a) Áp suất do nước tác dụng lên đáy cốc:

    \(p=d\cdot h=10^4\cdot20\cdot10^{-2}=2000\)N/m³

b) Áp lực lên đáy cốc:

   \(F=p\cdot S=2000\cdot\left(\pi\cdot0,02^2\right)=2,51N\)

em cám ơn chị nhiều ạ^^

Câu 10 : 

Ta có : AD là phan giác \(\widehat{ABC}\)

\(\Rightarrow\dfrac{AB}{AC}=\dfrac{BD}{DC}\)

=> Chọn C

Câu 11 :

Ta có : \(MN//BC\)

Theo hệ quả định lý Ta-lét ta có :
\(\dfrac{AM}{AB}=\dfrac{MN}{BC}\)

hay \(\dfrac{2}{5}=\dfrac{MN}{6,5}\)

\(\Rightarrow MN=\dfrac{2.6,5}{5}=2,6\left(cm\right)\)

=> CHọn D

Câu 12 :

Ta có : \(5m=50dm\)

\(\Rightarrow\dfrac{AB}{CD}=\dfrac{15}{50}=\dfrac{3}{10}\)

=> Chọn A

Bài 1:

a: \(\left(5-\frac23+\frac37\right):\left(24\frac{4}{21}-25\frac{8}{21}\right)\)

\(=\left(\frac{105}{21}-\frac{14}{21}+\frac{9}{21}\right):\left(24+\frac{4}{21}-25-\frac{8}{21}\right)\)

\(=\frac{100}{21}:\left(-1-\frac{4}{21}\right)=\frac{100}{21}:\frac{-25}{21}=\frac{100}{-25}=-4\)

b: \(\left(2\frac46-\frac{7}{15}\right):\left(\frac23-\frac25\right)^2\)

\(=\left(2+\frac23-\frac{7}{15}\right):\left(\frac{10}{15}-\frac{6}{15}\right)^2\)

\(=\left(2+\frac{10}{15}-\frac{7}{15}\right):\left(\frac{4}{15}\right)^2=\left(2+\frac{3}{15}\right):\frac{16}{225}=\frac{11}{5}\cdot\frac{225}{16}\)

\(=\frac{11}{16}\cdot45=\frac{495}{16}\)

c: \(\left(\frac{13}{18}-\frac{1}{72}\right):\frac{1}{18}-\left(\frac49-\frac{25}{16}\right)\cdot\frac92\)

\(=\left(\frac{52}{72}-\frac{1}{72}\right)\cdot18-\left(\frac{64}{144}-\frac{100}{144}\right)\cdot\frac92\)

\(=\frac{51}{72}\cdot18+\frac{36}{144}\cdot\frac92=\frac{51}{4}+\frac14\cdot\frac92=\frac{111}{8}\)

d: \(\frac35:\left(-\frac{1}{15}-\frac16\right)+\frac35:\left(-\frac13-1\frac{1}{15}\right)\)

\(=\frac35:\left(-\frac{2}{30}-\frac{5}{30}\right)+\frac35:\left(-\frac{5}{15}-\frac{16}{15}\right)\)

\(=\frac35:\frac{-7}{30}+\frac35:\frac{-21}{15}=\frac35\cdot\frac{-30}{7}+\frac35\cdot\frac{-5}{7}=\frac35\left(-\frac{30}{7}-\frac57\right)\)

\(=\frac35\cdot\left(-5\right)=-3\)

e: \(4\left(-\frac12\right)^3-2\left(-\frac12\right)^2+3\cdot\left(-\frac12\right)+\left(-1\right)^{2002}\)

\(=4\cdot\frac{-1}{8}-2\cdot\frac14-\frac32+1\)

\(=-\frac12-\frac12-\frac32+1=-\frac32\)

f: \(\frac{2^4\cdot2^6}{\left(2^5\right)^2}-\frac{2^5\cdot15^3}{6^3\cdot10^2}\)

\(=\frac{2^{10}}{2^{10}}-\frac{2^5\cdot5^3\cdot3^3}{2^3\cdot5^3\cdot2^2\cdot5^2}=1-\frac{3^3}{5^2}=1-\frac{27}{25}=-\frac{2}{25}\)

Bài 3:

a: |x-5|=8

=>\(\left[\begin{array}{l}x-5=8\\ x-5=-8\end{array}\right.\Rightarrow\left[\begin{array}{l}x=13\\ x=-5\end{array}\right.\)

b: \(\left|1-\frac23x\right|+\frac34-5\frac12=0\)

=>\(\left|\frac23x-1\right|+\frac34-\frac{11}{2}=0\)

=>\(\left|\frac23x-1\right|+\frac34-\frac{22}{4}=0\)

=>\(\left|\frac23x-1\right|=\frac{19}{4}\)

=>\(\left[\begin{array}{l}\frac23x-1=\frac{19}{4}\\ \frac23x-1=-\frac{19}{4}\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac23x=\frac{19}{4}+1=\frac{23}{4}\\ \frac23x=-\frac{19}{4}+1=-\frac{15}{4}\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{23}{4}:\frac23=\frac{23}{4}\cdot\frac32=\frac{69}{8}\\ x=-\frac{15}{4}:\frac23=-\frac{15}{4}\cdot\frac32=\frac{-45}{8}\end{array}\right.\)

c: |9-7x|+7=26

=>|7x-9|=26-7=19

=>\(\left[\begin{array}{l}7x-9=19\\ 7x-9=-19\end{array}\right.\Rightarrow\left[\begin{array}{l}7x=19+9=28\\ 7x=-19+9=-10\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-\frac{10}{7}\end{array}\right.\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)

\(\Leftrightarrow x=100\) vì \(\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)\ne0\)

\(\Leftrightarrow S=\left\{100\right\}\)

`7,`

`a, B+A=4x-2x^2+3`

`-> B=(4x-2x^2+3)-A`

`-> B=(4x-2x^2+3)-(x^2-2x+1)`

`B=4x-2x^2+3-x^2+2x-1`

`B=(-2x^2-x^2)+(4x+2x)+(3-1)`

`B=-3x^2+6x+2`

`b, C-A=-x+7`

`-> C=(-x+7)+A`

`-> C=(-x+7)+(x^2-2x+1)`

`-> C=-x+7+x^2-2x+1`

`C=x^2+(-x-2x)+(7+1)`

`C=x^2-3x+8`

`c,`

`A-D=x^2-2`

`-> D= A- (x^2-2)`

`-> D=(x^2-2x+1)-(x^2-2)`

`D=x^2-2x+1-x^2+2`

`D=(x^2-x^2)-2x+(1+2)`

`D=-2x+3`

`6,`

`a,`

`P+Q=4x-2x^2+3`

`-> Q=(4x-2x^2+3)-P`

`-> Q=(4x-2x^2+3)-(3x^2+x-1)`

`Q=4x-2x^2+3-3x^2-x+1`

`Q=(-2x^2-3x^2)+(4x-x)+(3+1)`

`Q=x^2+3x+4`

`b,`

`x^2-5x+2-P=H`

`-> H= (x^2-5x+2)-(3x^2+x-1)`

`H=x^2-5x+2-3x^2-x+1`

`H=(x^2-3x^2)+(-5x-x)+(2+1)`

`H=-4x^2-6x+3`

`c,`

`P-R=5x^2-3x-4`

`-> R= P- (5x^2-3x-4)`

`-> R=(3x^2+x-1)-(5x^2-3x-4)`

`R=3x^2+x-1-5x^2+3x+4`

`R=(3x^2-5x^2)+(x+3x)+(-1+4)`

`R=-2x^2+4x+3`

= 2263 + 1824 + 1326 + 1584

= 4087 + 2910

= 6997

TK MK NHA

bn à, mk ko nghĩ đây là bài tính nhanh đâu

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2R + 2nHCl → 2RCl_n + nH₂

Mol:     \(\dfrac{0,3}{n}\)                                      0,15

\(M_R=\dfrac{3,6}{\dfrac{0,3}{n}}=12n\left(g/mol\right)\)

Vì R là kim loại nên có hóa trị l,ll,lll

  ⇒ R là magie (Mg)

Bài 1

1.\(x\left(x+3\right)\)

\(=x^2+3x\)

2.\(3x\left(x+2\right)\)

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

3,\(x^2\left(3x-1\right)\)

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

4.\(-5x^3\left(3x^2-7\right)\)

\(=-15x^5+35x^3\)

5.\(3x\left(5x^2-2x-1\right)\)

\(=15x^3-6x^2-3x\)

6.\(-x^2\left(5x^3-x-\dfrac{1}{2}\right)\)

\(=-5x^5+x^3+\dfrac{x^2}{2}\)

7.\(\left(x^2+2x-3\right).\left(-x\right)\)

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

8.\(4x^3\left(-2x^2+4x^4-3\right)\)

\(=-8x^5+16x^7-12x^3\)

9.\(-5x^2\left(3x^2-2x+1\right)\)

\(=-15x^4+10x^3-5x^2\)

10.\(-4x^5\left(x^3-4x^2+7x-3\right)\)

\(=-4x^8+16x^7-28x^6+12x^5\)

11.\(\left(x+2\right)\left(x+3\right)\)

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

12.\(\left(x-7\right)\left(x-5\right)\)

\(=x^2-5x-7x+35\)

13.\(\left(3x+5\right)\left(2x-7\right)\)

\(=6x^2-21x+10x-35\)

14.\(\left(x-3\right)\left(x^2-2x-1\right)\)

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

15.\(\left(2x-1\right)\left(x^2-5x+3\right)\)

\(=2x^3-10x^2+6x-x^2+5x-3\)

16.\(\left(x-5\right)\left(-x^2+x-1\right)\)

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

17,\(\left(\dfrac{1}{2}x+3\right)\left(2x^2-4x-6\right)\)

\(=x^3-2x^2-3x+6x^2-12x-18\)

P/s:mình làm hơi tắt tại bài dài quá:))

anh chia ra 2 bài cho đỡ nhầm á em, giờ anh đang làm bài 2