a/ \(\frac{x-1}{9}=\frac{8}{3}\) 

\(\Leftrightarrow3\left(x-1\right)=72\)

\(\Leftrightarrow x-1=24\)

\(\Leftrightarrow x=25\)

Vậy ..

b/ \(\frac{-x}{4}=\frac{-9}{x}\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x^2=6^2=\left(-6\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

Vậy ..

c/ \(\frac{x}{4}=\frac{18}{x+1}\)

\(\Leftrightarrow x\left(x+1\right)=72\)

\(\Leftrightarrow x\left(x+1\right)=8.9\)

\(\Leftrightarrow x=8\)

Vậy ..

a, \(\frac{x-1}{9}=\frac{8}{3}\)

\(\Rightarrow\left(x-1\right).3=8.9\)

\(\Rightarrow\left(x-1\right).3=72\)

\(\Rightarrow x-1=72:3\)

\(\Rightarrow x-1=24\)

\(\Rightarrow x=24+1\)

\(\Rightarrow x=25\)

b, \(\frac{-x}{4}=\frac{-9}{x}\)

\(\Rightarrow-x.x=-9.4\)

\(\Rightarrow-\left(x^2\right)=-36\)

\(\Rightarrow x^2=36\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

c, \(\frac{x}{4}=\frac{18}{x+1}\)

\(\Rightarrow x\left(x+1\right)=4.18\)

\(\Rightarrow x.x+x.1=72\)

\(\Rightarrow x^2+x=72\)

\(\Rightarrow x^2+x-72=0\)

\(\Rightarrow x^2+x-8^2+8=0\)

\(\Rightarrow x=8\)

a) x-1=24

=>x=24+1=25

=> x=25

b)=>-(x^2)=-36

=>x=6

k mik nha

a) \(\frac{x-1}{9}=\frac{8}{3}\)

\(\Leftrightarrow3\left(x-1\right)=9\cdot8\)

\(\Leftrightarrow3x-3=72\)

\(\Leftrightarrow3x=75\)

\(\Leftrightarrow x=25\)

\(\frac{1}{15}<\frac{x}{12}<\frac{x}{9}<\frac{1}{4}\)

\(\Rightarrow\frac{2}{36}<\frac{3x}{36}<\frac{4y}{36}<\frac{9}{36}\)

Ta có:\(\frac{2}{36}<\frac{3x}{36}<\frac{9}{36}\)

\(\Rightarrow\)\(2<3x<9\)

\(\Rightarrow\)\(\frac{2}{3}\)<x<3

\(\Rightarrow1\le\)x\(<3\)

\(\Rightarrow x\in\left\{1,2,3\right\}\)

\(x=1\Rightarrow\frac{3}{36}<\frac{4y}{36}<\frac{9}{36}\)\(\Rightarrow\)\(3<4y<9\)

\(\Rightarrow\frac{3}{4}\)\(<\)x\(<\frac{9}{4}\)

\(\Rightarrow\)\(1\)\(\le\)x\(\le2\)

\(x=2\) và \(x=3\) tương tự

\(\frac{x}{-7}=\frac{5}{-35}\)

\(\frac{x.5}{-35}=\frac{5}{-35}\)

=> x . 5 = 5

x = 5 : 5 

x = 1

sao trả lời có một câu mấy dậy bạn giúp mình với

\(a,\frac{x-1}{9}=\frac{8}{3}\)

\(\Leftrightarrow x-1=24\)

\(\Rightarrow x=25\)

\(b,-\frac{x}{4}=-\frac{9}{x}\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

\(c,\frac{x}{4}=\frac{18}{x+1}\)

\(\Leftrightarrow x^2+x=72\)

\(\Leftrightarrow x\left(x+1\right)=72..\)

ấn nhầm: lm tiếp nhé!

\(x\left(x+1\right)=72\)

\(\text{Mà x thuộc Z nên }x\left(x+1\right)=8\left(8+1\right)\)

\(\Leftrightarrow x=8\)

a) \(\frac{x-1}{9}=\frac{8}{3}\)

\(\Rightarrow3x-3=72\)

\(\Rightarrow3x=75\)

\(\Rightarrow x=25\)

b) \(\frac{-x}{4}=\frac{-9}{x}\)

\(\Rightarrow\left(-x\right).\left(-x\right)=4.9\)

\(\Rightarrow x^2=36\)

\(\Rightarrow x=\pm6\)

c) \(\frac{x}{4}=\frac{18}{x+1}\)

\(\frac{x}{4}-\frac{18}{x+1}=0\)

\(\frac{x\left(x+1\right)}{4\left(x+1\right)}-\frac{72}{4\left(x+1\right)}=0\)

\(\frac{x^2+x-72}{4\left(x+1\right)}=0\)

đến đây chịu :))

a)ta có xy=7*9=7*3*3

vậy x =9;21 , y=7;3

b) xy=-2*5

mà x<0<y

nên x=-2 ,y=5

c)x-y=5 hay x=y+5

\(\frac{y+5+4}{y-5}=\frac{4}{3}\Rightarrow3y+27=4y-20\Rightarrow y=47\Rightarrow x=52\)

câu c mk nhầm đề sr bạn nha

\(\frac{y+5-4}{y-5}=\frac{4}{3}\Rightarrow3y+3=4y-5\Rightarrow y=8\Rightarrow x=13\)

Bài 3 

\(\frac{x-1}{9}=\frac{8}{3}\)

\(\Rightarrow\left(x-1\right).3=8.9\)

\(\Rightarrow\left(x-1\right).3=72\)

\(\Rightarrow x-1=24\)

\(\Rightarrow x=25\)

\(\frac{-x}{4}=\frac{-9}{x}\)

\(\Rightarrow\left(-x\right).x=\left(-9\right).4\)

\(\Rightarrow-x=-36\)

\(\Rightarrow x=36\)

\(\frac{x}{4}=\frac{18}{x+1}\)

\(\Rightarrow x.\left(x+1\right)=4.18\)

\(\Rightarrow x.\left(x+1\right)=72\)

Vì x và x + 1 là 2 số tự nhiên liên tiếp 

\(\Rightarrow x\left(x+1\right)=8.9\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=8\end{cases}}\)

Bài 4

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

Ta có :

\(x-y=5\)

\(\Rightarrow x=5+y\)

\(\Rightarrow\frac{y+5-4}{y-3}=\frac{4}{3}\)

\(\Rightarrow\frac{y+1}{y-3}=\frac{4}{3}\)\(\)

\(\Rightarrow\left(y+1\right).3=\left(y-3\right).4\)

\(\Rightarrow y.3+1.3=y.4-3.4\)

\(\Rightarrow y.3+3=y.4-12\)

\(\Rightarrow y.3-y.4=-12-3\)

\(\Rightarrow-1y=-15\)

\(\Rightarrow y=\left(-15\right):\left(-1\right)\)

\(\Rightarrow y=15\)

Vì x = y + 5

\(\Rightarrow x=15+4\)

\(\Rightarrow x=19\)

Vậy x = 19 , y = 15

\(\frac{-x}{4}=\frac{-9}{x}\)

\(\Rightarrow\left(-x\right).x=4.\left(-9\right)\)

\(\Rightarrow-x=-9;x=4\)

\(\Rightarrow x=9;x=4\)

\(\frac{-x}{4}=\frac{-9}{x}\)

\(\Rightarrow\left(-x\right).x=\left(-9\right).4\)

\(\Rightarrow\left(-x\right)^2=-36\)

\(\Rightarrow\left(-x\right)^2=-6^2\)

\(\Rightarrow-x=-6\)

\(\Rightarrow x\in\left\{6;-6\right\}\)

a, \(\frac{x}{y+z+1}=\frac{y}{x+z+3}=\frac{z}{x+y-4}=\frac{x+y+z}{y+z+1+x+z+3+x+y-4}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)

=>\(x+y+z=\frac{1}{2};\frac{x}{y+z+1}=\frac{1}{2};\frac{y}{x+z+3}=\frac{1}{2};\frac{z}{x+y-4}=\frac{1}{2}\)

=>\(\hept{\begin{cases}y+z+1=2x\\x+z+3=2y\\x+y-4=2z\end{cases}}\Rightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+3=3y\\x+y+z-4=3z\end{cases}\Rightarrow\hept{\begin{cases}3x=\frac{1}{2}+1\\3y=\frac{1}{2}+3\\3z=\frac{1}{2}-4\end{cases}}}\Rightarrow\hept{\begin{cases}3x=\frac{3}{2}\\3y=\frac{7}{2}\\3z=\frac{-7}{2}\end{cases}}\)

đến đây dễ rồi

b, =>(x-18)(x+16)=(x+4)(x-17)

=>x²+16x-18x-288=x²-17x+4x-68

=>x²-2x-288-x²+13x+68=0

=>11x-220=0

=>11x=220

=>x=20