a) \(x+2x+3x+...+100x=-213\)

\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)

\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)

\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)

d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)

                                                                    \(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)

c) \(3\left(x-2\right)+2\left(x-1\right)=10\)

\(\Rightarrow3x-6+2x-2=10\)

\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)

a) \(x+2x+3x+4x+...+100x=-213\)

\(x.\left(1+2+3+4+...+100\right)=-213\)

\(x.5050=-213\)

\(x=-\frac{213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)

\(\frac{1}{2}x=-\frac{43}{12}\)

\(x=\frac{-43}{6}\)

b: 2x^3-1=15

=>2x^3=16

=>x=2

\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)

=>y-25=32; z+9=50

=>y=57; z=41

d: 3/5x=2/3y

=>9x=10y

=>x/10=y/9=k

=>x=10k; y=9k

x^2-y^2=38

=>100k^2-81k^2=38

=>19k^2=38

=>k^2=2

TH1: k=căn 2

=>\(x=10\sqrt{2};y=9\sqrt{2}\)

TH2: k=-căn 2

=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)

\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)

\(\Leftrightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)

\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)

\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)

\(\Leftrightarrow x+1=0\)( \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\)) 

\(\Leftrightarrow x=-1\)

Vậy x=-1

mỗi phân số + 1 thì sẽ có tử chung là x + 1

chuyển vế có \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\))  =0

mà tổng các phân số kia khác 0 nên x+1 bằng 0 

=> x=-1

a: =>x/3=-5/2

hay x=-15/2

b: \(\Leftrightarrow\dfrac{7}{3}:x=\dfrac{1}{5}-\dfrac{4}{9}=\dfrac{9-20}{45}=\dfrac{-11}{45}\)

\(\Leftrightarrow x=\dfrac{7}{3}:\dfrac{-11}{45}=\dfrac{7}{3}\cdot\dfrac{-45}{11}=\dfrac{-105}{11}\)

c: \(\Leftrightarrow x=\dfrac{-7}{2}\cdot2=-7\)

d: =>x/27=-1/3+2/9=2/9-3/9=-1/9=-3/27

=>x=-3

Hay quá, mk cũng đang tìm câu này nè

a) ⇒ \(\dfrac{5}{3}x\) \(=\) \(\dfrac{5}{6}+\dfrac{1}{4}\)

⇒ \(\dfrac{5}{3}x=\dfrac{13}{12}\)

⇒ \(x=\dfrac{13}{12}:\dfrac{5}{3}\)

⇒\(x=\dfrac{13}{20}\)

3. Từ \(\dfrac{x-2}{27}=\dfrac{3}{x-2}\Rightarrow\left(x-2\right)^2=81\)

\(\Rightarrow\left(x-2\right)^2=\left(\pm9\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-2=-9\\x-2=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)

Vậy x = -7 hoặc x = 11

4. Từ \(\dfrac{2x+5}{9-2x}=\dfrac{2}{5}\)

\(\Rightarrow5\left(2x+5\right)=2\left(9-2x\right)\\ \Leftrightarrow10x+25=18-4x\\ \Leftrightarrow14x=-7\\ \Rightarrow x=-\dfrac{1}{2}\)

5. Từ \(\dfrac{x-7}{x+8}=\dfrac{x-8}{x+9}\)

\(\Rightarrow\left(x-7\right)\left(x+9\right)=\left(x-8\right)\left(x+8\right)\\ \Leftrightarrow x^2+2x-63=x^2-64\\ \Leftrightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)

b: Ta có: x/y=7/9

nên x/7=y/9

=>x/49=y/63

Ta có: y/z=7/3

nên y/7=z/3

=>y/63=z/27

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{49}=\dfrac{y}{63}=\dfrac{z}{27}=\dfrac{x-y+z}{49-63+27}=\dfrac{-15}{13}\)

Do đó: x=-735/13; y=-945/13; z=-405/13

c: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}=\dfrac{2x+5y-2z}{2\cdot7+5\cdot20-2\cdot32}=\dfrac{100}{50}=2\)

Do đó: x=14; y=40; z=64

d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}=\dfrac{x-y-z}{8-5-2}=3\)

Do đó: x=24; y=15; z=6

a) \(32< 2^x< 128\)

=> \(2^5< 2^x< 2^7\)

=> x = 6

b) \(2^{x-1}+4\cdot2^x=9\cdot2^5\)

=> \(2^{x-1}+2^2\cdot2^x=9\cdot2^5\)

=> \(2^{x-1}+2^{2+x}=9\cdot2^5\)

=> 9.2^x-1 = 9.2⁵

=> 2^x-1 = \(\frac{9\cdot2^5}{9}=2^5\)

=> x - 1 = 5 => x = 6

c) \(9\cdot27\le3^x\le243\)

=> \(243\le3^x\le243\)

=> x = 5

d) Giống câu b)

e) \(3^{x-1}+5\cdot3^{x-2}=216\)

=> 8.3^x-2 = 216

=> 3^x-2 = 27

=> 3^x-2 = 3³

=> x - 2 = 3 => x = 5

f) 27^x-3 = 9^x+3 

=> 27^x-3 = 9^x+3

=> (3³)^x-3 = (3²)^x+3

=> 3^3x-9 = 3^{2x + 6}

=> không thỏa mãn x vì x là phân số mà theo đề bài là số nguyên

g) x²⁰¹⁹ = x => x²⁰¹⁹ - x = 0 => x(x²⁰¹⁸ - 1) = 0 => x = 0 hoặc x = 1

a) 

\(2^5< 2^x< 2^7\) 

\(5< x< 7\) 

\(x=6\) 

b) 

\(2^{x-1}+2^2\cdot2^x=9\cdot2^5\) 

\(2^{x-1}+2^{2+x}=9\cdot2^5\) 

\(2^{x-1}\left(1+2^3\right)=9\cdot2^5\) 

\(2^{x-1}\cdot9=9\cdot2^5\) 

\(2^{x-1}=2^5\) 

\(x-1=5\) 

\(x=6\)

a, 2^4-x=32=2⁵

=> 4-x=5

<=> x=-1

b, (x+1,5)²+(y-2,5)¹⁰=0

Vì (x+1,5)2\(\ge\)0,   (y-2,5)¹⁰\(\ge\)0

\(\Rightarrow\hept{\begin{cases}x+1,5=0\\y-2,5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1,5\\y=2,5\end{cases}}}\)

a)\(2^{4-x}\)=32

=>\(2^{4-x}\)=32=\(2^5\)

=>4-x=5

=>x=4-5=-1

=>x=-1