a, \(3\left(x-1\right)-\left(2x+3\right)=5\)

\(\Rightarrow3x-3-2x-3=5\)

\(\Rightarrow x-6=5\)

\(\Rightarrow x=11\)

b, \(\frac{6}{13}=\frac{x}{7}\)

\(\Rightarrow x=\frac{6}{13}.7\)

\(\Rightarrow x=\frac{42}{13}\)

c, \(|x|=4\)

\(\Rightarrow x=4\)hoặc \(x=-4\)

d,, \(|x-2|+5=10\)

\(Th1:x-2>0\)

\(\Rightarrow x-2+5=10\)

\(\Rightarrow x+3=10\)

\(\Rightarrow x=7\)

\(Th2;x-2< 0\)

\(\Rightarrow2-x+5=10\)

\(\Rightarrow7-x=10\)

\(\Rightarrow x=-3\)

e, \(3|x-1|-6=9\)

\(Th1:x-1>0\)

\(\Rightarrow3\left(x-1\right)-6=9\)

\(\Rightarrow3x-3-6=9\)

\(\Rightarrow3x-9=9\)

\(\Rightarrow3x=18\)

\(\Rightarrow x=6\)

\(Th2:x-1< 0\)

\(\Rightarrow3\left(1-x\right)-6=9\)

\(\Rightarrow3-3x-6=9\)

\(\Rightarrow-3x-3=9\)

\(\Rightarrow-3x=12\)

\(\Rightarrow x=-4\)

sửa đề

\(a,5^x-3^2=\left(2^2\right)^2\)

\(\Rightarrow5^x-9=16\)

\(\Rightarrow5^x=16+9\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)

a) \(5^x-3^2=\left(2^2\right)^2\)(bài a sữa dấu cộng thành dấu trừ mới đúng nha)

\(5^x-9=16\)

\(5^x=25\)

\(5^x=5^2\)

\(x=2\)

b) \(50-7^x=1^9\)

\(50-7^x=1\)

\(7^x=49\)

\(7^x=7^2\)

\(x=2\)

c) \(15.3^x-10^2=45\)

\(15.3^x-100=45\)

\(15.3^x=145\)

\(3^x=\frac{29}{3}\)(đề sai)

d)\(2^3+3.2^x=56\)

\(8+3.2^x=56\)

\(3.2^x=48\)

\(2^x=16\)

\(2^x=2^4\)

\(x=4\)

e) \(5^4-2^3.5^x=5^2\)

\(625-8.5^x=25\)

\(8.5^x=600\)

\(5^x=75\)(Đề sai)

\(bbb:ab=a.b\)

\(\Rightarrow bbb:b:ab=a\)

\(\Rightarrow111:ab=a\)

\(\Rightarrow a=7\)

\(\Rightarrow b=777\)

Vậy \(\Rightarrow ab=37\)

mk làm qua rồi nhưng lâu ko làm nên ko chắc sẽ đúng ^^

có nè\

bạn Nhữ Thị Ngọc MaI có thể đao về cho mk đc ko

Bài 1

\(\frac{2017}{2018}+\frac{2018}{2019}\)và \(\left(\frac{2017+2018}{2018+2019}\right)\)mk chữa lại đề luôn đó 

Ta tách :

\(\frac{2017}{\left(2018+2019\right)+2018}\)

đến đây ta tách 

\(\frac{2017}{2018+2019}< \frac{2017}{2018}\)

vậy....

mấy câu khác tương tự 

2) \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{2}{2003}+\frac{2}{2004}+\frac{2}{2005}}\)

= \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{2.\frac{1}{2003}+2.\frac{1}{2004}+2.\frac{1}{2005}}\)

=\(\frac{1\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}{2.\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}\)

= \(\frac{1}{2}\)

3) \(2013+\left(\frac{2013}{1+2}\right)+\left(\frac{2013}{1+2+3}\right)+...+\left(\frac{2013}{1+2+3+...+2012}\right)\)

= \(2013.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}\right)\)

= \(2013.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{2025078}\right)\)

= \(2013.2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4050156}\right)\)

=\(4026.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)

= \(4026.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)

= \(4026.\left(1-\frac{1}{2013}\right)\)

= \(4026.\frac{2012}{2013}\)

=\(4024\)

TL:

a. Số 0

b. Số 1

hc tốt

\(\left(2+x\right).\left(7-x\right)=0\)

\(TH1:2+x=0\)

\(x=0-2\)

\(x=-2\)

\(TH2:7-x=0\)

\(x=7-0\)

\(x=7\)

Để ( 2 + x ) . ( 7 - x ) = 0 thì :

+) 2 + x = 0

          x = 0 - 2 

         x = -2

+) 7 - x = 0

         x = 7 - 0

         x = 7

_Vậy x \(\in\){ - 2 ; 7 }