y + y : 2/ 7 + y x 3/5 = 102
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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:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)
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
=>\(y=\frac{475}{50}=9,5\)
1)\(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)
\(\Rightarrow x=-\dfrac{3}{6}\cdot\left(-2\right)=1\)
\(\Rightarrow y=-18:\dfrac{-3}{6}=36\)
\(\Rightarrow z=-\dfrac{3}{6}\cdot\left(-24\right)=12\)
câu cuối làm tương tự
\(\frac{x}{2}=\frac{y}{3}\\ \Rightarrow\frac{x}{10}=\frac{y}{15}\)
\(\frac{y}{5}=\frac{z}{7}\\ \Rightarrow\frac{y}{10}=\frac{z}{14}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{14}=\frac{2x}{20}=\frac{3y}{45}=\frac{z}{14}=\frac{2x+3y+z}{20+45+14}=\frac{102}{79}\)
\(\Rightarrow x=\frac{1020}{79};y=\frac{1530}{79};z=\frac{1428}{79}\)
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
=>\(y=\frac{475}{50}=9,5\)
a/ 2x = 5y và x - 2y = -12
Ta có: 2x = 5y => \(\frac{x}{5}=\frac{y}{2}\)
Áp dụng: tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{5}=\frac{y}{2}=\frac{x-y}{5+2}=\frac{x-2y}{5+2.2}=\frac{-12}{9}=-\frac{4}{3}\)
\(\frac{x}{5}=-\frac{4}{3}\Rightarrow x=\frac{-4}{3}.5=-\frac{20}{3}\)
\(\frac{y}{2}=-\frac{4}{3}\Rightarrow y=-\frac{4}{3}.2=-\frac{8}{3}\)
Vậy:.................
b/ 2x = 3y = 4z và x + y + z =21
Ta có: 2x = 3y = 4z
=> \(\frac{2x}{12}=\frac{3y}{12}=\frac{4z}{12}\)
=> \(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)
Áp dụng: tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}=\frac{x+y+z}{6+4+3}=\frac{21}{13}\)
\(\frac{x}{6}=\frac{21}{13}\Rightarrow x=\frac{21}{13}.6=\frac{126}{13}\)
\(\frac{y}{4}=\frac{21}{13}\Rightarrow y=\frac{21}{13}.4=\frac{84}{13}\)
\(\frac{z}{3}=\frac{21}{13}\Rightarrow z=\frac{21}{13}.3=\frac{63}{13}\)
Vậy:...............
c/Áp dụng: tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{32}{8}=4\)
\(\frac{x}{3}=4\Rightarrow x=4.3=12\)
\(\frac{y}{5}=4\Rightarrow y=4.5=20\)
Vậy:................
d/ Ta có: 7x = 3y
=> \(\frac{7x}{21}=\frac{3y}{21}\)
=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng: tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{16}{-4}=-4\)
\(\frac{x}{4}=-4\Rightarrow x=\left(-4\right).4=-16\)
\(\frac{y}{7}=-4\Rightarrow y=\left(-4\right).7=-28\)
Vậy:................
Bài 9:
Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)
Vậy: (x,y,z,t)=(-10;6;34;-18)
Bài 11:
Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)
\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)
Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)
\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)
Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)
\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)
Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)
\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)
Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)
\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)
Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)
\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)
Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)
<=> y + y x 7/2 + y x 3/5 = 102
<=> y x 10/10 + y x 35/10 + y x 6/10 = 102
<=> y x 51/10 =102
<=> y = 20
Vậy ...