a/ đk: x khác -7/5 ; x khác -1/5

pt <=> \(\dfrac{\left(3x+2\right)\left(5x+1\right)}{\left(5x+7\right)\left(5x+1\right)}=\dfrac{\left(3x-1\right)\left(5x+7\right)}{\left(5x+7\right)\left(5x+1\right)}\)

\(\Rightarrow15x^2+13x+2=15x^2+16x-7\)

\(\Leftrightarrow15x^2+13x-15x^2-16x^2=-7-2\)

\(\Leftrightarrow-3x=-9\Leftrightarrow x=3\left(tm\right)\)

vậy x = 3

b/ đk: x khác -1/2; x khác -3

pt <=> \(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(2x+1\right)\left(x+3\right)}=\dfrac{\left(0,5x+2\right)\left(2x+1\right)}{\left(2x+1\right)\left(x+3\right)}\)

\(\Rightarrow x^2+4x+3=x^2+4,5x+2\)

\(\Leftrightarrow x^2+4x-x^2-4,5x=2-3\)

\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\left(tm\right)\)

vậy x = 2

a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}=\dfrac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x+1\right)}=\dfrac{3}{6}=\dfrac{1}{2}\)

\(\Rightarrow2\left(3x+2\right)=5x+7\)

\(\Rightarrow6x+4=5x+7\)

\(\Leftrightarrow x=3\)

Vậy x = 3

b) Ta có: \(\dfrac{0,5x+2}{x+3}=\dfrac{2\left(0,5x+2\right)}{2\left(x+3\right)}=\dfrac{x+4}{2x+6}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x+1}{2x+1}=\dfrac{0,5x+2}{x+3}=\dfrac{x+4}{2x+6}=\dfrac{\left(x+4\right)-\left(x+1\right)}{\left(2x+6\right)-\left(2x+1\right)}=\dfrac{3}{5}\)

\(\Rightarrow5\left(x+1\right)=3\left(2x+1\right)\)

\(\Rightarrow5x+5=6x+3\)

\(\Leftrightarrow x=2\)

\(\dfrac{3x+2}{5x-7}=\dfrac{3x-1}{5x+1}\)

\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x-7\right)\)

\(\Leftrightarrow15x^2+3x+10x+2=15x^2-21x-5x-7\)

\(\Leftrightarrow15x^2+13x+2=15x^2-26x-7\)

\(\Leftrightarrow15x^2-15x^2-13x-2=26x-7\)

\(\Leftrightarrow-13x-2=26x-7\)

\(\Leftrightarrow26x+13x=7+2\)

\(\Leftrightarrow39x=9\Leftrightarrow x=\dfrac{3}{13}\)

b tương tự

mn ơi giúp nhé

a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10

=1/2x^5-3x^4-5x^3

b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x

=-15x^5+12x^4-9x^3+9x^2

c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)

d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)

a: \(\dfrac{31-2x}{x+23}=\dfrac{9}{4}\)

=>121-8x=9x+207

=>-17x=86

hay x=-86/17

b: \(\dfrac{\left|2x-1\right|}{\dfrac{1}{2}}=\dfrac{18}{5}\)

=>|2x-1|=9/5

=>2x-1=9/5 hoặc 2x-1=-9/5

=>2x=14/5 hoặc 2x=-4/5

=>x=7/5 hoặc x=-2/5

a: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)

\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)

\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)

=>16x-7=13x+2

=>3x=9

hay x=3

b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)

\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)

=>x+2017=0

hay x=-2017

e: \(\left(2x-3\right)^2=144\)

=>2x-3=12 hoặc 2x-3=-12

=>2x=15 hoặc 2x=-9

=>x=15/2 hoặc x=-9/2

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

=>x+1=2 hoặc x+1=-2

=>x=1 hoặc x=-3

2: \(\Leftrightarrow7x-21=5x+25\)

=>2x=46

=>x=23

3: \(\Leftrightarrow x^2+4x+3=x^2+0.5x+4x+2\)

=>4,5x+2=4x+3

=>x=1

a) \(5^{x+3}+5^x=126\)

\(\Rightarrow5^x\cdot\left(5^3+1\right)=126\)

\(\Rightarrow5^x\cdot\left(125+1\right)=126\)

\(\Rightarrow5^x=\dfrac{126}{126}\)

\(\Rightarrow5^x=1\)

\(\Rightarrow5^x=5^0\)

\(\Rightarrow x=0\)

b) \(\dfrac{9-3x}{2}=\dfrac{5-2x}{3}\)

\(\Rightarrow\dfrac{3\cdot\left(9-3x\right)}{6}=\dfrac{2\cdot\left(5-2x\right)}{6}\)

\(\Rightarrow3\cdot\left(9-3x\right)=2\cdot\left(5-2x\right)\)

\(\Rightarrow27-9x=10-4x\)

\(\Rightarrow-4x+9x=27-10\)

\(\Rightarrow5x=17\)

\(\Rightarrow x=\dfrac{17}{5}\)

c) \(7-4\left(x+1\right)=5\)

\(\Rightarrow4\left(x+1\right)=7-5\)

\(\Rightarrow4\left(x+1\right)=2\)

\(\Rightarrow x+1=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}-1\)

\(\Rightarrow x=-\dfrac{1}{2}\)