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

\(\Leftrightarrow-5x-\frac{1}{5}-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)

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

\(\Leftrightarrow\left(\frac{-10}{2}x-\frac{1}{2}x\right)+\left(\frac{5}{15}-\frac{3}{15}\right)=\frac{3}{2}x-\frac{5}{6}\)

\(\Leftrightarrow\frac{-11}{2}x+\frac{2}{15}=\frac{3}{2}x-\frac{5}{6}\)

\(\Leftrightarrow\frac{-11}{2}x-\frac{3}{2}x=-\frac{5}{6}-\frac{2}{15}\)

\(\Leftrightarrow\frac{-14}{2}x=-\frac{25}{30}-\frac{4}{30}\)

\(\Leftrightarrow-7x=-\frac{29}{30}\)

\(\Leftrightarrow x=-\frac{29}{30}\times\frac{-1}{7}\)

\(\Leftrightarrow x=\frac{29}{210}\)

\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)

\(\Leftrightarrow3x-\frac{3}{2}-5x-3=\frac{1}{5}-x\)

\(\Leftrightarrow\left(3x-5x\right)-\left(\frac{3}{2}+3\right)=\frac{1}{5}-x\)

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

\(\Leftrightarrow-2x-\frac{9}{2}=\frac{1}{5}-x\)

\(\Leftrightarrow-2x+x=\frac{1}{5}+\frac{9}{2}\)

\(\Leftrightarrow-x=\frac{2}{10}+\frac{45}{10}\)

\(\Leftrightarrow-x=\frac{47}{10}\)

\(\Leftrightarrow x=\frac{-47}{10}\)

\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=0-\frac{1}{9}\)

\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\div3\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\times\frac{1}{3}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)

\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)

\(\Leftrightarrow3x=\frac{-1}{3}+\frac{1}{2}\)

\(\Leftrightarrow3x=\frac{-2}{6}+\frac{3}{6}\)

\(\Leftrightarrow3x=\frac{1}{6}\)

\(\Leftrightarrow x=\frac{1}{6}\div3\)

\(\Leftrightarrow x=\frac{1}{6}\times\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{18}\)

Nhiều thế vậy ...!!

Sao làm nổi?

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

b.

ĐKXĐ: \(x\ge2\)

\(\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=2\)

3.

ĐKXĐ: \(x\ge-1\)

\(x^2+x-12+12\left(\sqrt{x+1}-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\dfrac{12\left(x-3\right)}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4+\dfrac{12}{\sqrt{x+1}+2}\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

\(a.2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)

=> 2 / 1/2x - 1/3 / = 1/4 + 3/2 = 7/4

=> / 1/2x - 1/3 / = 7/4 : 2 = 7/8

=> \(\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\)

=> \(\hept{\begin{cases}\frac{1}{2}x=\frac{7}{8}+\frac{1}{3}=\frac{29}{24}\\\frac{1}{2}x=-\frac{7}{8}+\frac{1}{3}=-\frac{13}{24}\end{cases}}\)

Đến đây ez rồi

So sánh x với 1/3 Vx đk  : 60%x + 2/3x = 1/3.6

Ta có :

60%x + 2/3x = 1/3.6

=> 3/5x + 2/3x = 2

=> 19/15x = 2

=> x = 2 : 19/15

=> x = 30/19 = 1 + 11/19 > 1/3

<=> 60%x + 2/3x = 1/3 . 6 > 1/3

1,=\(x^2-3x-2x^2+6x=-x^2+3x\)

2,=\(3x^2-x-5+15x=3x^2+14x-5\)

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

4,=\(4x^2+12x-x-3=4x^2+11x-3\)

5: =>(x+5)^3=0

=>x+5=0

=>x=-5

6: =>(2x-3)^2=0

=>2x-3=0

=>x=3/2

7: =>(x-6)(x-10)=0

=>x=10 hoặc x=6

8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)

=>(x-4)^3=0

=>x-4=0

=>x=4

a, 60%x + 2/3x =1/3.6 1/3

3/5x +2/3x =1/3.19/3

x.(3/5+2/3)=19/9

x.(9/15+10/15)=19/9

x.19/15=19/9

x=19/9:19/15

x=15/9 

Vậy x=15/9

b,3.(3x-1/2)^3 +1/9=0

3.(3x-1/2)^3= -1/9

(3x-1/2)^3= -1/9:3

(3x-1/2)^3= -1/27

(3x-1/2)^3=(-1/3)^3

3x-1/2= -1/3

3x= -1/3-1/2

3x= -2/6+(-3/6)

3x= -5/6

x= -5/6 :3

x=-5/18 

Vậy x=-5/18

a)(9x+2)x3=60

        9x+2 =60:3

        9x+2  =20

              9x =20-2

              9x=18

                x=18:9

                 x=2

a,x=2

b,x=2

c,x=9

o phan a 

(9x+2)*3=60

9x+2=20

9x=18

x=2

o phan b

71+(26-3x)/5=75

(26-3x)/5=75-71=4

26-3x=4*5=20

3x=26-20=6

x=6/3=2

vay x=2

o phan c 

(x-6)^2=9

(x-6)^2=3^2

=>x-6=3

=>x=9

vay x =9

ở phần d 

1+3+.....+x=465

=>(x+1)*(x-1)/2+(x+1)=465

=>(x+1)*(x-1)/2

   2³.19 - 2³.14 + 1²⁰²⁰

= 2³.(19 - 14) + 1

= 8.5 + 1

= 41

    10² - [60: (5⁶: 5⁴ - 3.5)]

= 100 - [60: (5² - 15)]

= 100 - [60: (25 - 15)]

= 100 - [60 : 10]

= 100 - 6

= 94 

   160: {17 + [3².5 - ( 14 + 2⁷:2⁴)]}

= 160: {17 + [9.5 - (14 + 2³)]}

= 160: {17 + [45 - (14 + 8)]}

= 160: {17 + [45 - 22]}

= 160: {17 + 23}

= 160 : 40

= 4