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

`a)`

`A=-4x^5y^3+6x^4y^3-3x^2y^3z^2+4x^5y^3-x^4y^3+3x^2y^3z^2-2y^4+22`

`A=(-4x^5y^3+4x^5y^3)+(6x^4y^3-x^4y^3)-(3x^2y^3z^2-3x^2y^3z^2)-2y^4+22`

`A=5x^4y^3-2y^4+22`

        `->` Bậc: `7`

`b)B-5y^4=A`

`=>B=A+5y^4`

`=>B=5x^4y^3-2y^4+22+5y^4`

`=>B=5x^4y^3+3y^4+22`

Theo đề ra, ta có: \(\frac{x-1}{y+2}=\frac{3}{5}\)

\(\Rightarrow\frac{x-1}{3}=\frac{y+2}{5}=\frac{x-1+y+2}{8}=\frac{23-1+2}{8}=\frac{24}{8}=3\)

\(\frac{x-1}{3}=3\Rightarrow x=3.3+1=10\)

\(\frac{y+2}{5}=3\Rightarrow y=5.3-2=13\)

a: Ta có: xy=12

nên \(\left(x,y\right)\in\left\{\left(1;12\right);\left(-1;-12\right);\left(12;1\right);\left(-12;-1\right);\left(2;6\right);\left(-2;-6\right);\left(6;2\right);\left(-6;-2\right);\left(3;4\right);\left(4;3\right);\left(-3;-4\right);\left(-4;-3\right)\right\}\)

a) Ta có: \(\dfrac{x-1}{-4}=\dfrac{-4}{x-1}\)

\(\Leftrightarrow\left(x-1\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy: \(x\in\left\{5;-3\right\}\)

b) Ta có: \(\dfrac{x-4}{6}=\dfrac{-1}{3}\)

\(\Leftrightarrow x-4=-2\)

hay x=2

Vậy: x=2

a/ 

\(x-\dfrac{1}{-4}=-\dfrac{4}{x-1}\)

\(x+\dfrac{1}{4}+\dfrac{4}{x-1}=0\)

\(\dfrac{x\left(x-1\right)4}{4\left(x-1\right)}+\dfrac{16}{4\left(x-1\right)}=0\)

\(4x\left(x-1\right)+16=0\)(quy tắc khử mẫu lớp 8)

\(4x^2-4x+16=0\)

\(4x^2-2x-2x+16=0\)

\(\left(4x^2-2x\right)-\left(2x-16\right)=0\)

\(2x\left(2x-1\right)-2\left(x-16\right)=0\)

a,5/6

b,9/0

a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{7}=\dfrac{y}{6}=\dfrac{x-y}{7-6}=\dfrac{80}{1}=80\)

\(\Rightarrow\dfrac{x}{7}=80\Rightarrow x=80\cdot7=560\)

\(\Rightarrow\dfrac{y}{6}=80\Rightarrow y=80\cdot6=480\)

b) Áp dụng tính chất dãy tỉ số bằng nhau ta có::

\(\dfrac{x}{4}=\dfrac{y}{7}=\dfrac{x+y}{4+7}=\dfrac{12}{11}\)

\(\Rightarrow\dfrac{x}{4}=\dfrac{12}{11}\Rightarrow x=\dfrac{4\cdot12}{11}=\dfrac{48}{11}\)

\(\Rightarrow\dfrac{y}{7}=\dfrac{12}{11}\Rightarrow y=\dfrac{7\cdot12}{11}=\dfrac{84}{11}\)

Mình làm mẫu 2 câu thôi nhé

 x2=y3=z4x2=y3=z4 
\Leftrightarrow2x4=y3=z4=2x+y−z4+3−4=123=42x4=y3=z4=2x+y−z4+3−4=123=4 
\Rightarrowx=8
y=12
z=16
bài 2
x2=y5=z7x2=y5=z7
\Rightarrow2y=5x ;x=2,5y ;zx=3,5zx=3,5 ;2y=5x;z=3,5x
\RightarrowA = x-y+z/x+2y-z=x-2,5x+3,5+5x-3,5x=3,5

bn ghi rõ cho mik được ko