cậu viết chắc lâu lắm nhỉ

a)x=4, y=6 ,z=10                                   c)x=6,y=9,z=12                              e)x=-3,y=-5,z=154/3

b)x=12,y=16,z=28                                   d) y=-28, x=-42,z=-20                   f)x=36,y=24,z=9

g)nản                                                    h)x=1,y=2,z=3

        làm mất bao nhiêu lâu. k đúng giùm

a)  ko có " z" sao làm!!

b) áp dụng t/c dãy tỉ số bằng nhau ta có:

 \(\frac{x}{3}=\frac{y}{4}=\frac{z}{7}\) =\(\frac{z-x}{7-4}=\frac{16}{3}\)

=> x/3 = 16/3 => x = 16

=> y/4 = 16/3 => y = ...

=> z/7 = 16/3 => z = ...

Có ai trình bày chi tiết đc ko zậy? Mình chẳng hiểu gì hết trơn á!!!🤔

Đặt: \(\left\{{}\begin{matrix}x=a\\2y=b\\3z=c\end{matrix}\right.\Rightarrow a+b+c=18\)

Có: BDT

\(\Leftrightarrow\sum_{cyc}\left(\frac{b+c+5}{a+1}\right)\ge\frac{51}{7}\)

\(\Leftrightarrow\sum_{cyc}\left(\frac{a+b+c-a+5}{a+1}\right)\ge\frac{51}{7}\)(1)

Đặt tiếp tục: \(\left\{{}\begin{matrix}m=a+1\\n=b+1\\p=c+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=m-1\\b=n-1\\c=p-1\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sum_{cyc}\left(\frac{24-m}{m}\right)\ge\frac{51}{7}\)

\(\Leftrightarrow\sum_{cyc}\left(\frac{24}{m}-1\right)\ge\frac{51}{7}\)

\(\Leftrightarrow24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge\frac{72}{7}\)

\(\Leftrightarrow\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\ge\frac{3}{7}\)

\(\Leftrightarrow\left(m+n+p\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge21\cdot\frac{3}{7}=9\)

\(\left(\frac{m}{n}-2+\frac{n}{m}\right)+\left(\frac{p}{m}-2+\frac{m}{p}\right)+\left(\frac{n}{p}-2+\frac{p}{n}\right)\ge0\)

\(\Leftrightarrow\frac{\left(m-n\right)^2}{mn}+\frac{\left(p-m\right)^2}{pm}+\frac{\left(n-p\right)^2}{pn}\ge0\)(đúng)

Đặt: \(\left\{{}\begin{matrix}x=a\\2y=b\\3z=c\end{matrix}\right.\)

BĐT

\(\Leftrightarrow\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+b+5}{c+1}\ge\frac{51}{7}\)

\(\Leftrightarrow\frac{a+b+c-a+5}{a+1}+\frac{a+c+b-b+5}{b+1}+\frac{a+b+c-c+5}{c+1}\ge\frac{51}{7}\)

\(\Leftrightarrow\frac{24-\left(a+1\right)}{a+1}+\frac{24-\left(b+1\right)}{b+1}+\frac{24-\left(c+1\right)}{c+1}\ge\frac{51}{7}\)(1)

Đặt tiếp: \(\left\{{}\begin{matrix}a+1=m\\b+1=n\\c+1=p\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=m-1\\b=n-1\\c=p-1\end{matrix}\right.\)

(1)\(\Leftrightarrow\frac{24-m}{m}+\frac{24-n}{n}+\frac{24-p}{p}\ge\frac{51}{7}\)

\(\Leftrightarrow24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)-3\ge\frac{51}{7}\)

\(\Leftrightarrow24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge\frac{72}{7}\)

\(\Leftrightarrow\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\ge\frac{3}{7}\)

\(\Leftrightarrow\left(m+n+p\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge\frac{3}{7}\left(m+n+p\right)\)( do m+n+p>0)

\(\Leftrightarrow3+\frac{m}{n}+\frac{n}{m}+\frac{p}{n}+\frac{n}{p}+\frac{m}{p}+\frac{p}{m}\ge\frac{3}{7}\left[\left(a+b+c\right)+3\right]\)

\(\Leftrightarrow\frac{m}{n}+\frac{n}{m}+\frac{p}{n}+\frac{n}{p}+\frac{p}{m}+\frac{m}{p}-6\ge0\)

Tới đây chắc bn làm đc rồi

Mình chỉ hướng dẫn giải thôi nhá chứ nhiều bài quá

a) Đặt \(\frac{x}{5}=\frac{y}{7}=k\Rightarrow x=5k;y=7k\)

Thay x.y=315 => 5k.7k=315 <=> 35k²=315 => k²=9 => k=3

x=5.3=15 ; y=7.3=21

b) 5x=9y<=> \(\frac{x}{9}=\frac{y}{5}\)

Theo TCDTSBN ta có : \(\frac{x}{9}=\frac{y}{5}=\frac{2x+3y}{2.9+3.5}=\frac{-33}{33}=-1\)

x/9=-1=>x=-9 ; y/5=-1=>y=-5

các bài còn lại tương tự b 

Đặt \(\hept{\begin{cases}a=x\\b=2y\\c=3z\end{cases}}\) => a + b + c = 18

\(P=\frac{2y+3z+5}{1+x}+\frac{3z+x+5}{1+2y}+\frac{x+2y+5}{1+3z}=\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+b+5}{c+1}\)

Lại đặt \(\hept{\begin{cases}m=a+1\\n=b+1\\p=c+1\end{cases}}\Rightarrow\hept{\begin{cases}a=m-1\\b=n-1\\c=p-1\end{cases}}\) 

Ta có : \(\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+c+5}{c+1}=\frac{24-m}{m}+\frac{24-n}{n}+\frac{24-p}{p}\)

\(=24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)-3\ge\frac{24.9}{m+n+p}-3=\frac{24.9}{\left(a+1\right)+\left(b+1\right)+\left(b+1\right)}-3\)

                                                       \(=\frac{24.9}{18+3}-3=\frac{51}{7}\)

Đặt \(\frac{x}{8}=\frac{y}{3}=\frac{z}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=8k\\y=3k\\z=5k\end{matrix}\right.\)

Ta có: \(A=\frac{x+y-3z}{x+y+3z}\)

\(=\frac{8k+3k-3\cdot5k}{8k+3k+3\cdot5k}=\frac{11k-15k}{11k+15k}=\frac{-4k}{16k}=\frac{-1}{4}\)

Vậy: \(A=-\frac{1}{4}\)

Mấy bài còn lại tương tự nhé cậu