị là số nguyên thì

\(\Leftrightarrow2⋮x-2\)

\(\Leftrightarrow x-2\in U\left(2\right)=\left\{\pm2,\pm1\right\}\)

\(\Rightarrow x-2=2\Rightarrow x=4\)

\(x-2=-2\Rightarrow x=0\)

\(x-2=1\Rightarrow x=3\)

\(x-2=-1\Rightarrow x=1\)

Vậy \(S=\left\{4;0;3;1\right\}\)

Câu 2:

a) \(A=\left(x+5\right)\left(2x-3\right)-2x\left(x+3\right)-\left(x-15\right)\)

\(=x\left(2x-3\right)+5\left(2x-3\right)-2x^2-6x-x+15\)

\(=2x^2-3x+10x-15-2x^2-6x-x+15\)

\(=0\)

b) \(B=2\left(x-5\right)\left(x+1\right)+\left(x+3\right)-\left(x-15\right)\)

\(=2\left[x\left(x+1\right)-5\left(x+1\right)\right]+x+3-x+15\)

\(=2.\left[\left(x^2+x\right)-\left(5x+5\right)\right]+x+3-x+15\)

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

\(=2x^2+2x-10x-10+x+3-x+15\)

\(=2x^2-8x+8\)

\(=2x\left(x-4\right)+8\)

Thay: \(x=\frac{3}{4}\) vào B ta đc:

\(2.\frac{3}{4}\left(\frac{3}{4}-4\right)+8\)

\(=\frac{3}{2}.\frac{-13}{4}+8\)

\(=\frac{25}{8}\)

c) \(C=5x^2\left(3x-2\right)-\left(4x+7\right)\left(6x^2-x\right)-\left(7x-9x^3\right)\)

\(=5x^23x-5x^22-\left[4x\left(6x^2-x\right)+7\left(6x^2-x\right)\right]-7x+9x^3\)

\(=15x^3-10x^2-\left[4x6x^2-4x^2+42x^2-7x\right]-7x+9x^3\)

\(=15x^3-10x^2-24x^3+4x^2-42x^2+7x-7x+9x^3\)

\(=-48x^2\)

P/s: Ko chắc!

Bài 4: 

a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)

\(\Leftrightarrow3x-40=2\)

=>3x=42

hay x=14

b: \(\Leftrightarrow x^3+8-x^3-2x=0\)

=>-2x+8=0

=>-2x=-8

hay x=4

c: \(x\left(x-2\right)+\left(x-2\right)=0\)

=>(x-2)(x+1)=0

=>x=2 hoặc x=-1

d: \(5x\left(x-3\right)-x+3=0\)

=>5x(x-3)-(x-3)=0

=>(x-3)(5x-1)=0

=>x=3 hoặc x=1/5

e: \(3x\left(x-5\right)-\left(x-1\right)\left(3x+2\right)=30\)

\(\Leftrightarrow3x^2-15x-3x^2-2x+3x+2=30\)

=>-14x=28

hay x=-2

f: \(\Leftrightarrow\left(x+2\right)\left(x+30-x-5\right)=0\)

=>x+2=0

hay x=-2

1 .

a) 2155-(174+2155)+(-68+174)

=2155-174-2155-68+174

=(2155-2155)+(-174+174)-68

=-68

b) 35(14-23)-23(14-35)

=35.(-9) - 23.(-21)

=-315+483

=168

c) -1911-(1234-1911)

= -1911-1234+1911

=(-1911+1911)-1234

=-1234

d) 32.(-39)+16.(-22)

=16[2.(-39)-22]

=16(-78-22)

=16.(-100)

=-1600

2. Tìm x

a) 3x+17=2

=>3x=-15

=>x=-5

b) -5x -(-3)=13

=>-5x+3=13

=>-5x=10

=>x=-2

c) 45-(x-9)=-35

=>x-9=80

=>x=89

d) 15-(x-7)= -21

=>x-7=36

=>x=43

e) (7-x).(x+19)=0

=>7-x=0 hoặc x+19=0

+)TH1: 7-x=0=>x=7

+)TH2: x+19=0=>x=-19

P/s: Tự KL nhé

bai1;

a)(3x-1)^2+(x+3).(2x-1)

=3x^2-6x+1+2x^2-1x+6x

=x^2-1x+1

b)(x-2).(x^2+2x+4)-x(x^2-2)

=x^3+2x^2+4x-2x^2-4x-8-x^3+2x

=2x-8

Bài 2:

a. \(x^3-27+3x\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+3x+9\right)+3x\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+3x+9+3x\right)\)

\(=\left(x-3\right)\left(x^2+6x+9\right)\)

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

b. \(5x^3-7x^2+10x-14\)

\(=5x\left(x^2+2\right)-7\left(x^2+2\right)\)

\(=\left(x^2+2\right)\left(5x-7\right)\)