`#\text{nn}`

`1.`

`a)`

V của SO₂:

\(\text{V}_{\text{SO}_2}=\text{n}_{\text{SO}_2}\cdot24,79=0,1\cdot24,79=2,479\left(\text{l}\right)\)

Vậy, `0,1` mol SO₂ có V là `2,479` l

`b)`

n của CO₂:

\(\text{n}_{\text{CO}_2}=\dfrac{\text{V}_{\text{CO}_2}}{24,79}=\dfrac{4,958}{24,79}=0,2\left(\text{mol}\right)\)

Vậy, có `0,2` mol CO₂ với V là `4,958` l

`2.`

`a)`

M của BaCO₃:

\(\text{M}_{\text{BaCO}_3}=137+12+16\cdot3=197\) (amu)

`b)`

m của BaCO₃:

\(\text{m}_{\text{BaCO}_3}=\text{n}_{\text{BaCO}_3}\cdot\text{M}_{\text{BaCO}_3}=49,25\left(\text{g}\right).\)

 `#\text{nn}`

`1)`

`A=2(3+x)+x^2` tại `x=-1`

Thay `x = -1` vào A:

`A = 2(3 - 1) + (-1)^2 = 2*2 + 1 = 4 + 1 = 5`

Vậy, `A = 5` tại `x = -1`

`2)`

`B= 3(x+5)+x^2` tại `x=1`

Thay `x = 1` vào B:

`B = 3(1 + 5) + 1^2 = 3 * 6 + 1 = 18 + 1 = 19`

Vậy, `B = 19` tại `x = 1`

`3)`

`C=5-4x(x-2)` tại `x=-1`

Thay `x = -1` vào C:

`C = 5 - 4 * (-1) * (-1 - 2)`

`= 5 + 4 * (-3) = 5 - 12 = -7`

Vậy, `C = -7` tại `x = -1`

`4)`

`D= 4x(2x-3) * 5x(x-2)` tại `x=2`

Thay `x = 2` vào D:

`D = 4 * 2 * (2 * 2 - 3) * 5 * 2 * (2 - 2) = 8 * (4 - 3) * 10 * 0 = 0`

Vậy, `D = 0` tại `x = 2`

`5)`

`E= x(x-2022)+2023(x+1) - x^2` tại `x=2024`

`E = x(x - 2022) + 2023(x + 1) - x^2`

`= x^2 - 2022x + 2023x + 2023 - x^2 = 2023 + x`

Thay `x = 2024` vào E:

`E = 2023 + 2024`

`= 4047`

Câu c mình sửa lại kq là \(\dfrac{4}{3}\), sr bạn mình nhầm.

`#\text{nn}`

`a)`

`1/5 \times 2/3 + 2/6 \times 1/5`

`= 1/5 \times (2/3 + 1/3)`

`= 1/5 \times 3/3`

`= 1/5 \times 1`

`= 1/5`

`b)`

\(\dfrac{4}{5}+\dfrac{1}{5}\times\left(-\dfrac{5}{5}\right)+\dfrac{9}{5}\times\dfrac{1}{5}\\ =\dfrac{3}{5}+\dfrac{1}{5}\times\left(1-1+\dfrac{9}{5}\right)\\ =\dfrac{3}{5}+\dfrac{1}{5}\times\dfrac{9}{5}\\ =\dfrac{3}{5}+\dfrac{9}{25}\\ =\dfrac{24}{25}\)

`c)`

\(\dfrac{34}{15}\div\dfrac{3}{2}-\dfrac{4}{15}\div\dfrac{3}{2}\\ =\dfrac{34}{15}\cdot\dfrac{2}{3}-\dfrac{4}{15}\cdot\dfrac{2}{3}\\ =\dfrac{2}{3}\cdot\left(\dfrac{34}{15}-\dfrac{4}{15}\right)\\ =\dfrac{2}{3}\cdot\dfrac{30}{15}\\ =\dfrac{2}{3}\cdot2=3\)

`d)`

\(\dfrac{5}{17}+\dfrac{12}{5}\cdot\dfrac{4}{17}+\dfrac{12}{17}\cdot\dfrac{1}{5}\\ =\dfrac{5}{17}+\dfrac{12}{17}\cdot\dfrac{4}{5}+\dfrac{12}{17}\cdot\dfrac{1}{5}\\ =\dfrac{5}{17}+\dfrac{12}{17}\cdot\left(\dfrac{4}{5}+\dfrac{1}{5}\right)\\ =\dfrac{5}{17}+\dfrac{12}{17}\cdot\dfrac{5}{5}\\ =\dfrac{5}{17}+\dfrac{12}{17}\cdot1\\ =\dfrac{5}{17}+\dfrac{12}{17}\\ =\dfrac{17}{17}=1\)

`#\text{nn}`

`a,`

\(\dfrac{8}{5}\cdot\left(-\dfrac{3}{4}\right)\\ =\dfrac{2}{5}\cdot\left(-3\right)\\ =-\dfrac{6}{5}\)

`b,`

\(\dfrac{7}{3}\cdot\dfrac{15}{14}\\ =1\cdot\dfrac{5}{2}\\ =\dfrac{5}{2}\)

`c,`

\(\left(-2,6\right)\cdot\dfrac{14}{5}\\ =7,28\)

- Làm thêm mấy câu sau ha =)))

\(65^2-35^2\\ =\left(65-35\right)\left(65+35\right)\\ =30\cdot100\\ =3000\)

____

\(\dfrac{77^2+17^2-34\cdot77}{77^2-17^2}\\ =\dfrac{77^2+2\cdot17\cdot77+17^2}{\left(77-17\right)\left(77+17\right)}\\ =\dfrac{\left(77+17\right)^2}{60\cdot94}\\ =\dfrac{94^2}{60\cdot94}\\ =\dfrac{94}{60}\\ =\dfrac{47}{30}\)

____

\(2024^2-24^2\\ =\left(2024-24\right)\left(2024+24\right)\\ =2000\cdot2048\\ =4096000\)

____

\(75^2+65^2+55^2-45^2-35^2-25^2\\ =\left(75^2-25^2\right)+\left(65^2-35^2\right)+\left(55^2-45^2\right)\\ =\left(75-25\right)\left(75+25\right)+\left(65-35\right)\left(65+35\right)+\left(55-45\right)\left(55+45\right)\\ =50\cdot100+30\cdot100+10\cdot100\\ =\left(50+30+10\right)\cdot100\\ =90\cdot100\\ =9000\)

`4,`

- Bài này bạn sử dụng các HĐT để làm bài nhé, mình gợi ý 1 số:

`A^2 - B^2 = (A - B)(A + B)`

`(A - B)^2 = A^2 - 2AB + B^2`

(Mình làm 1 số vd thui, chứ mình sắp đi học rồi =))

`39*41 = (40 - 1)(40 + 1) = 40^2 - 1^2 = 1600 - 1 = 1599`

`101^2 = (100 + 1)^2 = 100^2 + 2*100*1 + 1^2 = 10000 + 200 + 1 = 10201`

`5,`

`a,`

`(2x + 1)^2 - 4(x + 2)^2 = 9`

`=> 4x^2 + 4x + 1 - 4(x^2 + 4x + 4) = 9`

`=> 4x^2 + 4x + 1 - 4x^2 - 16x - 16 = 9`

`=> (4x^2 - 4x^2) + (4x - 16x) + (1 - 16) = 9`

`=> -12x - 15 = 9`

`=> -12x = 24`

`=> x = -2`

Vậy, `x = -2`

`b,`

`(20 - 2x)^2 = 256`

`=> (20 - 2x)^2 = +-16^2`

\(\Rightarrow\left[{}\begin{matrix}20-2x=16\\20-2x=-16\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=20-16\\2x=20-\left(-16\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=36\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=18\end{matrix}\right.\)

Vậy, `x \in {2; 18}.`

`3,`

`a,`

`(x^2 + y^2)^2 - 4x^2y^2`

`= (x^2)^2 + 2*x^2*y^2 + (y^2)^2 - 4x^2y^2`

`= x^4 + 2x^2y^2 + y^4 - 4x^2y^2`

`= x^4 + y^4 - 2x^2y^2`

`b,`

`2(x - y)(x + y) + (x + y)^2 + (x - y)^2`

`= 2(x^2 - y^2) + x^2 + 2xy + y^2 + x^2 - 2xy + y^2`

`= 2x^2 - 2y^2 + 2x^2 + 2y^2`

`= 4x^2`

`c,`

`2(x - 1)^2 - 4(3 + x)^2 + 2x(x - 5)`

`= 2(x^2 - 2x + 1) - 4(9 + 6x + x^2) + 2x^2 - 10x`

`= 2x^2 - 4x + 2 - 36 - 24x - 4x^2 + 2x^2 - 10x`

`= (2x^2 - 4x^2 + 2x^2) + (-4x -24x - 10x) + (2 - 36)`

`= -38x - 34`

`d,`

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

`= 2(4x^2 + 20x + 25) - 3(16x^2 - 1)`

`= 8x^2 + 40x + 50 - 48x^2 + 3`

`= -40x^2 + 40x + 53`