\(6-2\left(x-1\right)=4\)

\(\Rightarrow2\left(x-1\right)=6-4\)

\(\Rightarrow2\left(x-1\right)=2\)

\(\Rightarrow x-1=1\)

\(\Rightarrow x=1+1=2\)

________________

\(2\cdot\left(x-2\right)+1=7\)

\(\Rightarrow2\cdot\left(x-2\right)=7-1\)

\(\Rightarrow2\cdot\left(x-2\right)=6\)

\(\Rightarrow x-2=3\)

\(\Rightarrow x=3+2=5\)

_______________

\(\left(2\cdot x-3\right)+4=9\)

\(\Rightarrow2\cdot x-3=5\)

\(\Rightarrow2\cdot x=3+5\)

\(\Rightarrow2\cdot x=8\)

\(\Rightarrow x=\dfrac{8}{2}=4\)

________________

\(\left(3\cdot x-2\right)-1=3\)

\(\Rightarrow3\cdot x-2=3+1\)

\(\Rightarrow3\cdot x-2=4\)

\(\Rightarrow3\cdot x=6\)

\(\Rightarrow x=\dfrac{6}{3}=2\)

a: =>2(x-1)=2

=>x-1=1

=>x=2

b: =>2(x-2)=6

=>x-2=3

=>x=5

c; =>2x-3=5

=>2x=8

=>x=4

d: =>3x-2=4

=>3x=6

=>x=2

e: =>2(6-x)=4

=>6-x=2

=>x=4

f: =>x-2=5

=>x=7

g: =>10-2x=4

=>2x=6

=>x=3

h: =>2x+4=3

=>2x=-1

=>x=-1/2

j: =>x+2=12

=>x=10

l: =>2x+3=3

=>2x=0

=>x=0

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

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

\(\left(4+2x\right)-1=8\\ 3\left(4+2x\right)=8+1=9\\ 4+2x=\dfrac{9}{3}=3\\ 2x=3-4=-1\\ x=-\dfrac{1}{2}\)

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

\(20+\left(x-2\right)=25\\ x-2=25-20=5\\ x=5+2=7\)

\(\left(x+2\right)-5=7\\ x+2=7+5=12\\ x=12-2=10\)

\(\left(2x-3\right)+4=9\\ 2x-3=9-4=5\\ 2x=5+3=8\\ x=\dfrac{8}{2}=4\)

\(\left(10-\dfrac{2}{x}\right)-1=3\\ 10-\dfrac{2}{x}=3+1=4\\ \dfrac{2}{x}=10-4=6\\ x=\dfrac{2}{6}=\dfrac{1}{3}\)

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

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

1) x = 4/5 - 1/3

x = 7/15

2) 5/3.x=1/21

x=1/35

3) -12/13.x = 1/13

x=-1/12

7) th1: x-1=2/3

x = 5/3

Th2: x - 1 = -2/3

x=1/3

\(\left(2x-1\right)^3-4x^2\left(2x-3\right)=5.\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)

\(\Leftrightarrow6x-1=5\)

\(\Leftrightarrow6x=6\)

\(\Leftrightarrow x=1\)

\(\left(x+4\right)^3-x^2\left(x+12\right)=15.\)

\(\Leftrightarrow x^3+12x^2+48x+64-x^3-12x^2=15\)

\(\Leftrightarrow48x+64=15\)

\(\Leftrightarrow48x=-49\)

\(\Leftrightarrow x=\frac{-49}{48}\)

a: \(\dfrac{96}{\left(x-4\right)\left(x+4\right)}+\dfrac{7+x}{4-x}=\dfrac{2x-1}{x+4}-3\)

\(\Leftrightarrow\dfrac{96}{\left(x-4\right)\left(x+4\right)}-\dfrac{\left(x+7\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\dfrac{3\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)

Suy ra: \(96-x^2-11x-28=2x^2-9x+4-3\left(x^2-16\right)\)

\(\Leftrightarrow-x^2-11x+68=2x^2-9x+4-3x^2+48\)

\(\Leftrightarrow-x^2-11x+68=-x^2-9x+52\)

=>-11x+68=-9x+52

=>-2x=-16

hay x=8(nhận)

b: \(\dfrac{2}{x-1}+\dfrac{3}{x-2}=\dfrac{3}{x-3}\)

\(\Leftrightarrow2\left(x-2\right)\left(x-3\right)+3\left(x-1\right)\left(x-3\right)=3\left(x-1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(x^2-5x+6\right)+3\left(x^2-4x+3\right)=3\left(x^2-3x+2\right)\)

\(\Leftrightarrow2x^2-10x+12+3x^2-12x+9=3x^2-9x+6\)

\(\Leftrightarrow5x^2-22x+21-3x^2+9x-6=0\)

\(\Leftrightarrow2x^2-13x+15=0\)

\(\Leftrightarrow2x^2-10x-3x+15=0\)

=>(x-5)(2x-3)=0

=>x=5(nhận) hoặc x=3/2(nhận)

Vậy \(( - 32{x^5} + 1):( - 2x + 1) = 16{x^4} + 8{x^3} + 4{x^2} + 2x + 1\).

1) 2^{2x + 1} = 32

=> 2^{2x + 1} = 2⁵

=> 2x + 1 = 5

=> 2x = 5 - 1

=> 2x = 4

=> x = 2

(2) 3.x³ - 100 = 275

=> 3x³ = 275 + 100

=> 3x³ = 375

=> x³ = 375 : 3

=> x³ = 125

=> x³ = 5³

=> x = 5

(4) (x - 1)³ - 2⁵ = 7²

=> (x - 1)³ = 49 + 32

=> (x - 1)³ = 81

(xem lại đề)

5) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

  \(\frac{x}{3}=\frac{y}{5}=\frac{x-y}{3-5}=\frac{-4}{-2}=2\)

=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{5}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.5=10\end{cases}}\)

Vậy ...

6) Ta có: \(\frac{x}{2}=\frac{y}{3}\) => \(\frac{x}{10}=\frac{y}{15}\)

       \(\frac{y}{5}=\frac{z}{4}\) => \(\frac{y}{15}=\frac{z}{12}\)

=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

 \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{10+15+12}=\frac{-49}{37}\)

=> \(\hept{\begin{cases}\frac{x}{10}=-\frac{49}{37}\\\frac{y}{15}=-\frac{49}{37}\\\frac{z}{12}=-\frac{49}{37}\end{cases}}\) => \(\hept{\begin{cases}x=-\frac{49}{37}\cdot10=\frac{-490}{37}\\y=-\frac{49}{37}\cdot15=-\frac{735}{37}\\z=-\frac{49}{37}\cdot12=-\frac{588}{37}\end{cases}}\)

Vậy ...

mk lm bài mà mk cho là ''khó'' nhất thôi nha 

\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)và \(x+y+z=-49\)

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{15}=\frac{z}{12}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

ADTC dãy tỉ số bằng nhau ta có 

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{10+15+12}=-\frac{49}{37}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=-\frac{49}{37}\\\frac{y}{15}=-\frac{49}{37}\\\frac{z}{12}=-\frac{49}{37}\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{49}{37}.10=-\frac{490}{37}\\y=-\frac{49}{37}.15=-\frac{735}{37}\\z=-\frac{49}{37}.12=-\frac{588}{37}\end{cases}}}\)

Hoctot

a: Ta có: \(100-7\left(x-5\right)=58\)

\(\Leftrightarrow7\left(x-5\right)=42\)

\(\Leftrightarrow x-5=6\)

hay x=11

b: Ta có: \(12\left(x-1\right):3=4^3+2^3\)

\(\Leftrightarrow12\left(x-1\right)=216\)

\(\Leftrightarrow x-1=18\)

hay x=19

6: =>x=9/10+1/5=9/10+2/10=11/10

7: =>x=3/8-5/12=36/96-40/96=-1/24

8: =>x=7/6-5/4=14/12-15/12=-1/12

9: =>x=1/35+2/7=1/35+10/35=11/35

10: =>x=2/7-7/10=20/70-49/70=-29/70

1, \(x=-\dfrac{5}{3}-\dfrac{1}{2}=-\dfrac{10}{6}-\dfrac{3}{6}=-\dfrac{13}{6}\)

2, \(x=\dfrac{1}{3}-\dfrac{3}{5}=\dfrac{5-9}{15}=-\dfrac{4}{15}\)

4, \(x=-\dfrac{7}{9}+\dfrac{4}{3}=\dfrac{-7+12}{9}=\dfrac{5}{9}\)

5, \(x=\dfrac{5}{6}-\dfrac{7}{3}=\dfrac{5-14}{6}=-\dfrac{9}{6}=-\dfrac{3}{2}\)

6, \(x=\dfrac{9}{10}+\dfrac{1}{5}=\dfrac{9+2}{10}=\dfrac{11}{10}\)

7, \(x=\dfrac{3}{8}-\dfrac{5}{12}=\dfrac{9-10}{12}=-\dfrac{1}{12}\)

8, \(x=\dfrac{7}{6}-\dfrac{5}{4}=\dfrac{14-15}{12}=-\dfrac{1}{12}\)

9, \(x=\dfrac{1}{35}+\dfrac{2}{7}=\dfrac{1+10}{35}=\dfrac{11}{35}\)

10, \(x=-\dfrac{7}{10}+\dfrac{2}{7}=\dfrac{-49+20}{70}=\dfrac{-29}{70}\)

a) ĐKXĐ: \(x\ge0\)

Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)

\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)

(12000+X):376=X

12000+X=376 x X

12000=376 x X - X

12000=375 x X

X=12000:375= 32