a)\(\left(x+8\right)-11=20-15\)

\(\left(x+8\right)-11=5\)

\( x+8=5+11\)

\(x+8=16\)

\(x=8\)

b) \(2x-\left(3+x\right)=5-7\)

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

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

\(x=1\)

c) \( \left(3x-2^4\right)\times7^5=2\times7^6\)

\(3x-2^4=2\times\left(7^6:7^5\right) \)

\(\left(3x-2^4\right)=2\times7^2\)

\(3x-2^4=2\times49\)

\(3x-16=98\)

\(3x=114\)

\(x=38\)

(2x+1)(y-5)=12

Vì x,y \(\in N\)

=> 2x+1;y-5 \(\in N\)

=> 2x+1, y-5 \(\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

Vì 2x+1 là số lẻ => \(2x+1\in\left\{\pm1;\pm3\right\}\)

Xét bảng

Vậy các cắp (x,y) tm là (0;17), (1;9)

cảm ơn bn nha

\(\left(2x+1\right)\cdot\left(y-5\right)=12\)

<=>\(x=\frac{17-y}{2y-10}\)

thay x vào phương trình 

=>\(\left(\frac{17-y+y-5}{y-5}\right)\cdot\left(y-5\right)=12\)

<=>\(\frac{12}{y-5}\cdot\left(y-5\right)=12\)

<=>\(12=12\)(Luôn đúng khi và chỉ khi y khác 5 )\(y\ne5,y\inℝ\)

giả sử thay y=1 ta có 

=>\(2x=\frac{12}{1-5}-1\)

<=>\(2x=-4\)

=>\(x=-2\)

Vậy \(x=-2\)và \(y=1\)

5.(3x+8)-7.(2x+3)=16

=> 15x+40-14x-21=16

=>(15x-14x)+(40-21)

=> x+19=16

=>x=-3. Vậy x=-3

k cho mik nha

Chúc bạn hok tốt

\(5\left(3x+8\right)-7\left(2x+3\right)=16\)

\(15x+40-14x-21=16\)

\(x+19=16\)

\(x+19-16=0\)

\(x+3=0\)

\(x=-3\)

a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\) (1)

\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)=16\)

\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2+x-5\right)=16\)

\(\Leftrightarrow6x^2+21x-2x-7-6x^2-x+5=16\)

\(\Leftrightarrow18x-2=16\)

\(\Leftrightarrow18x=16+2\)

\(\Leftrightarrow18x=18\)

\(\Leftrightarrow x=1\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{1\right\}\)

b) \(\left(10x+9\right)\cdot x-\left(5x-1\right)\left(2x+3\right)=8\) (2)

\(\Leftrightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)

\(\Leftrightarrow10x^2+9x-\left(10x^2+13x-3\right)=8\)

\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)

\(\Leftrightarrow-4x+3=8\)

\(\Leftrightarrow-4x=8-3\)

\(\Leftrightarrow-4x=5\)

\(\Leftrightarrow x=-\dfrac{5}{4}\)

Vậy tập nghiệm phương trình (2) là \(S=\left\{-\dfrac{5}{4}\right\}\)

c) \(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\) (3)

\(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

\(\Leftrightarrow42x-41=0\)

\(\Leftrightarrow42x=41\)

\(\Leftrightarrow x=\dfrac{41}{42}\)

Vậy tập nghiệm phương trình (3) là \(S=\left\{\dfrac{41}{42}\right\}\)

d) \(x\left(x+1\right)\left(x+6\right)-x^3=5x\) (4)

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

\(\Leftrightarrow x^3+6x^2+x^2+6x-x^3=5x\)

\(\Leftrightarrow7x^2+6x=5x\)

\(\Leftrightarrow7x^2+6x-5x=0\)

\(\Leftrightarrow7x^2+x=0\)

\(\Leftrightarrow x\left(7x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\7x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{7}\end{matrix}\right.\)

Vậy tập nghiệm phương trình (4) là \(S=\left\{-\dfrac{1}{7};0\right\}\)

a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)

A = \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}\)...\(\dfrac{9999}{10000}\)

A = \(\dfrac{1.3.2.4..3.5......99.101}{2.2.3.3.4.4....100.100}\)

A = \(\dfrac{1.2.3..4.5.....99}{2.3.4.5.....99.100}\).\(\dfrac{3.4.5....100.101}{2.3.4.5...100}\)

A = \(\dfrac{1}{100}\).\(\dfrac{101}{2}\)

A = \(\dfrac{101}{200}\)

2; B = (1 - \(\dfrac{1}{2}\)).(1 - \(\dfrac{1}{8}\))...(1 - \(\dfrac{1}{n+1}\))

   Xem lại đề bài.

a) 5x.(x+3/4) = 0

=> x = 0

x+3/4 = 0 => x = -3/4

b) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}.\)

\(\Rightarrow\frac{x+7}{2010}+\frac{x+6}{2011}-\frac{x+5}{2012}-\frac{x+4}{2013}=0\)

\(\frac{x+7}{2010}+1+\frac{x+6}{2011}+1-\frac{x+5}{2012}-1-\frac{x+4}{2013}-1=0\)

\(\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)-\left(\frac{x+5}{2012}+1\right)-\left(\frac{x+4}{2013}+1\right)=0\)

\(\frac{x+2017}{2010}+\frac{x+2017}{2011}-\frac{x+2017}{2012}-\frac{x+2017}{2013}=0\)

\(\left(x+2017\right).\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)

=> x + 2017 = 0

x = -2017

a) để 2x - 3 > 0

=> 2x > 3

x > 3/2

b) 13-5x < 0

=> 5x < 13

x < 13/5

c) \(\frac{x+3}{2x-1}>0\)

=> x + 3 > 0

x > -3

d) \(\frac{x+7}{x+3}=\frac{x+3+4}{x+3}=1+\frac{4}{x+3}\)

Để x+7/x+3 < 1

=> 1 + 4/x+3 < 1

=> 4/x+3 < 0

=> không tìm được x thỏa mãn điều kiện