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 2: (x-3).(y+2) = -5

    Vì x, y \(\in\)Z   => x-3 \(\in\)Ư(-5) = {5;-5;1;-1}

Ta có bảng: 

bài 3: a(a+2)<0

TH1 : \(\orbr{\begin{cases}a< 0\\a+2>0\end{cases}}\)=>\(\orbr{\begin{cases}a< 0\\a>-2\end{cases}}\)=> -2<a<0 ( TM)

TH2: \(\orbr{\begin{cases}a>0\\a+2< 0\end{cases}}\Rightarrow\orbr{\begin{cases}a>0\\a< -2\end{cases}}\Rightarrow loại\)
 

           Vậy -2<a<0

Bài 5: \(\left(x^2-1\right)\left(x^2-4\right)< 0\)

TH 1 : \(\hept{\begin{cases}x^2-1>0\\x^2-4< 0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2>1\\x^2< 4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x>1\\x< 2\end{cases}}\)\(\Rightarrow\)1 < a < 2

TH 2: \(\hept{\begin{cases}x^2-1< 0\\x^2-4>0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2< 1\\x^2>4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x< 1\\x>2\end{cases}}\)\(\Rightarrow\)loại

                         Vậy 1<a<2

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)

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

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

a\(\left(x-3\right)^2-\left(x+2\right)^2-5\left(\frac{1}{5}-7\right)=-30\)

=>(x-3-x-2)(x-3+x+2)-x+35=-30

=>-5(2x-1)-x+35=-30

=>-10x+5-x+35=-30

=>-11x+40=-30

=>-11x=-70 =>x=70/11

d)\(\left(x+3\right)^2-\left(x+5\right)\left(x-5\right)=2\)

\(=>\left(x+3\right)^2-x^2+25=2\)

\(=>\left(z+3-z\right)\left(z+3+z\right)+25=2\)

\(=>3\left(2z+3\right)+25-2=0\)

\(=>6z+9+23=0\)

\(=>6x+32=0=>6x=-32=>x=-\frac{16}{3}\)

e)\(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)

\(=>3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\)

\(=>3x^2+12x+12+4x^2-4x+1-7x^2+63\)

\(=>8x+76=36=>8x=36-76=>x=-40\div8=-5\)

g)\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

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

\(=>4x-1=5=>4x=6=>x=\frac{3}{2}\)

\(\frac{6-2x}{5}=\frac{-y-4}{4}=\frac{3z-15}{2}\)

\(\Rightarrow\hept{\begin{cases}4\left(6-2x\right)=5\left(-y-4\right)\\2\left(-y-4\right)=4\left(3z-15\right)\\2\left(6-2x\right)=5\left(3z-15\right)\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}8x-5y=44\\y+6z=26\\4x+15z=87\end{cases}}\Leftrightarrow\hept{\begin{cases}y=\frac{8x-44}{5}\\z=\frac{87-4x}{15}\end{cases}}\)

Mà x-y+z=-1 nên \(x-\frac{8x-44}{5}+\frac{87-4x}{15}=-1\)\(\Leftrightarrow15x-3\left(8x-44\right)+87-4x+15=0\)

\(\Leftrightarrow\)\(13x=234\Rightarrow x=18\Rightarrow y=\frac{8.18-44}{5}=20\Rightarrow z=\frac{87-4.18}{15}=1\)

Vậy x=18 ; y=20 ; z=1

dễ ợt bài này còn không biết làm mình học lớp 11 trường lê hồng phong

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