Ta có:

p + n + e = 40

2p + n = 40 (nguyên tử trung hòa về điện)

n = 40 - 2p

Ta có:

p ≤ n ≤ 1,5p

p ≤ 40 - 2p ≤ 1,5p

3p ≤ 40 ≤ 3,5p

\(\rArr\left[\begin{array}{l}3p\le40\\ 3,5p\ge40\end{array}\rArr\left[\begin{array}{l}p\le13,\left(3\right)\\ p\ge11,4286\end{array}\right.\right.\)

Mà p là số tự nhiên

\(\rArr p\in\left\lbrace12;13\right\rbrace\)

- Trường hợp 1: Nếu p = 12 ⇒ Nguyên tử nguyên tố Mg

⇒ n = 40 - 2p = 40 - 2 x 12 = 40 - 24 = 16

Mà trong nguyên tử Mg có 12n

⇒ Loại trường hợp này

- Trường hợp 2: Nếu p = 13 ⇒ Nguyên tử nguyên tố Al

⇒ n = 40 - 13 x 2 = 40 - 26 = 14

Trong nguyên tử Al có 14n

⇒ Chọn trường hợp này

Vậy trong nguyên tử A có: p = e = 13; n = 14

*Đây là bài làm không xét đến trường hợp của đồng vị. Theo bảng tuần hoàn thì Al sẽ là đáp án phù hợp nhất nhé. Chúc bạn học tốt!*

a: \(A=-\frac38\cdot x^2y\cdot\frac23xy^2z^2\cdot\frac45x^3y\)

\(=\left(-\frac38\cdot\frac23\cdot\frac45\right)\cdot x^2\cdot x\cdot x^3\cdot y\cdot y^2\cdot y\cdot z^2=-\frac15x^6y^4z^2\)

b: Khi x=-1;y=-2;z=3 thì \(A=-\frac15\cdot\left(-1\right)^6\cdot\left(-2\right)^4\cdot3^2=-\frac15\cdot1\cdot16\cdot9=-\frac{144}{5}\)

c: \(x^6\ge0\forall x\)
\(y^4\ge0\forall y\)

\(z^2\ge0\forall z\)

Do đó: \(x^6y^4z^2\ge0\forall x,y,z\)

=>\(A=-\frac15x^6y^4z^2\le0\forall x,y,z\)

=>A không thể nhận giá trị dương

a: \(x^3\left(x+1\right)+\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3+1\right)\)

\(=\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)

b: \(x^2+7x-8\)

\(=x^2+8x-x-8\)

=x(x+8)-(x+8)

=(x+8)(x-1)

c: \(3x^2-5x+2\)

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

=3x(x-1)-2(x-1)

=(x-1)(3x-2)

d: \(x^3-8x^2+9x\)

\(=x\cdot x^2-x\cdot8x+x\cdot9\)

\(=x\left(x^2-8x+9\right)\)

e: \(2x^2+5x-3\)

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

=2x(x+3)-(x+3)

=(x+3)(2x-1)

Giải:

Giá tiền của \(x+9\) cuốn truyện tranh là:

\(\left(x+9\right)\times15=15x+135\) (nghìn đồng)

Giá tiền của \(x+6\) cuốn sách tham khảo là:

(\(x+6\)) x 12 = 12\(x+72\) (nghìn đồng)

Giá tiền của \(x\) cuốn sách khoa học là:

\(x\) x 21 = 21\(x\) (nghìn đồng)

Tổng số tiền cần trả là:

15\(x\)+135+12\(x\)+72+21\(x\) = 48\(x\)+207(nghìn đồng)

Chọn C.48\(x\) +207

Số tiền phải trả cho x cuốn sách khoa học là 21x(nghìn đồng)

Số tiền phải trả cho x+6 cuốn sách tham khảo là:

12(x+6)(nghìn đồng)

Số tiền phải trả cho x+9 cuốn truyện tranh là:

15(x+9)(nghìn đồng)

Tổng số tiền phải trả là:

21x+12(x+6)+15(x+9)

=21x+12x+72+15x+135

=48x+207(nghìn đồng)

=>Chọn C

Bài 1:

a: \(\left(\frac{9}{25}-2^2\right):\left(-0,2\right)\)

\(=\left(\frac{9}{25}-4\right):\left(\frac{-1}{5}\right)=\frac{-91}{25}\cdot\frac{-5}{1}=\frac{91}{5}\)

b: \(\left(-\frac15\right)^2+\frac15-2\cdot\left(-\frac12\right)^3-\frac12\)

\(=\frac{1}{25}+\frac15-2\cdot\frac{-1}{8}-\frac12\)

\(=\frac{1}{25}+\frac{5}{25}+\frac14-\frac12=\frac{6}{25}-\frac14=\frac{24}{100}-\frac{25}{100}=-\frac{1}{100}\)

c: \(\left(3-\frac14+\frac23\right)^2:2022^0\)

\(=\left(\frac{36}{12}-\frac{3}{12}+\frac{8}{12}\right)^2=\left(\frac{41}{12}\right)^2=\frac{1681}{144}\)

d: \(2^2\cdot9:\left(3\frac45+0,2\right)\)

\(=4\cdot9:\left(3,8+0,2\right)\)

\(=\frac{36}{4}=9\)

e: \(\left(\frac14+\frac23\right)^2-1\frac13=\left(\frac{3}{12}+\frac{8}{12}\right)^2-\frac43\)

\(=\left(\frac{11}{12}\right)^2-\frac43=\frac{121}{144}-\frac{192}{144}=-\frac{71}{144}\)

f: \(1:\left(-1\frac52+0,5\right)^2\)

\(=1:\left(-\frac72+\frac12\right)^2\)

\(=1:\left(-3\right)^2=\frac19\)

Bài 2:

a: \(-\frac{5}{14}+\frac38-\frac{2}{14}-\frac38+\frac12\)

\(=\left(-\frac{5}{14}-\frac{2}{14}+\frac12\right)+\left(\frac38-\frac38\right)\)

\(=\left(-\frac{7}{14}+\frac{7}{14}\right)+0=0+0=0\)

b: \(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)-\frac35+\left(\frac57-\frac57\right)\)

\(=\frac{30}{15}-\frac35=2-\frac35=\frac75\)

c: \(-\frac25\cdot\frac57+\frac{-2}{5}\cdot\frac97\)

\(=-\frac25\left(\frac57+\frac97\right)=-\frac25\cdot2=-\frac45\)

d: \(\frac{55}{27}+\frac{-21}{5}+\frac{-55}{27}-\frac{-21}{5}\)

\(=\left(\frac{55}{27}-\frac{55}{27}\right)+\left(-\frac{21}{5}+\frac{21}{5}\right)\)

=0+0=0

e: \(\frac57:\left(\frac{15}{8}-\frac14\right)-\frac57:\left(\frac14+\frac12\right)\)

\(=\frac57:\left(\frac{15}{8}-\frac28\right)-\frac57:\left(\frac14+\frac24\right)\)

\(=\frac57:\frac{13}{8}-\frac57:\frac34\)

\(=\frac57\cdot\frac{8}{13}-\frac57\cdot\frac43=\frac57\left(\frac{8}{13}-\frac43\right)=\frac57\cdot\left(\frac{24}{39}-\frac{52}{39}\right)\)

\(=\frac57\cdot\frac{-28}{39}=\frac{5\cdot\left(-4\right)}{39}=-\frac{20}{39}\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27\right)\cdot\frac{-5}{3}-\left(28+\frac27\right)\cdot\frac{-5}{3}\)

\(=-\frac53\left(16+\frac27-28-\frac27\right)=-\frac53\cdot\left(-12\right)=20\)

Bài 1:

a: \(\left(\frac{9}{25}-2^2\right):\left(-0,2\right)\)

\(=\left(\frac{9}{25}-4\right):\left(\frac{-1}{5}\right)=\frac{-91}{25}\cdot\frac{-5}{1}=\frac{91}{5}\)

b: \(\left(-\frac15\right)^2+\frac15-2\cdot\left(-\frac12\right)^3-\frac12\)

\(=\frac{1}{25}+\frac15-2\cdot\frac{-1}{8}-\frac12\)

\(=\frac{1}{25}+\frac{5}{25}+\frac14-\frac12=\frac{6}{25}-\frac14=\frac{24}{100}-\frac{25}{100}=-\frac{1}{100}\)

c: \(\left(3-\frac14+\frac23\right)^2:2022^0\)

\(=\left(\frac{36}{12}-\frac{3}{12}+\frac{8}{12}\right)^2=\left(\frac{41}{12}\right)^2=\frac{1681}{144}\)

d: \(2^2\cdot9:\left(3\frac45+0,2\right)\)

\(=4\cdot9:\left(3,8+0,2\right)\)

\(=\frac{36}{4}=9\)

e: \(\left(\frac14+\frac23\right)^2-1\frac13=\left(\frac{3}{12}+\frac{8}{12}\right)^2-\frac43\)

\(=\left(\frac{11}{12}\right)^2-\frac43=\frac{121}{144}-\frac{192}{144}=-\frac{71}{144}\)

f: \(1:\left(-1\frac52+0,5\right)^2\)

\(=1:\left(-\frac72+\frac12\right)^2\)

\(=1:\left(-3\right)^2=\frac19\)

Bài 2:

a: \(-\frac{5}{14}+\frac38-\frac{2}{14}-\frac38+\frac12\)

\(=\left(-\frac{5}{14}-\frac{2}{14}+\frac12\right)+\left(\frac38-\frac38\right)\)

\(=\left(-\frac{7}{14}+\frac{7}{14}\right)+0=0+0=0\)

b: \(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)-\frac35+\left(\frac57-\frac57\right)\)

\(=\frac{30}{15}-\frac35=2-\frac35=\frac75\)

c: \(-\frac25\cdot\frac57+\frac{-2}{5}\cdot\frac97\)

\(=-\frac25\left(\frac57+\frac97\right)=-\frac25\cdot2=-\frac45\)

d: \(\frac{55}{27}+\frac{-21}{5}+\frac{-55}{27}-\frac{-21}{5}\)

\(=\left(\frac{55}{27}-\frac{55}{27}\right)+\left(-\frac{21}{5}+\frac{21}{5}\right)\)

=0+0=0

e: \(\frac57:\left(\frac{15}{8}-\frac14\right)-\frac57:\left(\frac14+\frac12\right)\)

\(=\frac57:\left(\frac{15}{8}-\frac28\right)-\frac57:\left(\frac14+\frac24\right)\)

\(=\frac57:\frac{13}{8}-\frac57:\frac34\)

\(=\frac57\cdot\frac{8}{13}-\frac57\cdot\frac43=\frac57\left(\frac{8}{13}-\frac43\right)=\frac57\cdot\left(\frac{24}{39}-\frac{52}{39}\right)\)

\(=\frac57\cdot\frac{-28}{39}=\frac{5\cdot\left(-4\right)}{39}=-\frac{20}{39}\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27\right)\cdot\frac{-5}{3}-\left(28+\frac27\right)\cdot\frac{-5}{3}\)

\(=-\frac53\left(16+\frac27-28-\frac27\right)=-\frac53\cdot\left(-12\right)=20\)

Bài 1:

a: \(\left(\frac{9}{25}-2^2\right):\left(-0,2\right)\)

\(=\left(\frac{9}{25}-4\right):\left(\frac{-1}{5}\right)=\frac{-91}{25}\cdot\frac{-5}{1}=\frac{91}{5}\)

b: \(\left(-\frac15\right)^2+\frac15-2\cdot\left(-\frac12\right)^3-\frac12\)

\(=\frac{1}{25}+\frac15-2\cdot\frac{-1}{8}-\frac12\)

\(=\frac{1}{25}+\frac{5}{25}+\frac14-\frac12=\frac{6}{25}-\frac14=\frac{24}{100}-\frac{25}{100}=-\frac{1}{100}\)

c: \(\left(3-\frac14+\frac23\right)^2:2022^0\)

\(=\left(\frac{36}{12}-\frac{3}{12}+\frac{8}{12}\right)^2=\left(\frac{41}{12}\right)^2=\frac{1681}{144}\)

d: \(2^2\cdot9:\left(3\frac45+0,2\right)\)

\(=4\cdot9:\left(3,8+0,2\right)\)

\(=\frac{36}{4}=9\)

e: \(\left(\frac14+\frac23\right)^2-1\frac13=\left(\frac{3}{12}+\frac{8}{12}\right)^2-\frac43\)

\(=\left(\frac{11}{12}\right)^2-\frac43=\frac{121}{144}-\frac{192}{144}=-\frac{71}{144}\)

f: \(1:\left(-1\frac52+0,5\right)^2\)

\(=1:\left(-\frac72+\frac12\right)^2\)

\(=1:\left(-3\right)^2=\frac19\)

Bài 2:

a: \(-\frac{5}{14}+\frac38-\frac{2}{14}-\frac38+\frac12\)

\(=\left(-\frac{5}{14}-\frac{2}{14}+\frac12\right)+\left(\frac38-\frac38\right)\)

\(=\left(-\frac{7}{14}+\frac{7}{14}\right)+0=0+0=0\)

b: \(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)-\frac35+\left(\frac57-\frac57\right)\)

\(=\frac{30}{15}-\frac35=2-\frac35=\frac75\)

c: \(-\frac25\cdot\frac57+\frac{-2}{5}\cdot\frac97\)

\(=-\frac25\left(\frac57+\frac97\right)=-\frac25\cdot2=-\frac45\)

d: \(\frac{55}{27}+\frac{-21}{5}+\frac{-55}{27}-\frac{-21}{5}\)

\(=\left(\frac{55}{27}-\frac{55}{27}\right)+\left(-\frac{21}{5}+\frac{21}{5}\right)\)

=0+0=0

e: \(\frac57:\left(\frac{15}{8}-\frac14\right)-\frac57:\left(\frac14+\frac12\right)\)

\(=\frac57:\left(\frac{15}{8}-\frac28\right)-\frac57:\left(\frac14+\frac24\right)\)

\(=\frac57:\frac{13}{8}-\frac57:\frac34\)

\(=\frac57\cdot\frac{8}{13}-\frac57\cdot\frac43=\frac57\left(\frac{8}{13}-\frac43\right)=\frac57\cdot\left(\frac{24}{39}-\frac{52}{39}\right)\)

\(=\frac57\cdot\frac{-28}{39}=\frac{5\cdot\left(-4\right)}{39}=-\frac{20}{39}\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27\right)\cdot\frac{-5}{3}-\left(28+\frac27\right)\cdot\frac{-5}{3}\)

\(=-\frac53\left(16+\frac27-28-\frac27\right)=-\frac53\cdot\left(-12\right)=20\)

Bài 1:

a: \(\left(\frac{9}{25}-2^2\right):\left(-0,2\right)\)

\(=\left(\frac{9}{25}-4\right):\left(\frac{-1}{5}\right)=\frac{-91}{25}\cdot\frac{-5}{1}=\frac{91}{5}\)

b: \(\left(-\frac15\right)^2+\frac15-2\cdot\left(-\frac12\right)^3-\frac12\)

\(=\frac{1}{25}+\frac15-2\cdot\frac{-1}{8}-\frac12\)

\(=\frac{1}{25}+\frac{5}{25}+\frac14-\frac12=\frac{6}{25}-\frac14=\frac{24}{100}-\frac{25}{100}=-\frac{1}{100}\)

c: \(\left(3-\frac14+\frac23\right)^2:2022^0\)

\(=\left(\frac{36}{12}-\frac{3}{12}+\frac{8}{12}\right)^2=\left(\frac{41}{12}\right)^2=\frac{1681}{144}\)

d: \(2^2\cdot9:\left(3\frac45+0,2\right)\)

\(=4\cdot9:\left(3,8+0,2\right)\)

\(=\frac{36}{4}=9\)

e: \(\left(\frac14+\frac23\right)^2-1\frac13=\left(\frac{3}{12}+\frac{8}{12}\right)^2-\frac43\)

\(=\left(\frac{11}{12}\right)^2-\frac43=\frac{121}{144}-\frac{192}{144}=-\frac{71}{144}\)

f: \(1:\left(-1\frac52+0,5\right)^2\)

\(=1:\left(-\frac72+\frac12\right)^2\)

\(=1:\left(-3\right)^2=\frac19\)

Bài 2:

a: \(-\frac{5}{14}+\frac38-\frac{2}{14}-\frac38+\frac12\)

\(=\left(-\frac{5}{14}-\frac{2}{14}+\frac12\right)+\left(\frac38-\frac38\right)\)

\(=\left(-\frac{7}{14}+\frac{7}{14}\right)+0=0+0=0\)

b: \(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)-\frac35+\left(\frac57-\frac57\right)\)

\(=\frac{30}{15}-\frac35=2-\frac35=\frac75\)

c: \(-\frac25\cdot\frac57+\frac{-2}{5}\cdot\frac97\)

\(=-\frac25\left(\frac57+\frac97\right)=-\frac25\cdot2=-\frac45\)

d: \(\frac{55}{27}+\frac{-21}{5}+\frac{-55}{27}-\frac{-21}{5}\)

\(=\left(\frac{55}{27}-\frac{55}{27}\right)+\left(-\frac{21}{5}+\frac{21}{5}\right)\)

=0+0=0

e: \(\frac57:\left(\frac{15}{8}-\frac14\right)-\frac57:\left(\frac14+\frac12\right)\)

\(=\frac57:\left(\frac{15}{8}-\frac28\right)-\frac57:\left(\frac14+\frac24\right)\)

\(=\frac57:\frac{13}{8}-\frac57:\frac34\)

\(=\frac57\cdot\frac{8}{13}-\frac57\cdot\frac43=\frac57\left(\frac{8}{13}-\frac43\right)=\frac57\cdot\left(\frac{24}{39}-\frac{52}{39}\right)\)

\(=\frac57\cdot\frac{-28}{39}=\frac{5\cdot\left(-4\right)}{39}=-\frac{20}{39}\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27\right)\cdot\frac{-5}{3}-\left(28+\frac27\right)\cdot\frac{-5}{3}\)

\(=-\frac53\left(16+\frac27-28-\frac27\right)=-\frac53\cdot\left(-12\right)=20\)

1-x = (1/3 ) mũ 2
1-x = +-1/3
1-x =1/3
x=0,6
1-x=-1/3
x=1,3

Ta có: \(\left(1-x\right)^2=\frac19\)

=>\(\left(x-1\right)^2=\frac19\)

=>\(\left[\begin{array}{l}x-1=\frac13\\ x-1=-\frac13\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13+1=\frac43\\ x=-\frac13+1=\frac23\end{array}\right.\)

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