a)\(\frac{-5}{6}\).\(\frac{120}{25}\)<x<\(\frac{-7}{15}\).\(\frac{9}{14}\)

       -4                 <x<\(\frac{-3}{10}\)

\(\frac{-40}{10}\)<      x   <\(\frac{-3}{10}\)=>x E {-39:-38:-37:.....:-4}

b)\(\left(\frac{-5}{3}\right)^3\)<x<\(\frac{-24}{35}.\frac{-5}{6}\)

\(\frac{-875}{189}< x< \frac{108}{189}\)

=> x  E {\(\frac{-874}{189},\frac{-873}{189},......,\frac{107}{189}\)}

bạn tự quy đồng lên mà giải

Hình như sai ở đâu thì phải 

lay x phần 7 - 1 phần 4 khác ra

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 5:

Theo đề ra, ta có:

\(\frac{x}{y}=\frac{2}{5}\Rightarrow\frac{x}{2}=\frac{y}{5}\)

Ta đặt: \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=5k\end{cases}}\)

\(\Rightarrow k^2=4\Rightarrow k=\pm2\)

Trường hợp 1: Với \(k=2\)

\(\Rightarrow\frac{x}{2}=2\Rightarrow x=2.2=4\)

\(\Rightarrow\frac{y}{5}=2\Rightarrow y=5.2=10\)

Trường hợp 2: Với \(k=-2\)

\(\Rightarrow\frac{x}{2}=-2\Rightarrow x=2.\left(-2\right)=-4\)

\(\Rightarrow\frac{y}{5}=-2\Rightarrow y=5.\left(-2\right)=-10\)

Bài 4:

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\)

\(\Rightarrow\frac{3\left(x-1\right)}{3.2}=\frac{4\left(y+3\right)}{4.4}=\frac{5\left(z-5\right)}{5.6}\Rightarrow\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{5z-25}{30}\)

\(=\frac{-\left(3x-3\right)-\left(4y+12\right)+\left(5z-25\right)}{-6-16+30}=\frac{\left(-3x-4y+5z\right)+3-12-25}{8}=\frac{50-34}{8}=2\)

\(\Rightarrow\frac{3x-3}{6}=2\Rightarrow3x-3=12\Rightarrow x=15\)

\(\Rightarrow\frac{4y+12}{16}=2\Rightarrow4y+12=32\Rightarrow y=5\)

\(\Rightarrow\frac{5z-25}{30}=2\Rightarrow5x-25=60\Rightarrow z=17\)

Bài 3:

Theo đề ra, ta có: \(x:y:z=3:8:5\Rightarrow\frac{x}{3}=\frac{y}{8}=\frac{z}{5}\)

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\frac{x}{3}=\frac{y}{8}=\frac{z}{5}=\frac{3x+y-2z}{3.3+8-2.5}=2\)

\(\Rightarrow\frac{x}{3}=2\Rightarrow x=3.2=6\)

\(\Rightarrow\frac{y}{8}=2\Rightarrow y=8.2=16\)

\(\Rightarrow\frac{z}{5}=2\Rightarrow z=5.2=10\)

a) \(\frac{1}{2}-|\frac{5}{4}-2x|=\frac{1}{3}\Leftrightarrow|\frac{5}{4}-2x|=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{6}\\\frac{5}{4}-2x=-\frac{1}{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{5}{4}-\frac{1}{6}=\frac{13}{12}\\2x=\frac{5}{4}+\frac{1}{6}=\frac{17}{12}\end{cases}}}\)

Tự làm nốt và kết luận 

b) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)

Vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)\ne0\forall x\Rightarrow x+1=0\Leftrightarrow x=-1\)

Vậy ....

c) \(\frac{x}{y}=\frac{10}{9}\Leftrightarrow\frac{x}{10}=\frac{y}{9};\frac{y}{z}=\frac{3}{4}\Leftrightarrow\frac{y}{3}=\frac{z}{4}\Leftrightarrow\frac{y}{9}=\frac{x}{12}\)

\(\Leftrightarrow\frac{x}{10}=\frac{y}{9}=\frac{z}{12}\). Mà \(x-y+z=78\). Áp dụng t/c dãy tỉ số bằng nhau 

\(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)

\(\Rightarrow x=6.10=60;y=6.9=54;z=6.12=72\)

Vậy..........

\(x-\left(\frac{5}{6}-x\right)=x-\frac{2}{3}\)

\(x-\frac{5}{6}+x-x=-\frac{2}{3}\)

\(x=\frac{-2}{3}+\frac{5}{6}\)

\(x=\frac{-4}{6}+\frac{5}{6}\)

\(x=\frac{1}{6}\)

\(x-\left(\frac{5}{6}-x\right)=x-\frac{2}{3}\)

\(x-\frac{5}{6}+x=x-\frac{2}{3}\)

\(\Rightarrow x+x-\frac{5}{6}=x-\frac{2}{3}\Rightarrow x+x-x=-\frac{2}{3}+\frac{5}{6}\)

\(\Rightarrow x=\frac{1}{6}\Rightarrow\)x ko tồn tại

\(x-\left(\frac{5}{6}-x\right)=x-\frac{2}{3}\)

\(x-\frac{5}{6}+x=x-\frac{2}{3}\)

\(2x-x=\frac{-2}{3}+\frac{5}{6}\)

\(x=\frac{1}{6}\)

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

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

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

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

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

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

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

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

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

\(\Rightarrow-x=-6\)

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