Tham khảo

KHO THE

\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)

\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)

\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)

Câu 1:

a: A={x∈R|x>=1}

=>A=[1;+∞)

B={x∈R|x<=3}

=>B=(-∞;3]

A\(\cap\) B=[1;+∞)\(\cap\) (-∞;3]

=[1;3]

A\(\cup\) B=[1;+∞)\(\cup\) (-∞;3]

=(-∞;+∞)

b: A={x∈R|x<=1}

=>A=(-∞;1]

B={x∈R|x>=3}

=>B=[3;+∞)

A\(\cap\) B=(-∞;1]\(\cap\) [3;+∞)

=∅

A\(\cup\) B=(-∞;1]\(\cup\) [3;+∞)

Câu 2:

A=(-3;5]\(\cup\) [8;10]\(\cup\) (2;8)

=(-3;10]

B=[0;2]\(\cup\) (-∞;5]\(\cup\) (1;+∞)

=(-∞;+∞)

C=[-4;7]\(\cup\) (0;10)

=[-4;10)

D=(-∞;3]\(\cup\) (-5;+∞)

=(-∞;+∞)

E=(3;+∞)\(-∞;1]

=(3;+∞)

F=(1;3]\[0;4)

=∅

Câu 5:

Điện trở tương đương:

R₂₃ = R₂ + R₃ = 6 + 4 = 10\(\Omega\)

R₂₃₄ = (R₂₃.R₄) : (R₂₃ + R₄) = (10.10) : (10 + 10) = 5Ω

R_tđ = R₁ + R₂₃₄ = 2 + 5 = 7Ω

Tham khảo:

Câu 6:

5/6 : x= 1/6 - 4/5 

5/6 : x= -19/30

x = 5/6 : (-19)/30

x = -25/19

\(x=\dfrac{5}{6}:\left(\dfrac{1}{6}-\dfrac{4}{5}\right)=\dfrac{5}{6}:\left(-\dfrac{19}{30}\right)=-\dfrac{25}{19}\)

Bài 5: 

a: Xét ΔBEC và ΔADC có 

\(\widehat{C}\) chung

\(\widehat{EBC}=\widehat{DAC}\)

Do đó: ΔBEC\(\sim\)ΔADC

a) \(\dfrac{1}{4}-3\left(\dfrac{1}{12}+\dfrac{3}{8}\right)=\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{9}{8}=-\dfrac{9}{8}\)

b) \(\left(-\dfrac{2}{3}+\dfrac{3}{5}\right):\dfrac{1}{50}-30=\left(-\dfrac{2}{3}+\dfrac{3}{5}\right).50-30=-\dfrac{100}{3}+30-30=-\dfrac{100}{3}\)

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