Câu hỏi đâu em ơi

Bài 9:

6: \(\Leftrightarrow x+\dfrac{4}{3}-\dfrac{12}{5}=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{7}{8}=\dfrac{20-2+7}{8}=\dfrac{25}{8}\)

=>x=25/8-4/3+12/5=503/120

7: \(\Leftrightarrow\dfrac{1}{7}-\dfrac{1}{5}+x-\dfrac{3}{2}=\dfrac{16}{70}-\dfrac{9}{70}=\dfrac{1}{10}\)

=>x=1/10-1/7+1/5+3/2=58/35

8: \(\Leftrightarrow\dfrac{5}{4}-x+\dfrac{1}{5}-\dfrac{19}{20}=1-\dfrac{1}{2}+\dfrac{7}{10}=\dfrac{6}{5}\)

=>1/2-x=6/5

=>x=1/2-6/5=-7/10

9: \(\Leftrightarrow5-\dfrac{1}{3}+\dfrac{17}{4}-x=1+\dfrac{2-11}{12}=1-\dfrac{9}{12}=\dfrac{1}{4}\)

=>x=14/3+4=26/3

10: \(\Leftrightarrow\dfrac{-1}{12}+x-\dfrac{5}{12}=3-\dfrac{7}{6}+\dfrac{1}{2}\)

=>x-1/2=11/6+1/2

=>x=11/6+1=17/6

\(#nKaizz\)

`9,`

`6, 1 1/3 - 2 2/5 + x= 5/2 - (1/4 - 7/8)`

`4/3 -12/5 + x = 5/2 - (-5/8)`

`4/3 - 12/5 + x=25/8`

`-16/15 + x=25/8`

`x=25/8 - (-16/15)`

`x=25/8 + 16/15 = 503/120`

`7, 1/7 - (1/5 - x + 1 1/2)= 8/35 - 9/70`

`1/7 - (1/5 - x + 3/2)= 1/10`

`1/5 - x + 3/2 = 1/7 -1/10`

`1/5 -x + 3/2 = 3/70`

`1/5 -x=3/70 - 3/2`

`1/5-x=-51/35`

`x=1/5 - (-51/35)`

`x=1/5 + 51/35 = 58/35`

`8, 5/4 - x +(1/5 - 19/20)= 1- (1/2 - 7/10)`

`5/4 - x+ -3/4 = 1- (-1/5)`

`5/4 - x+ -3/4 = 6/5`

`5/4 - x=6/5 - (-3/4)`

`5/4 - x=39/20`

`x=5/4 - 39/20`

`x=-7/10`

`9, 5- (1/3 - 4 1/4+x) = 1+ (1/6 - 11/12)`

`5- (1/3- 4 1/4+x)=1+ -3/4 `

`5- (1/3 - 4 1/4+x) = 1/4`

`1/3- 4 1/4 +x=5-1/4`

`-47/12 +x=19/4`

`x=19/4 - (-47/12)`

`x=26/3`

`10, 1/4 - (1/3-x+5/12)=3-(1 1/6 - 1/2)`

`1/4-(1/3-x+5/12)=3-2/3`

`1/4 - (1/3-x+5/12) =7/3`

`1/3 - x+5/12 = 1/4 - 7/3`

`1/3-x+5/12 = -25/12`

`1/3-x = (-25/12)-5/12`

`1/3-x = -5/2`

`x=1/3 - (-5/2)`

`x=17/6`

Em đăng ký tham gia cuộc thi này

Em đăng ký tham gia cuộc thi này

Gì Dậy

b: Tọa độ là:

\(\left\{{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{1}{3}x+2\\y=\dfrac{2}{3}x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{4}{3}\end{matrix}\right.\)

Câu 7:

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\\\%m_{CuO}=44\%\end{matrix}\right.\)

c, \(n_{CuO}=\dfrac{10-0,1.56}{80}=0,055\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Fe}+n_{CuO}=0,155\left(mol\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,155.98}{100}.100\%=15,19\%\)

d, Theo PT: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,055\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1.152=15,2\left(g\right)\\m_{CuSO_4}=0,055.160=8,8\left(g\right)\end{matrix}\right.\)

Câu 8:

a, \(CuCO_3+2HCl\rightarrow CuCl_2+CO_2+H_2O\)

b, \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{CuCO_3}=n_{CO_2}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuCO_3}=\dfrac{0,15.124}{20}.100\%=93\%\\\%m_{CuCl_2}=7\%\end{matrix}\right.\)

c, \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

Bài 6: 

a, \(SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3+H_2O\)

b, \(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaSO_3}=n_{SO_2}=0,05\left(mol\right)\)

\(\Rightarrow C\%_{Ca\left(OH\right)_2}=\dfrac{0,05.74}{100}.100\%=3,7\%\)

c, \(m_{CaSO_3}=0,05.120=6\left(g\right)\)

Para 1 - b

Para 2 - a

Para 3 - c

T - F - T - T - NG

1 B

2 A

3 D

4 D

5 A

a) Xét ΔMNI vuông tại M và ΔHPI vuông tại P có

\(\widehat{MIN}=\widehat{HIP}\)(hai góc đối đỉnh)

Do đó: ΔMNI\(\sim\)ΔHPI(g-g)

b) Ta có: ΔMNI\(\sim\)ΔHPI(cmt)

nên \(\widehat{MNI}=\widehat{HPI}\)(hai góc tương ứng)

hay \(\widehat{MNI}=\widehat{MPK}\)

Xét ΔMNI vuông tại M và ΔMPK vuông tại M có

\(\widehat{MNI}=\widehat{MPK}\)(cmt)

Do đó: ΔMNI\(\sim\)ΔMPK(g-g)

Suy ra: \(\dfrac{MN}{MP}=\dfrac{MI}{MK}\)(Các cặp cạnh tương ứng tỉ lệ)

hay \(\dfrac{MN}{MI}=\dfrac{MP}{MK}\)

Xét ΔMNP vuông tại M và ΔMIK vuông tại M có

\(\dfrac{MN}{MI}=\dfrac{MP}{MK}\)(cmt)

Do đó: ΔMNP\(\sim\)ΔMIK(c-g-c)