6-8+10-12+...-y=-200
Thanks m.n :m.n giải thích hộ mik nhé cảm ưn
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Ta thấy rằng từ 6 đến y có:\(\frac{y-6}{2}+1=\frac{y-4}{2}\)số
Nên sẽ có: \(\frac{y-4}{4}\)cặp
Theo đề bài ta có:
\(6-8+10-12+...-y=-200\)
\(\Rightarrow\left(6-8\right)+\left(10-12\right)+...+\left(y-2-y\right)=-200\)
\(\Rightarrow-2-2-...-2=-200\) (có \(\frac{y-4}{4}\)số -2)
\(\Rightarrow-2.\frac{y-4}{4}=-200\)
\(\Rightarrow y-4=400\)
\(\Rightarrow y=404\)
\(-23.63+23.21-58.23\\ =23.\left(-63\right)+23.21-58.23\\ =23\left(-63+21-58\right)\\ =23.\left(-100\right)=-2300\)
-23 . 63 + 23 . 21 - 58 . 23
= 23 . (-63 + 21 - 58) = 23 . (-100) = -2300
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
=>\(y=\frac{475}{50}=9,5\)
\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10+6\cdot12}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20+18\cdot24}\)
\(A=\frac{2\cdot3\left[1\cdot2\right]+2\cdot3\left[2\cdot4\right]+2\cdot3\left[3\cdot6\right]+2\cdot3\left[4\cdot8\right]+2\cdot3\left[5\cdot10\right]}{3\cdot4\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}\)
\(A=\frac{\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}{2\cdot3\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}=\frac{1}{2\cdot3}=\frac{1}{6}\)