không có x y nào vì không có số tự nhiên nào có tích lũy thừa bằng 0

2x - 7y = 0

x = 7y/2 (1)

x+3y -26 = 0 thay (1) vào có

7y/2 +3y - 26 = 0 => y = 4

thay y = 4 vào (1) có

x = 7.4/2 = 14

kl: x = 14 ; y= 4

Vì: \(\left(2x-7y\right)^{2010}\ge0;\left|x+3y-26\right|^{2010}\ge0\)

Nên: \(\left(2x-7y\right)^{2010}+\left|x+3y-26\right|^{2010}=0\)

\(\Leftrightarrow\begin{cases}2x-7y=0\\x+3y-26=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=14\\y=4\end{cases}\)

bạn xem lại đề bài lại hộ mình 3x+5x-7y+30 nhé

1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)

2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)

3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)

Áp dụng t/c dtsbn:

\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)

b) \(2x=3y=6z\) và \(x+y+z=1830\)

Ta có: \(2x=3y=6z\Rightarrow\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{6}}\) và \(x+y+z=1830\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

\(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{6}}=\frac{x+y+z}{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=\frac{1830}{1}=1830\) 

\(\Rightarrow x=1830.\frac{1}{2}=915\)

\(y=1830.\frac{1}{3}=610\)

\(z=1830.\frac{1}{6}=305\)

a)  \(\left(a-2009\right)^2+\left(b+2010\right)^2=0\)

Ta có: \(\left(a-2009\right)^2\ge0\)

\(\left(b+2010\right)^2\ge0\)

Để \(\left(a-2009\right)^2+\left(b+2010\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}a-2009=0\Rightarrow a=2009\\b+2010=0\Rightarrow b=-2010\end{cases}}\)

Vậy \(a=2009\)

\(b=-2010\)

- Bà P _ _ _ _ giỏi nhể =)