Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
Giải:
Ta có: \(a+b=-3\)
\(\Rightarrow a+b+c=-3+c\)
\(\Rightarrow a+\left(-5\right)=-3+c\)
\(\Rightarrow a-c=\left(-3\right)-\left(-5\right)\)
\(\Rightarrow a-c=2\)
Mà \(c+a=-4\)
\(\Rightarrow a=\left(-4+2\right):2=-1\)
\(\Rightarrow c=\left(-4\right)-\left(-1\right)=-3\)
Lại có: \(b+c=-5\)
\(\Rightarrow b+\left(-3\right)=-5\)
\(\Rightarrow b=-2\)
Vậy bộ số \(\left(a;b;c\right)\) là \(\left(-1;-3;-2\right)\)
Bài 2:
\(P=78+\left|78-129\right|+\left(-29\right)\)
\(\Rightarrow P=78+\left[-\left(78-129\right)\right]+\left(-29\right)\)
\(\Rightarrow P=78+\left(-78\right)+129+\left(-29\right)\)
\(\Rightarrow P=100\)
a) 18.17−3.6.7
= 18.17 - 18 . 7
= 18 ( 17 - 7 )
= 18 . 10 = 180
b) 54−6.(17+9)
= 54 - 102 - 54
= - 102
c) 33.(17−5)−17.(33−5)
= 33. 17 - 33. 5 - 17 .33 + 17 . 5
= 17 . 5 - 33 . 5
= 5 ( 17 - 33 )
= 5. (-16)
= - 80
a) 18.17 - 3.6.7
= 18.17 - 18.7
= 18.(17 - 7 )
= 18 . 10
= 180
b) 54 - 6.(17 + 9)
= 6.9 - 6.17 + 6.9
= (6.9 - 6.9) - 6.17
= 0 - 6.17
= 0 - 102
= -102
c) 33.(17 - 5) - 17.(33 - 5)
= 33.17 - 33.5 - 17 . 33 - 17.5
= 33.(17 - 17) - 5.(33 - 17 )
= 33. 0 - 5.16
= 0 - 80
= -80
\(\frac{8}{15}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{4242}+\frac{3333}{5656}\right)\)
= \(\frac{8}{15}.\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{14}+\frac{33}{56}\right)\)
= \(\frac{8}{15}.\frac{231}{40}\)
= \(\frac{77}{25}\)
a) \(\frac{-8}{18}-\frac{15}{27}=\frac{-4}{9}-\frac{5}{9}=\frac{-4-5}{9}=\frac{-9}{9}=-1\)
b) \(\frac{19}{24}-\left(-\frac{1}{2}+\frac{7}{24}\right)\)
\(=\frac{19}{24}+\frac{1}{2}-\frac{7}{24}=\left(\frac{19}{24}-\frac{7}{24}\right)+\frac{1}{2}=\frac{1}{2}+\frac{1}{2}=1\)
c) \(\frac{3^{11}\cdot11+3^{11}\cdot21}{3^9\cdot2^5}=\frac{3^{11}\left(11+21\right)}{3^9\cdot2^5}\)
\(=\frac{3^{11}\cdot32}{3^9\cdot32}=3^2=9\)
a) \(-\frac{8}{18}-\frac{15}{27}=-\frac{4}{9}-\frac{5}{9}=\frac{-9}{9}=-1\)
b) \(\frac{19}{24}-\left(-\frac{1}{2}+\frac{7}{24}\right)\)
\(=\frac{19}{24}+\frac{12}{24}-\frac{7}{24}=\frac{24}{24}=1\)
c) \(P=\frac{3^{11}.11+3^{11}.21}{3^9.2^5}\)
\(P=\frac{3^{11}.\left(11+21\right)}{2^9.2^5}=\frac{3^{11}.32}{2^9.32}=3^2=9\)
d) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}=\frac{99}{50}\)
\(\frac{3}{5}-\frac{7}{45}+\frac{11}{9.13}-\frac{15}{13.17}+\frac{19}{17.21}-\frac{23}{21.25}\)
\(=\frac{3}{5}-\frac{7}{45}+\frac{11}{117}-\frac{15}{221}+\frac{19}{357}-\frac{23}{42}\)
\(=\left(\frac{-7}{45}-\frac{15}{221}-\frac{23}{42}\right)+\left(\frac{3}{5}+\frac{11}{117}+\frac{19}{357}\right)\)
đến đây bạn tự tính nhé.
sua lai
A) \(\left(44.52.60\right):\left(11.13.15\right)\)
\(=\left(4.11.4.13.4.15\right):\left(11.13.15\right)\)
\(=4^3\left(11.13.15\right):\left(11.13.15\right)\)
\(=64\)
B) \(123.456456-456.123123\)
\(=123.456.1001-456.123.1001\)
\(=56088.1001-56088.1001\)
\(=1001.\left(56088-56088\right)\)
\(=1001.0\)
\(=0\)
C) \(\left(98.7676-9898.76\right):\left(2001.2002....2020\right)\)
\(=\left(98.76.101-98.101.76\right):\left(2001.2002....2020\right)\)
\(=0:\left(2001.2002....2020\right)\)
\(=0\)
\(a.\) \(50\cdot374\cdot2=\left(50\cdot2\right)\cdot347=100\cdot347=34700\)
\(b.\)\(36\cdot97+97\cdot64=97\cdot\left(36+64\right)=97\cdot100=9700\)
\(c.157\cdot289-289\cdot57=289\cdot\left(157-57\right)=289\cdot100=28900\)