a) 3 x 1/5 + 2/5 = 3/5 + 2/5 = 1

b) 3 : 1/5 - 2/5 = 15 - 2/5 = 73/5

bài 2 :
a) X x 1/2 = 1 - 1/5

    X x 1/2 = 4/5 

           X   = 4/5 : 1/2

           X   = 8/5 

vậy x =...

b) x : 1/3 = 1 + 1/5

    x : 1/3 = 6/5

          x   = 6/5 x 1/3

          x   = 2/5

vậy x =...

a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)

a)x.(5-2x)-2x.(1-x)=15
   x [ 5 - 2x -2.(1-x) ] = 15
   x ( 5 - 2x -2 + 2x ) =15
   x . 3 =15
   x = 5
b)(3x+2)²+(1+3x).(1-3x)=2
   9x²+12x+4+1-9x²=2
   12x + 5 = 2
    12x = -3
        x = -1/4

a)\(\Leftrightarrow\)\(5x-2x^2-2x+2x^2=15\)

\(\Leftrightarrow\)\(3x=15\)

\(\Leftrightarrow\)\(x=5\)

b)\(\Leftrightarrow\)\(9x^2+12x+4+1-9x^2-2=0\)

\(\Leftrightarrow\)\(12x+3=0\)

\(\Leftrightarrow\)\(x=-0,25\)

a) \(x^2-x+x=4\)

\(x^2=4\)

\(x=\pm2\)

b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)

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

\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

c) Ta có: \(a+b+c=5-3-2=0\)

\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)

d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :

\(t^2-11t+18=0\)

\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)

\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)

\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)

a.

$x^4-25x^3=0$

$\Leftrightarrow x^3(x-25)=0$

\(\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.\)

b.

$(x-5)^2-(3x-2)^2=0$

$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$

$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix} -2x-3=0\\ 4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{7}{4}\end{matrix}\right.\)

c.

$x^3-4x^2-9x+36=0$

$\Leftrightarrow x^2(x-4)-9(x-4)=0$

$\Leftrightarrow (x-4)(x^2-9)=0$

$\Leftrightarrow (x-4)(x-3)(x+3)=0$

\(\Leftrightarrow \left[\begin{matrix} x-4=0\\ x-3=0\\ x+3=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=4\\ x=3\\ x=-3\end{matrix}\right.\)

d. ĐK: $x\neq 0$

$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$

$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$

$\Leftrightarrow -2(-x^2+3x-4)=0$

$\Leftrightarrow x^2-3x+4=0$

$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)

Vậy pt vô nghiệm.

làm ơn giúp 🙏🙏🙏

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9

a)3(x-2)+2(x-3)=5

=>3x-6+2x-6=5

=>5x=17

=>x=17/5

b)(2x-8)^2=16

TH1:2x-8=4=>x=6

TH2:2x-8=-4=>x=2

\(a,\Leftrightarrow3x-6+2x-6=5\\ \Leftrightarrow5x-12=5\\ \Leftrightarrow x=\dfrac{17}{5}\\ b,\left(2x-8\right)^2-16=0\\ \Leftrightarrow\left(2x-12\right)\left(2x-4\right)=0\\ \Leftrightarrow4\left(x-6\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-4x+1-4x^2+11x+3=3\\ \Leftrightarrow7x=-1\\ \Leftrightarrow x=-\dfrac{1}{7}\)