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B+4=(191/210+1)+(161/240+1)+(129/272+1)+(95/306+1)
B+4=401/210+401/240+401/272+401/306
B+4=401x(1/210+1/240+1/272+1/306)
B+4=401X 1/63
B=401/63-4
B=149/63
Hoc tot!
1.
\(\dfrac{191}{210}+\dfrac{161}{240}+\dfrac{129}{272}+\dfrac{95}{306}=\dfrac{191}{14\cdot15}+\dfrac{161}{15\cdot16}+\dfrac{129}{16\cdot17}+\dfrac{95}{17\cdot18}=\dfrac{191}{14}-\dfrac{191}{15}+\dfrac{161}{15}-\dfrac{161}{16}+\dfrac{129}{16}-\dfrac{129}{17}+\dfrac{95}{17}-\dfrac{95}{18}=\dfrac{191}{14}-\left(\dfrac{191}{15}+\dfrac{161}{15}\right)-\left(\dfrac{161}{16}+\dfrac{129}{16}\right)-\left(\dfrac{129}{17}+\dfrac{95}{17}\right)-\dfrac{95}{18}=\dfrac{191}{14}-\dfrac{352}{15}-\dfrac{145}{8}-\dfrac{224}{17}-\dfrac{95}{18}=\dfrac{584460}{42840}-\dfrac{1005312}{42840}-\dfrac{776475}{42840}-\dfrac{564480}{42840}-\dfrac{226100}{42840}=\dfrac{-1987907}{42840}\)
2.
a) \(\left(x-5\right)\cdot y=3\)
Ta lập bảng sau:
| x-5 | 1 | 3 |
| x | 6 | 8 |
| y | 3 | 1 |
Vậy các cặp (x;y) là: (6;3) ; (8;1)
b) x+1=2
x=2-1
x=1
Vậy x=1.
mình xin lỗi nha, câu b bài 2 đề là (x + 1)y + y mới đúng
a)Ta có:
\(\frac{419}{-723}< 0< \frac{-697}{-313}\)
\(\Rightarrow\frac{419}{-723}< \frac{-697}{-313}\)
c)\(\frac{17}{215}>\frac{17}{314}\)
d)Ta có:
\(\frac{11}{54}< \frac{22}{54}< \frac{22}{37}\)
\(\Rightarrow\frac{11}{54}< \frac{22}{37}\)
e)Ta có:
\(\frac{-385}{-126}>0>\frac{-57}{3461}\)
\(\Rightarrow\frac{-385}{-126}>\frac{-57}{3461}\)
f)Ta có:
\(\frac{123}{109}>1>\frac{556}{789}\)
\(\Rightarrow\frac{123}{109}>\frac{556}{789}\)
g)Ta có:
\(\frac{-56}{57}>-1>\frac{-49}{47}\)
\(\Rightarrow\frac{-56}{57}>\frac{-49}{47}\)
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:
\(P=78+\left|78-129\right|+\left(-29\right)\)
\(\Rightarrow P=78+\left[-\left(78-129\right)\right]+\left(-29\right)\)
\(\Rightarrow P=78+\left(-78\right)+129+\left(-29\right)\)
\(\Rightarrow P=100\)
129-5[29-(6-1)2]
=129-5(29-52)
=129-5(29-25)
=129-5.4
=129-20
=109
129-5[29-(6-1)2] =129-5[29-42]
= 129-5[29-16]
= 129-5.13
= 129-65
= 64
nha