Lời giải:
Gọi số thứ nhất là $a$ và số thứ hai là $b$.

Theo bài ra ta có:
\(\left\{\begin{matrix} 4b+5a=18040\\ 3a-2b=2002\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 5a+4b=18040\\ 6a-4b=4004\end{matrix}\right.\)

\(\Rightarrow 5a+6a=18040+4004\)

\(\Leftrightarrow 11a=22044\Leftrightarrow a=2004\)

\(b=\frac{3a-2002}{2}=2005\)

3.1:

Ô tô đi hết 1h50'+40'=2h30'

3.2:

Ô tô đi hết:

17h50'-15h15'

=2h35'

a) 4Na + O₂ --t^o--> 2Na₂O

b) \(n_{Na_2O}=\dfrac{24,8}{62}=0,4\left(mol\right)\)

4Na + O₂ --t^o--> 2Na₂O

0,8<--0,2<-------0,4

=> m_Na = 0,8.23 = 18,4 (g)

c) m_O2 = 0,2.32 = 6,4 (g)

V_O2 = 0,2.24,79 = 4,958(l)

a) \(k=-5\)

b) 

Tìm x xong rồi tìm y

3 thì làm kiểu gì cũng được

Bài 3:

Hình 1:

Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(x+60^0+40^0=180^0\)

=>\(x=180^0-100^0=80^0\)

Hình 2:

Xét ΔABD có \(\widehat{ADC}\) là góc ngoài tại đỉnh D

nên \(\widehat{ADC}=\widehat{ABD}+\widehat{BAD}=110^0\)

=>\(y=110^0\)

ΔDAC cân tại D

=>\(\widehat{DAC}=\widehat{DCA}=\dfrac{180^0-\widehat{ADC}}{2}\)

=>\(x=\dfrac{180^0-110^0}{2}=35^0\)

Hình 3:

Ta có: \(\widehat{CAB}+60^0=180^0\)

=>\(\widehat{CAB}=180^0-60^0=120^0\)

Xét ΔCAB có \(\widehat{CAB}+\widehat{ABC}+\widehat{ACB}=180^0\)

=>\(x+2x+120^0=180^0\)

=>\(3x=60^0\)

=>\(x=20^0\)

Hình 4:

Ta có: \(\widehat{ADC}+\widehat{ADB}=180^0\)

=>\(\widehat{ADB}=180^0-80^0=100^0\)

Xét ΔADB có

\(\widehat{ADB}+\widehat{ABD}+\widehat{BAD}=180^0\)

=>\(x+60^0+100^0=180^0\)

=>\(x=20^0\)

ta có: \(\widehat{ACD}+135^0=180^0\)

=>\(\widehat{ACD}=180^0-135^0=45^0\)

Xét ΔACD có \(\widehat{ACD}+\widehat{ADC}+\widehat{DAC}=180^0\)

=>\(y+80^0+45^0=180^0\)

=>\(y=55^0\)

A= 3/4 +2/5-7/5+5/4

  = (3/4 + 5/4) + (2/5-7/5)

  = 2 + (-1)

  = 1

1 poverty

2 decision

3 worried

4 awake

5 singers

7 northen

8 disappeared

9 difference

10 unimportant

11 windy

12 portable

13 sensibly

14 painting

15 mainly

SOS😥👉🏳👈

Câu 1.

\(M=\left(-\dfrac{2a^3b^2}{3}xy^2z\right)^3.\left(-\dfrac{3}{4}ab^{-3}x^2yz^2\right)^2.\left(-xy^2z^2\right)^2\)

\(=\left(-\dfrac{8}{27}a^9b^6x^3y^6z^3\right).\left(\dfrac{9}{16}a^2b^{-6}x^4y^2z^4\right).\left(x^2y^4z^4\right)\)

\(=-\dfrac{8}{27}.\dfrac{9}{16}.a^{11}x^9y^{12}z^{11}\)

\(=-\dfrac{1}{6}a^{11}x^9y^{12}z^{11}\)

Hệ số: \(-\dfrac{1}{6}\)

Bậc: \(43\)

Câu 2.

a) \(A\left(x\right)=\dfrac{1}{2}x^5+\dfrac{3}{4}x-12x^4-1\dfrac{2}{3}x^3+5+x^2+\dfrac{5}{3}x^3-\dfrac{11}{4}x+1\dfrac{1}{2}x^5+4x\)

\(=\left(\dfrac{1}{2}x^5+\dfrac{3}{2}x^5\right)+\left(-12x^4\right)+\left(-\dfrac{5}{3}x^3+\dfrac{5}{3}x^3\right)+x^2+\left(\dfrac{3}{4}x-\dfrac{11}{4}x+4x\right)+5\)

\(=2x^5-12x^4+x^2+2x+5\)

\(B\left(x\right)=-2x^5+\dfrac{3}{7}x+12x^4-\dfrac{7}{3}x^3-3-6x^2+\dfrac{13}{3}x^3+3\dfrac{4}{7}x\)

\(=\left(-2x^5\right)+12x^4+\left(-\dfrac{7}{3}x^3+\dfrac{13}{3}x^3\right)-6x^2+\left(\dfrac{3}{7}x+\dfrac{25}{7}x\right)-3\)

\(=-2x^5+12x^4+2x^3-6x^2+4x-3\)

b) \(C\left(x\right)=A\left(x\right)+B\left(x\right)=\left(2x^5-12x^4+x^2+2x+5\right)+\left(-2x^5+12x^4+2x^3-6x^2+4x-3\right)\)

\(=\left(2x^5-2x^5\right)+\left(-12x^4+12x^4\right)+2x^3+\left(x^2-6x^2\right)+\left(2x+4x\right)+\left(5-3\right)\)

\(=2x^3-5x^2+6x+2\)

\(D\left(x\right)=A\left(x\right)-B\left(x\right)=\left(2x^5-12x^4+x^2+2x+5\right)-\left(-2x^5+12x^4+2x^3-6x^2+4x-3\right)\)

\(=\left(2x^5+2x^5\right)+\left(-12x^4-12x^4\right)-2x^3+\left(x^2+6x^2\right)+\left(2x-4x\right)+\left(5+3\right)\)

\(=4x^5-24x^4-2x^3+7x^2-2x+8\)

c) \(2x^3-5x^2+6x+2-2x^3+5x^2=-4\)

\(\Rightarrow\left(2x^3-2x^3\right)+\left(-5x^2+5x^2\right)+6x+2\)

\(\Rightarrow6x+2=-4\)

\(\Rightarrow6x=-6\)

\(\Rightarrow x=-1\)

Câu 3.

1) \(M-3xy^2+2xy-x^3+2x^2y=2xy-3x^3+3x^2y-xy^2\)

\(\Rightarrow M=\left(3xy^2+2xy-x^3+2x^2y\right)+\left(2xy-3x^3+3x^2y-xy^2\right)\)

\(=\left(3xy^2-xy^2\right)+\left(2xy+2xy\right)+\left(-x^3-3x^3\right)+\left(2x^2y+3x^2y\right)\)

\(=2xy^2+4xy-4x^3+5x^2y\)

2)

Để cho \(f\left(x\right)\) có nghiệm thì \(6-3x=0\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

Để cho \(g\left(x\right)\) có nghiệm thì \(x^2-1=0\)

\(\Rightarrow x^2=1\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)