\(\frac{x}{5}=\frac{y}{9}\Rightarrow\hept{\begin{cases}x=5k\\y=9k\end{cases}}\)

\(\Rightarrow xy=45k^2\)  mà xy = 405

\(\Rightarrow45k^2=405\)

\(\Rightarrow k^2=9\)

\(\Rightarrow k=\pm3\)

thay k vào là được nhaa

Đặt \(\frac{x}{5}=\frac{y}{9}=k\Rightarrow x=5k,y=9\)\(k\)

\(\Rightarrow xy=45k^2\)mà \(xy=405\Rightarrow k^2=9\Rightarrow\hept{\begin{cases}k=9\\k=-9\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=45,y=81\\x=-45,y=-81\end{cases}}\)

                                  Vậy\(\hept{\begin{cases}x=45,y=81\\x=-45,y=-81\end{cases}}\)

Đặt : \(\frac{x}{5}=\frac{y}{9}=k\Rightarrow x=5k;y=9k\)

\(\Rightarrow x\cdot y=5k\cdot9k=45k^2=405\Rightarrow k^2=\frac{405}{45}=9\Rightarrow k=\pm3\)

TH1 : \(k=3\Rightarrow\hept{\begin{cases}\frac{x}{5}=3\Rightarrow x=3\cdot5=15\\\frac{y}{9}=3\Rightarrow y=3\cdot9=27\end{cases}}\)

TH2 : \(k=-3\Rightarrow\hept{\begin{cases}\frac{x}{5}=-3\Rightarrow x=-3\cdot5=-15\\\frac{y}{9}=-3\Rightarrow y=-3\cdot9=-27\end{cases}}\)

Vậy x = 15 thì y = 27; x = -15 thì y = -27

Chúc bạn học tốt!

Lời giải:

a, Ta có: \(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}\Rightarrow\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}\). Mà theo đề bài: 5x + y - 2z = 28

=> Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{5x}{50}=\frac{x}{10}=2\Leftrightarrow x=20\\\frac{y}{6}=2\Leftrightarrow y=12\\\frac{2z}{42}=\frac{z}{21}=2\Leftrightarrow z=42\end{matrix}\right.\)(TMĐK)

Vậy: \(x=20;y=12;z=42\)

b, Ta có: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\) ; \(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{20}=\frac{z}{28}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\Rightarrow\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}\). Mà theo đề bài: 2x+3y - z = 124

=> Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{124}{62}=2\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{2x}{30}=\frac{x}{15}=2\Leftrightarrow x=30\\\frac{3y}{60}=\frac{y}{20}=2\Leftrightarrow y=40\\\frac{z}{28}=2\Leftrightarrow z=56\end{matrix}\right.\)(TMĐK)

Vây:\(x=30;y=40;z=56\)

c, Ta có: \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x.x}{2}=\frac{x.y}{3}\). Mà x.y = 54

\(\Rightarrow\frac{x.x}{2}=\frac{x.y}{3}=\frac{54}{3}=18\)

\(\Rightarrow\frac{x^2}{2}=18\Rightarrow x^2=36\Rightarrow x\in\left\{6;-6\right\}\)

Nếu \(x=6\Rightarrow\frac{6.y}{3}=18\Rightarrow6.y=54\Rightarrow y=9\)

Nếu \(x=-6\Rightarrow\frac{-6.y}{3}=18\Rightarrow-6.y=54\Rightarrow y=-9\)

Vậy: \(\left(x;y\right)\in\left\{\left(6;9\right),\left(-6;-9\right)\right\}\)

d, Ta có: \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Rightarrow\frac{12x}{18}=\frac{12y}{16}=\frac{12z}{15}\). Mà theo đề bài, ta có: x + y + z = 49

=> Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{12x}{18}=\frac{12y}{16}=\frac{12z}{15}=\frac{12.\left(x+y+z\right)}{18+16+15}=\frac{12.49}{49}=12\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{12x}{18}=\frac{2x}{3}=12\Rightarrow x=18\\\frac{12y}{16}=\frac{3y}{4}=12\Rightarrow y=16\\\frac{12z}{15}=\frac{4z}{5}=12\Rightarrow z=15\end{matrix}\right.\)(TMĐK)

Vậy:\(x=18;y=16;z=15\)

e, Ta có: \(\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{5^2}=\frac{y^2}{3^2}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\).Mà theo đề bài, ta có: x² - y² = 4

=> Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{4}{16}=\frac{1}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x^2}{25}=\frac{1}{4}\Rightarrow x^2=\frac{25}{4}\Rightarrow x\in\left\{\frac{5}{2};-\frac{5}{2}\right\}\\\frac{y^2}{9}=\frac{1}{4}\Rightarrow x^2=\frac{9}{4}\Rightarrow x\in\left\{\frac{3}{2};-\frac{3}{2}\right\}\end{matrix}\right.\)(TMĐK)

Vậy:..................................

Ta có :

1) 45^10 . 5^30= (5.9)^10 . 5^30 = 5^10 . 5^30 . 9^10 = 5^40 . 3^20 = 25^20 . 3^20=75^20

2)\(\sqrt{40+2}=\sqrt{42}<\sqrt{49}=7=6+1=\sqrt{36}+\sqrt{1}<\sqrt{40}+\sqrt{2}\)

Vậy \(\sqrt{40+2}<\sqrt{40}+\sqrt{2}\)

3)\(Cho\frac{x}{3}=\frac{y}{4}=k\Rightarrow x=3k;y=4k\)

Ta lại có:

\(xy=12\Rightarrow3k.4k=12\)

\(12.k^2=12\Rightarrow k^2=1\Rightarrow k=1:-1\)

\(Vơik=1\Rightarrow x=1.3=3;y=1.4=4\)

\(k=-1\Rightarrow x=-1.3=-3;y=-1.4=-4\)

*Bài làm:

a)*Ta có : \(\frac{x}{10}\) = \(\frac{y}{6}\) = \(\frac{z}{21}\)

\(\Rightarrow\) \(\frac{5x}{50}\) = \(\frac{y}{6}\) = \(\frac{2z}{42}\) . \(và5x+y-2z=28\)

\(\Rightarrow\) Áp dụng tính chất dãy tỉ số bằng nhau ta được :

\(\frac{5x}{50}\) = \(\frac{y}{6}\) = \(\frac{2z}{42}\) = \(\frac{5x+y-2z}{50+6-42}\) = \(\frac{28}{14}\) = \(2\)

\(\Rightarrow\left\{{}\begin{matrix}5x=2.50=100\\y=2.6=12\\2z=2.42=84\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=20\\y=12\\z=42\end{matrix}\right.\)

*Vậy \(\left(x;y;z\right)=\left(20;12;42\right)\) .

b)*Ta có: \(\frac{x}{3}\) = \(\frac{y}{4}\) ; \(\frac{y}{5}\) = \(\frac{z}{7}\)

\(\Rightarrow\) \(\frac{x}{15}\) = \(\frac{y}{20}\) ; \(\frac{y}{20}\) = \(\frac{z}{28}\)

\(\Rightarrow\) \(\frac{x}{15}\) = \(\frac{y}{20}\) = \(\frac{z}{28}\)

\(\Rightarrow\) \(\frac{2x}{30}\) = \(\frac{3y}{60}\) = \(\frac{z}{28}\) .\(và2x+3y-z=124\)

\(\Rightarrow\) Áp dụng tính chất dãy tỉ số bằng nhau ta được :

\(\frac{2x}{30}\) = \(\frac{3y}{60}\) = \(\frac{z}{28}\) = \(\frac{2x+3y-z}{30+60-28}\) = \(\frac{124}{62}\) = \(2\)

\(\Rightarrow\left\{{}\begin{matrix}2x=2.30=60\\3y=2.60=120\\z=2.28=56\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=30\\y=40\\z=56\end{matrix}\right.\)

*Vậy \(\left(x;y;z\right)=\left(30;40;56\right)\) .

c) *Ta có: \(\frac{2x}{3}\) = \(\frac{3y}{4}\) = \(\frac{4z}{5}\)

\(\Rightarrow\) \(\frac{40x}{60}\) = \(\frac{45y}{60}\) = \(\frac{48z}{60}\)

\(\Rightarrow40x=45y=48z\)

\(\Rightarrow\) \(\frac{40x}{720}\) = \(\frac{45y}{720}\) = \(\frac{48z}{720}\)

\(\Rightarrow\) \(\frac{x}{18}\) = \(\frac{y}{16}\) = \(\frac{z}{15}\) .\(vàx+y+z=49\)

\(\Rightarrow\) Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{x}{18}\) = \(\frac{y}{16}\) = \(\frac{z}{15}\) = \(\frac{x+y+z}{18+16+15}\) =\(\frac{49}{49}\) = \(1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.18=18\\y=1.16=16\\z=1.15=15\end{matrix}\right.\)

*Vậy \(\left(x;y;z\right)=\left(18;16;15\right)\) .

d) *Ta có: Đặt: \(\frac{x}{2}\) = \(\frac{y}{3}\) = \(k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

\(Mà\) \(xy=54\) (theo đề bài)

\(\Rightarrow\) \(xy=2k.3k=54\)

\(\Rightarrow\) \(xy=6k^2=54\)

\(\Rightarrow\) \(k^2=9\)

\(\Rightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)

~ Với \(k=3\) thì: \(\left\{{}\begin{matrix}x=2.3=6\\y=3.3=9\end{matrix}\right.\)

~ Với \(k=-3\) thì: \(\left\{{}\begin{matrix}x=2.\left(-3\right)=-6\\y=3.\left(-3\right)=-9\end{matrix}\right.\)

*Vậy \(\left(x;y\right)=\left\{\left(6;9\right),\left(-6;-9\right)\right\}\) .

*Chúc bạn hok tốt!

Mình thấy bạn hỏi dạng bài này nhiều rồi mà. nguyen ngoc son

\(A=\frac{\left(x-y\right)^2}{x-y}+\frac{2xy}{x-y}=\left(x-y\right)+\frac{2}{x-y}\ge2\sqrt{\left(x-y\right).\frac{2}{x-y}}=2\sqrt{2}\)

khi x -y = căn 2 ; xy =1 => x =\(\frac{\sqrt{6}+\sqrt{2}}{2}\)y= thay dấu + = -

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\Rightarrow x=2k;y=3k;z=5k\)

\(xy+yz+zx=6k^2+15k^2+10k^2\)

\(\Rightarrow31k^2=31\Rightarrow k^2=1\)\(\Rightarrow k=1\Rightarrow x=2;y=3;z=5\)

Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow x=2k;y=3k;z=5k\)

\(\Rightarrow xy+yz+xz=31\)

\(\Rightarrow6k^2+15k^2+10k^2=31\)

\(\Rightarrow31k^2=31\)

\(\Rightarrow k=\hept{\begin{cases}-1\\1\end{cases}}\)

Với k = 1 => x = 2;y=3;z=5

Với k = -1=> x=-2;y=-3;z=-5