\(\left\{{}\begin{matrix}x+y=52\\4x+5y=233\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+4y=208\\4x+5y=233\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-5y=208-233\\4x+5y=233\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-y=25\\4x=233-5y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=25\\4x=233-5.25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=25\\4x=108\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=25\\x=27\end{matrix}\right.\)

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

Áp dụng t.c của dãy tỉ só bằng nhau,ta có:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y}{3+4}=\dfrac{16}{7}\)

=>\(x=\dfrac{16}{7}.3=\dfrac{48}{7}\)

\(y=\dfrac{16}{7}.4=\dfrac{64}{7}\)

\(z=\dfrac{16}{7}.5=\dfrac{80}{7}\)

Vậy...

Các câu sau tương tự

ADTC dãy tỉ số bằng nhau ta cs

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x^2+y^2}{2^2+3^2}=\frac{52}{13}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=4\\\frac{y}{3}=4\\\frac{z}{4}=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=12\\z=16\end{matrix}\right.\)

ko biết

a: \(\frac{52}{17}>\frac{51}{17}=3\)

\(3=\frac{121}{41}>\frac{120}{41}\)

Do đó: \(\frac{52}{17}>\frac{120}{41}\)

b: \(\frac34+\frac14:\left(\frac{7}{12}-\frac16\right)\)

\(=\frac34+\frac14:\left(\frac{7}{12}-\frac{2}{12}\right)\)

\(=\frac34+\frac14:\frac{5}{12}\)

\(=\frac34+\frac14\times\frac{12}{5}=\frac34+\frac35=\frac{15}{20}+\frac{12}{20}=\frac{27}{20}\)

c: \(372,463\cdot998+744,926\)

\(=372,463\cdot998+372,463\cdot2\)

\(=372,463\times\left(998+2\right)=372,463\times1000=372463\)

d: Số số hạng trong dãy số 2;4;6;...;100 là:

\(\left(100-2\right):2+1=98:2+1=49+1=50\) (số)

\(2-4+6-8+10-12+\cdots+98-100+102\)

\(=\left(2-4\right)+\left(6-8\right)+\cdots+\left(98-100\right)+102\)

=(-2)+(-2)+...+(-2)+102

\(=-2\cdot\frac{50}{2}+102=-50+102=52\)

e: (y+112)-113=79

=>y+112-113=79

=>y-1=79

=>y=79+1=80

f: \(\frac34-y=\frac12\)

=>\(y=\frac34-\frac12=\frac14\)

g: \(\left(\frac45-2\times y\right)+\frac16=\frac56\)

=>\(\frac45-2\times y=\frac56-\frac16=\frac46=\frac23\)

=>\(2\times y=\frac45-\frac23=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}\)

=>\(y=\frac{2}{15}:2=\frac{1}{15}\)

h: (y+1)+(y+2)+...+(y+50)=1750

=>50y+(1+2+...+50)=1750

=>\(50y+50\times\frac{51}{2}=1750\)

=>50y+1275=1750

=>50y=1750-1275=475

=>\(y=\frac{475}{50}=9,5\)

giúp mình nhé

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a: \(\frac{52}{17}>\frac{51}{17}=3\)

\(3=\frac{121}{41}>\frac{120}{41}\)

Do đó: \(\frac{52}{17}>\frac{120}{41}\)

b: \(\frac34+\frac14:\left(\frac{7}{12}-\frac16\right)\)

\(=\frac34+\frac14:\left(\frac{7}{12}-\frac{2}{12}\right)\)

\(=\frac34+\frac14:\frac{5}{12}\)

\(=\frac34+\frac14\times\frac{12}{5}=\frac34+\frac35=\frac{15}{20}+\frac{12}{20}=\frac{27}{20}\)

c: \(372,463\cdot998+744,926\)

\(=372,463\cdot998+372,463\cdot2\)

\(=372,463\times\left(998+2\right)=372,463\times1000=372463\)

d: Số số hạng trong dãy số 2;4;6;...;100 là:

\(\left(100-2\right):2+1=98:2+1=49+1=50\) (số)

\(2-4+6-8+10-12+\cdots+98-100+102\)

\(=\left(2-4\right)+\left(6-8\right)+\cdots+\left(98-100\right)+102\)

=(-2)+(-2)+...+(-2)+102

\(=-2\cdot\frac{50}{2}+102=-50+102=52\)

e: (y+112)-113=79

=>y+112-113=79

=>y-1=79

=>y=79+1=80

f: \(\frac34-y=\frac12\)

=>\(y=\frac34-\frac12=\frac14\)

g: \(\left(\frac45-2\times y\right)+\frac16=\frac56\)

=>\(\frac45-2\times y=\frac56-\frac16=\frac46=\frac23\)

=>\(2\times y=\frac45-\frac23=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}\)

=>\(y=\frac{2}{15}:2=\frac{1}{15}\)

h: (y+1)+(y+2)+...+(y+50)=1750

=>50y+(1+2+...+50)=1750

=>\(50y+50\times\frac{51}{2}=1750\)

=>50y+1275=1750

=>50y=1750-1275=475

=>\(y=\frac{475}{50}=9,5\)