câu a ghi đề lại cho anh đc ko ko thấy gì hết

ko hình dung được nó

b)

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{5x-5}{10}=\dfrac{3y+9}{12}=\dfrac{4z-20}{24}\)

\(\Rightarrow\dfrac{\left(5x-3y-4z\right)-\left(5+9-20\right)}{10-12-24}=\dfrac{46+6}{-26}=-2\)

\(\Rightarrow x-1=-4\Rightarrow x=-3\)

\(\Rightarrow y+3=-8\Rightarrow y=-11\)

\(\Rightarrow z-5=-12\Rightarrow-7\)

\(\left\{{}\begin{matrix}xy=\dfrac{1}{2}\\yz=\dfrac{3}{5}\\zx=\dfrac{27}{10}\end{matrix}\right.\Rightarrow xyyzzx=\dfrac{1}{2}\cdot\dfrac{3}{5}\cdot\dfrac{27}{10}\Leftrightarrow\left(xyz\right)^2=\dfrac{81}{100}\)

\(\Rightarrow\left[{}\begin{matrix}xyz=-\dfrac{9}{10}\\xyz=\dfrac{9}{10}\end{matrix}\right.\)

+ Khi \(xyz=-\dfrac{9}{10}\)

\(\Rightarrow\left\{{}\begin{matrix}z=-\dfrac{9}{10}:\dfrac{1}{2}=-\dfrac{9}{5}\\x=-\dfrac{9}{10}:\dfrac{3}{5}=-\dfrac{3}{2}\\y=-\dfrac{9}{10}:\dfrac{27}{10}=-\dfrac{1}{3}\end{matrix}\right.\)

+ Khi \(xyz=\dfrac{9}{10}\)

\(\Rightarrow\left\{{}\begin{matrix}z=\dfrac{9}{10}:\dfrac{1}{2}=\dfrac{9}{5}\\x=\dfrac{9}{10}:\dfrac{3}{5}=\dfrac{3}{2}\\y=\dfrac{9}{10}:\dfrac{27}{10}=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(\dfrac{3}{2};\dfrac{1}{3};\dfrac{9}{5}\right);\left(-\dfrac{3}{2};-\dfrac{1}{3};-\dfrac{9}{5}\right)\)

\(\left(x.y\right).\left(y.z\right)\left(z.x\right)=\dfrac{1}{2}.\dfrac{3}{5}.\dfrac{27}{10}\\ \Rightarrow\left(x.y.z\right)^2=\dfrac{81}{100}\\ \Rightarrow\left[{}\begin{matrix}x.y.z=\dfrac{9}{10}\\x.y.z=-\dfrac{9}{10}\end{matrix}\right.\)

Nếu x.y.z=9/10

\(\Rightarrow z=\dfrac{9}{10}:\dfrac{1}{2}=\dfrac{9}{5};x=\dfrac{9}{10}:\dfrac{3}{5}=\dfrac{3}{2};y=\dfrac{9}{10}:\dfrac{27}{10}=\dfrac{1}{3}\)

Nếu x.y.z=-9/10

\(\Rightarrow z=-\dfrac{9}{5};x=-\dfrac{3}{2};y=-\dfrac{1}{3}\)

b: Ta có: x/y=7/9

nên x/7=y/9

=>x/49=y/63

Ta có: y/z=7/3

nên y/7=z/3

=>y/63=z/27

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

\(\dfrac{x}{49}=\dfrac{y}{63}=\dfrac{z}{27}=\dfrac{x-y+z}{49-63+27}=\dfrac{-15}{13}\)

Do đó: x=-735/13; y=-945/13; z=-405/13

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

\(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}=\dfrac{2x+5y-2z}{2\cdot7+5\cdot20-2\cdot32}=\dfrac{100}{50}=2\)

Do đó: x=14; y=40; z=64

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

\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}=\dfrac{x-y-z}{8-5-2}=3\)

Do đó: x=24; y=15; z=6

Cậu không làm được hay cần gấp con nào nhỉ ?

Bài 1:

a: \(\Leftrightarrow\dfrac{x+2}{2}=x-5\)

=>2x-10=x+2

=>x=12

b: \(\Leftrightarrow\left(x+2\right)^2=100\)

=>x+2=10 hoặc x+2=-10

=>x=-12 hoặc x=8

c: \(\Leftrightarrow\left(2x-5\right)^3=27\)

=>2x-5=3

=>2x=8

=>x=4

b: 2x^3-1=15

=>2x^3=16

=>x=2

\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)

=>y-25=32; z+9=50

=>y=57; z=41

d: 3/5x=2/3y

=>9x=10y

=>x/10=y/9=k

=>x=10k; y=9k

x^2-y^2=38

=>100k^2-81k^2=38

=>19k^2=38

=>k^2=2

TH1: k=căn 2

=>\(x=10\sqrt{2};y=9\sqrt{2}\)

TH2: k=-căn 2

=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)

Bài 1:

\(a,\dfrac{x}{3}=\dfrac{y}{7}\) và \(x+y=20\)

\(=\dfrac{x+y}{3+7}=\dfrac{20}{10}=2\)

\(\Rightarrow x=2.3=6\)

\(y=2.7=14\)

Vậy \(x=6\) và \(y=14\)

\(b,\dfrac{x}{5}=\dfrac{y}{2}\) và \(x-y=6\)

\(=\dfrac{x-y}{5-2}=\dfrac{6}{3}=2\)

\(\Rightarrow x=2.5=10\)

\(y=2.2=4\)

Vậy \(x=10\) và \(y=4\)

\(c,\dfrac{x}{7}=\dfrac{18}{14}\)

Từ tỉ lệ thức trên ta có:

\(14x=7.18\)

\(x=\dfrac{7.18}{14}\)

\(x=9\)

Vậy \(x=9\)

\(d,6:x=1\dfrac{3}{4}:5\)

\(6:x=\dfrac{7}{20}\)

\(x=6:\dfrac{7}{20}\)

\(x=\dfrac{120}{7}\)

Vậy \(x=\dfrac{120}{7}\)

\(e,\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\) và \(x-y+z=8\)

\(=\dfrac{x-y+z}{2-4+6}=\dfrac{8}{4}=2\)

\(\Rightarrow x=2.2=4\)

\(y=2.4=8\)

\(z=2.6=12\)

Vậy \(x=4;y=8;z=12\)

a, \(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x+y}{3+7}=\dfrac{1}{2}\)

Từ đó suy ra x=1,5; y=3,5

b,\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x-y}{5-2}=\dfrac{1}{2}\)

Từ đó suy ra x=2,5; y=1

c,\(\dfrac{x}{7}=\dfrac{18}{14}\Leftrightarrow\dfrac{x}{7}=\dfrac{9}{7}\Rightarrow x=9\)

d,\(\dfrac{6}{x}=\dfrac{\dfrac{7}{4}}{5}\Leftrightarrow\dfrac{6}{x}=\dfrac{24}{7}\left(\dfrac{\dfrac{7}{4}}{5}\right)\Leftrightarrow\dfrac{6}{x}=\dfrac{6}{\dfrac{120}{7}}\Rightarrow x=\dfrac{120}{7}\)

e,\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{8}=\dfrac{x-y+z}{2-4+8}=\dfrac{4}{3}\)

Từ đó suy ra x=\(\dfrac{8}{3}\); y=\(\dfrac{16}{3}\); z=\(\dfrac{32}{3}\)

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

\(\Rightarrow\dfrac{2\left(x-2\right)}{8}=\dfrac{y+1}{5}=\dfrac{2\left(z+3\right)}{14}\)

\(\Rightarrow\dfrac{2x-4}{8}=\dfrac{y+1}{5}=\dfrac{2z+6}{14}\)

Dựa vào tính chất dãy tỉ số bằng nhau ta có:

\(=\dfrac{2x-4+y+1-2z-6}{8+5-14}\)

\(=\dfrac{2x+y-2z-9}{-1}\)

\(=\dfrac{7-9}{-1}=2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-2}{4}=2\Rightarrow x-2=8\Rightarrow x=10\\\dfrac{y+1}{5}=2\Rightarrow y+1=10\Rightarrow y=9\\\dfrac{z+3}{7}=2\Rightarrow z+3=14\Rightarrow z=11\end{matrix}\right.\)