\(35-5\left(x-1\right)=10\\ \Leftrightarrow35-5x+5=10\\ \Rightarrow40-5x=10\)

\(\Rightarrow-5x=10-40\\ \Rightarrow-5x=-30\\ \Rightarrow x=\dfrac{-30}{-5}=6\)

c) 

\(24\left(x-16\right)=12^2\)

\(\Rightarrow24x-384=144\\ \Rightarrow24x=144+384\\ \Rightarrow24x=528\\ \Rightarrow x=\dfrac{528}{24}=22\)

d) 

\(\left(x^2-10\right)\div5=3\\ \Rightarrow\left(x^2-10\right)=3\times5\\ \Rightarrow x^2-10=15\)

\(\Rightarrow x^2=15+10\\ \Rightarrow x^2=25\\ \Rightarrow x^2=5^2\Rightarrow x=5\)

Ơ, bn ơi @Kenny sai sai ở đâu thì phải.

giúp mình với ạ mình cần gấp

a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)

⇒\(\dfrac{y-x}{5-2}=\dfrac{6}{3}=2\)

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

\(\dfrac{y}{5}=2\Rightarrow y=10\)

\(\dfrac{z}{10}=2\Rightarrow z=20\)

b) Ta có: \(\dfrac{x}{8}=\dfrac{2y}{6}=\dfrac{z}{7}\)

\(\dfrac{x-2y+z}{8-6+7}=\dfrac{18}{9}=2\)

\(\dfrac{x}{8}=2\Rightarrow x=16\)

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

\(\dfrac{z}{7}=2\Rightarrow z=14\)

Bài 2:

Với x,y,z,t là số tự nhiên khác 0

Có \(\dfrac{x}{x+y+z+t}< \dfrac{x}{x+y+z}< \dfrac{x}{x+y}\)

\(\dfrac{y}{x+y+z+t}< \dfrac{y}{x+y+t}< \dfrac{y}{x+y}\)

\(\dfrac{z}{x+y+z+t}< \dfrac{z}{y+z+t}< \dfrac{z}{z+t}\)

\(\dfrac{t}{x+y+z+t}< \dfrac{t}{x+z+t}< \dfrac{t}{z+t}\)

Cộng vế với vế \(\Rightarrow1< M< \dfrac{x+y}{x+y}+\dfrac{z+t}{z+t}=2\)

=> M không là số tự nhiên.

Bài 1:

Ta có:

\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\) 

\(B=\left(1+\dfrac{2007}{2}\right)+\left(1+\dfrac{2006}{3}\right)+...+\left(1+\dfrac{2}{2007}\right)+\left(1+\dfrac{1}{2008}\right)+1\) 

\(B=\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2007}+\dfrac{2009}{2008}+\dfrac{2009}{2009}\) 

\(B=2009.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)\) 

\(\Rightarrow\dfrac{A}{B}=\dfrac{2009.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}=2009\)

sai rồi kìa \(\frac{A}{B}\)chớ không phải \(\frac{B}{A}\)

bằng \(\frac{1}{2009}\)mới dúng

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

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

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

\(\Leftrightarrow3x=4\)

hay \(x=\dfrac{4}{3}\)

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

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)

B=2x^2+6x-3x^2-3=x+1-x(x-2)=0

  =-x^2+6x-3=x+1-x^2+x=0

  =4x-3=0

x=3/4

1:

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}=\dfrac{2x-y}{2\cdot1,1-1,3}=\dfrac{5.5}{0.9}=\dfrac{55}{9}\)

=>x=121/18; y=143/18; z=77/9

bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 =>  x-1/3=y-2/4=z-3/5 

áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1

do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự

Bài 1: 

a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )

cũng dễ thôi

Bài 1 :

\(3x+5=2\left(x-\frac{1}{4}\right)\)

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

\(\Leftrightarrow5+\frac{1}{2}=2x-3x\)

\(\Leftrightarrow\frac{11}{2}=-x\)

\(\Leftrightarrow\frac{-11}{2}=x\)

Vậy \(x=\frac{-11}{2}\)

Bài 2:

a, \(\left|x+\frac{19}{5}\right|+\left|y+\frac{2018}{2019}\right|+\left|z-3\right|=0\)

Vì \(\hept{\begin{cases}\left|x+\frac{19}{5}\right|\ge0\\\left|y+\frac{2018}{2019}\right|\ge0\\\left|z-3\right|\ge0\end{cases}}\)

       Mà \(\left|x+\frac{19}{5}\right|+\left|y+\frac{2018}{2019}\right|+\left|z-3\right|=0\)

\(\Rightarrow+,\left|x+\frac{19}{5}\right|=0\)

\(\Leftrightarrow x+\frac{19}{5}=0\)

\(\Leftrightarrow x=\frac{-19}{5}\)

\(\Rightarrow+,\left|y+\frac{2018}{2019}\right|=0\)

\(\Leftrightarrow y+\frac{2018}{2019}=0\)

\(\Leftrightarrow y=\frac{-2018}{2019}\)

\(\Rightarrow+,\left|z-3\right|=0\)

\(\Leftrightarrow z-3=0\)

\(\Leftrightarrow z=3\)

Vậy \(\hept{\begin{cases}x=\frac{-19}{5}\\y=\frac{-2018}{2019}\\z=3\end{cases}}\)

b, Ta có : \(\left|x-\frac{1}{2}\right|+\left|2y+4\right|+\left|z-5\right|\ge0\)

Vì : \(\hept{\begin{cases}\left|x-\frac{1}{2}\right|\ge0\\\left|2y+4\right|\ge0\\\left|z-5\right|\ge0\end{cases}}\)

Mà : \(\left|x-\frac{1}{2}\right|+\left|2y+4\right|+\left|z-5\right|\ge0\)

\(\Rightarrow+,\left|x-\frac{1}{2}\right|\ge0\)

\(\Rightarrow x\inℚ\)

\(\Rightarrow+,\left|2y+4\right|\ge0\)

\(\Rightarrow y\inℚ\)

\(\Rightarrow+,\left|z-5\right|\ge0\)

\(\Rightarrow z\inℚ\)

Vậy chỉ cần \(\hept{\begin{cases}x\inℚ\\y\inℚ\\z\inℚ\end{cases}}\)thì thỏa mãn.

234*(-26)+134*26

a) \(|x+\frac{3}{4}|+|y-\frac{1}{5}|+|x+y+z|=0\)

\(\Rightarrow|x+\frac{3}{4}|=|y-\frac{1}{5}|=|x+y+z|=0\)

\(\Rightarrow|x+\frac{3}{4}|=0\)                           \(\Rightarrow|y-\frac{1}{5}|=0\)                                \(\Rightarrow|x+y+z|=0\)

\(\Rightarrow x+\frac{3}{4}=0\)                              \(\Rightarrow y-\frac{1}{5}=0\)                                      \(\Rightarrow x+y+z=0\)

\(x=\frac{-3}{4}\)                                                \(y=\frac{1}{5}\)                                                 thay x=-3/4; y=1/5 vào biểu thức trên

                                                                                                                                          ta có \(\frac{-3}{4}+\frac{1}{5}+z=0\)

                                                                                                                                                        \(z=0-\frac{-3}{4}-\frac{1}{5}\)

      VẬY X=-3/4; Y=1/5; Z=11/20

B) \(|3x-4|+\left|3y-5\right|=0\)

\(\Rightarrow\left|3x-4\right|=\left|3y-5\right|=0\)

\(\Rightarrow\left|3x-4\right|=0\)                                    \(\Rightarrow\left|3y-5\right|=0\)

\(3x-4=0\)                                                    \(3y-5=0\)

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

VẬY X= 4/3; Y=5/3

C) \(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|< 0\)

ĐỂ \(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|< 0\)

\(\Rightarrow\left|x+\frac{3}{4}\right|;\left|y-\frac{2}{5}\right|;\left|z+\frac{1}{2}\right|< 0\)

MÀ GIÁ TRỊ TUYỆT ĐỐI LUÔN MANG SỐ NGUYÊN DƯƠNG

\(\Rightarrow x;y;z\in\varnothing\)

d) \(\left|x+\frac{1}{5}\right|+\left|3-y\right|=0\)

\(\Rightarrow\left|x+\frac{1}{5}\right|=\left|3-y\right|=0\)

\(\Rightarrow\left|x+\frac{1}{5}\right|=0\)                                \(\Rightarrow\left|3-y\right|=0\)

\(x+\frac{1}{5}=0\)                                                 \(3-y=0\)

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

VẬY X= -1/5; Y=3

CHÚC BN HỌC TỐT!!!!!!!

Ta có : 

\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x+\frac{3}{4}=0\\y-\frac{1}{5}=0\\x+y+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-3}{4}\\y=\frac{1}{5}\\z=0-\frac{-3}{4}-\frac{1}{5}\end{cases}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x=\frac{-3}{4}\\y=\frac{1}{5}\\z=\frac{11}{20}\end{cases}}\)

Vậy \(x=\frac{-3}{4};y=\frac{1}{5};z=\frac{11}{20}\)

\(b)\) Ta có : 

\(\left|3x-4\right|+\left|3y-5\right|=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x-4=0\\3y-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=4\\3y=5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\y=\frac{5}{3}\end{cases}}}\)

Vậy \(x=\frac{4}{3}\) và \(y=\frac{5}{3}\)

1: \(\dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z+7}{5}\)

mà x+y-z=8

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

\(\dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z+7}{5}=\dfrac{x-1+y-2-z-7}{3+4-5}=\dfrac{8-3-7}{2}=\dfrac{-2}{2}=-1\)

=>\(\left\{{}\begin{matrix}x-1=-1\cdot3=-3\\y-2=-1\cdot4=-4\\z+7=-1\cdot5=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-2\\y=-2\\z=-12\end{matrix}\right.\)

2: \(\dfrac{x+1}{3}=\dfrac{y+2}{-4}=\dfrac{z-3}{5}\)

mà 3x+2y=47-42=5

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

\(\dfrac{x+1}{3}=\dfrac{y+2}{-4}=\dfrac{z-3}{5}=\dfrac{3x+3+2y+4}{3\cdot3+2\left(-4\right)}=\dfrac{5+7}{9-8}=12\)

=>\(\left\{{}\begin{matrix}x+1=12\cdot3=36\\y+2=-12\cdot4=-48\\z-3=12\cdot5=60\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=35\\y=-48-2=-50\\z=60+3=63\end{matrix}\right.\)

Bài 2:

\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)

\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)