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Dạng 3:
Bài 1:
a) Số lượng số hạng là:
\(\left(999-1\right):1+1=999\) (số hạng)
Tổng dãy là:
\(A=\left(999+1\right)\cdot999:2=499500\)
b) Số lượng số hạng là:
\(\left(100-7\right):3+1=32\) (số hạng)
Tổng dãy là:
\(S=\left(100+7\right)\cdot32:2=1712\)
a) \(7^2-7\left(13-x\right)=14\)
\(7\left(13-x\right)=49-14=35\)
\(13-x=5\)
\(x=13-5=8\)
b) \(5x-5^2=10\)
\(5x=10+25=35\)
\(x=7\)
c) \(4\left(x-5\right)-2^3=2^4.3=48\)
\(4\left(x-5\right)=48+8=56\)
\(x-5=14\)
\(x=19\)
4:
a: \(\Leftrightarrow49+7\left(x-13\right)=14\)
=>7(x-13)=35
=>x-13=5
=>x=18
b: \(5x-5^2=10\)
=>\(5x=10+25=35\)
=>x=7
c: \(4\left(x-5\right)-2^3=2^4\cdot3\)
=>\(4\left(x-5\right)=16\cdot3+8=56\)
=>x-5=14
=>x=19
Bài 3:
a) \(2\cdot5^2+3:71^0-54:3^3\)
\(=2\cdot25+3:1-54:27\)
\(=50+3-2\)
\(=50+1\)
\(=51\)
b) \(150+50:5-2\cdot3^2\)
\(=150+10-2\cdot9\)
\(=160-18\)
\(=142\)
2:
c: \(30=2\cdot3\cdot5;48=2^4\cdot3\)
=>\(ƯCLN\left(30;48\right)=2\cdot3=6\)
=>\(x\inƯ\left(6\right)\)
=>\(x\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
d: \(25=5^2;80=2^4\cdot5\)
=>\(ƯCLN\left(25;80\right)=5\)
\(x\inƯC\left(25;80\right)\)
=>\(x\inƯ\left(5\right)\)
=>\(x\in\left\{1;-1;5;-5\right\}\)
e: \(36=2^2\cdot3^2;18=3^2\cdot2\)
=>\(ƯCLN\left(36;18\right)=2\cdot3^2=18\)
=>x=18
f: \(49=7^2;84=2^2\cdot3\cdot7\)
=>\(ƯCLN\left(49;84\right)=7\)
=>x=7
a) Ta có: \(A=\dfrac{7}{12}+\dfrac{5}{12}:6-\dfrac{11}{36}\)
\(=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}\)
\(=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}\)
\(=\dfrac{25}{36}\)
b) Ta có: \(B=\left(\dfrac{4}{5}+\dfrac{1}{2}\right):\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\)
\(=\left(\dfrac{8}{10}+\dfrac{5}{10}\right):\dfrac{-5}{13}\)
\(=\dfrac{13}{10}\cdot\dfrac{13}{-5}\)
\(=-\dfrac{169}{50}\)
c) Ta có: \(C=\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)\)
\(=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{115}{132}\cdot\dfrac{132}{103}=\dfrac{115}{103}\)
2^3+x=32
2^3+x=2^4
Ta có:
3+x=4
X=4-3
X=1