đặt: \(\frac{x}{4}=\frac{2y}{5}=\frac{5z}{6}=k\)

\(\Rightarrow\begin{cases}x=4k\\ y=\frac{5k}{2}\\ z=\frac{6k}{5}\end{cases}\) (1)

thay (1) vào biểu thức \(x^2-3y^2+2z^2=325\) ta được:

\(\left(4k\right)^2-3\cdot\left(\frac{5k}{2}\right)^2+2\cdot\left(\frac{6k}{5}\right)^2=325\)

\(16k^2-\frac{75k^2}{4}_{}+\frac{72k^2}{25}=325\)

\(\frac{1600k^2}{100}-\frac{1875k^2}{100}+\frac{288k^2}{100}=325\)

\(\frac{13k^2}{100}=325\Rightarrow13k^2=32500\)

\(=>k^2=2500\Rightarrow k=\pm50\)

\(\left[\begin{array}{l}\begin{cases}x=4k=4\cdot50=200\\ y=\frac{5k}{2}=\frac{5\cdot50}{2}=125\\ z=\frac{6k}{5}=\frac{6\cdot50}{5}=60\end{cases}\\ \begin{cases}x=4k=4\cdot\left(-50\right)=-200\\ y=\frac{5k}{2}=\frac{5\cdot\left(-50\right)}{2}=-125\\ z=\frac{6k}{5}=\frac{6\cdot\left(-50\right)}{5}=-60\end{cases}\end{array}\right.\)

kết luận: \(\left(x;y;z\right)=\left[\begin{array}{l}\left(200;125;60\right)\\ \left(-200;-125;-60\right)\end{array}\right.\)

số mol khí NO2 là:

\(n_{NO2}=\frac{V_{NO2}}{24,79}=\frac{6,1975}{24,79}=0,25\left(mol\right)\)

khối lượng khí NO2 là:

\(m_{NO2}=n_{NO2}\cdot M_{NO2}=0,25\cdot46=11,5\left(g\right)\)

câu 1: \(\left(2x+y\right)^3-2\left(y-x\right)^3\)

\(=8x^3+12x^2y+6xy^2+y^3-\left(2y^3-6y^2x+6x^2y-2x^3\right)\)

\(=8x^3+12x^2y+6xy^2+y^3-2y^3+6y^2x-6x^2y+2x^3\)

\(=10x^3+\left(12x^2y-6x^2y\right)+\left(6xy^2+6y^2x\right)+\left(y^3-2y^3\right)\)

\(=10x^3+6x^2y+12xy^2-y^3\)

câu 2: \(\left(2x-3\right)^3-2x\left(2x+1\right)^2\)

\(=8x^3-36x^2+54x-27-\left(8x^3+8x^2+2x\right)\)

\(=\left(8x^3-8x^3\right)+\left(-36x^2-8x^2\right)+\left(54x-2x\right)-27\)

\(=-44x^2+52x-27\)

câu 3: \(\left(3x-1\right)^3-27x^2\left(x+1\right)\)

\(=27x^3-27x^2+9x-1-\left(27x^3+27x^2\right)\)

\(=\left(27x^3-27x^3\right)+\left(-27x^2-27x^2\right)+9x-1\)

\(=-54x^2+9x-1\)

câu 4: \(\left(2x+1\right)^3-8x\left(x-1\right)^2\)

\(=8x^3+12x^2+6x+1-\left(8x^3-16x^2+8x\right)\)

\(=28x^2-2x+1\)

câu a: 15 - (13 + x) - x = - (23 - 17)

15 - 13 - x - x = -23 + 17

2 - 2x = -6

-2x = -6 - 2

-2x = -8

x = (-8): (-2) = 4

vậy x = 4

câu b: x - (35 - x) = 43 - 48

x - 35 + x = -5

2x = -5 + 35

2x = 30

x = 30 : 2

x = 15

vậy x = 15

câu c: - (35 - x) - (37 - x) + x = 33

-35 + x - 37 + x + x = 33

3x = 33 + 37 + 35

3x = 105

x = 105 : 3 = 35

vậy x = 35

câu d: - (x - 6 + 85) - (x + 51) = -54

-x + 6 - 85 - x - 51 = -54

-2x = -54 - 6 + 85 + 51

-2x = 76

x = 76 : (-2) = -38

vậy x = -38

câu 1: \(\left(2x+y\right)^3-2\left(y-x\right)^3\)

\(=8x^3+12x^2y+6xy^2+y^3-\left(2y^3-6y^2x+6x^2y-2x^3\right)\)

\(=8x^3+12x^2y+6xy^2+y^3-2y^3+6y^2x-6x^2y+2x^3\)

\(=10x^3+\left(12x^2y-6x^2y\right)+\left(6xy^2+6y^2x\right)+\left(y^3-2y^3\right)\)

\(=10x^3+6x^2y+12xy^2-y^3\)

câu 2: \(\left(2x-3\right)^3-2x\left(2x+1\right)^2\)

\(=8x^3-36x^2+54x-27-\left(8x^3+8x^2+2x\right)\)

\(=\left(8x^3-8x^3\right)+\left(-36x^2-8x^2\right)+\left(54x-2x\right)-27\)

\(=-44x^2+52x-27\)

câu 3: \(\left(3x-1\right)^3-27x^2\left(x+1\right)\)

\(=27x^3-27x^2+9x-1-\left(27x^3+27x^2\right)\)

\(=\left(27x^3-27x^3\right)+\left(-27x^2-27x^2\right)+9x-1\)

\(=-54x^2+9x-1\)

câu 4: \(\left(2x+1\right)^3-8x\left(x-1\right)^2\)

\(=8x^3+12x^2+6x+1-\left(8x^3-16x^2+8x\right)\)

\(=28x^2-2x+1\)

\(6HNO_3+Fe_2O_3\to2Fe\left(NO_3\right)_3+3H_2O\)

\(CO_2+2NaOH\to Na_2CO_3+H_2O\)

\(FeO+H_2\to Fe+H_2O\)

\(BaO+H_2O\to Ba\left(OH\right)_2\)

câu a: barium oxide

câu b: sulfur dioxide

câu c: diphosphorus pentoxide

8 : (a - 1) + 12 : (a - 1) = 5

(8 + 12) : (a - 1) = 5

20 : (a - 1) = 5

a - 1 = 20 : 5

a - 1 = 4

a = 1 + 4

a = 5

vậy a = 5

\(A=\frac{2^{30}\cdot5^7+2^{13}\cdot5^{27}}{2^{27}\cdot5^7+2^{10}\cdot5^{27}}=\frac{2^3\cdot\left(2^{27}\cdot5^7+2^{10}\cdot5^{27}\right)}{2^{27}\cdot5^7+2^{10}\cdot5^{27}}=2^3=8\)

\(\frac{5}{2x}-3=\frac{1}{x}\)

\(\frac{5}{2x}-\frac{1}{x}=3\)

\(\frac{5}{2x}-\frac{2}{2x}=3\)

\(\frac{3}{2x}=3\)

\(\frac{1}{2x}=\frac11\)

\(2x=1\)

\(x=\frac12\)

vậy x = \(\frac12\)

\(2H_2+O_2\to2H_2O\)

0,3 0,15 0,3

số mol khí H2 đã đốt cháy là:

\(n_{H2}=\frac{V_{H2}}{24,79}=\frac{7,437}{24,79}=0,3\left(mol\right)\)

a. thể tích O2 đã dùng là:

\(V_{O2}=24,79\cdot n_{O2}=24,79\cdot0,15=3,7185\left(L\right)\)

b. khối lượng nước H2O tạo thành là:

\(m_{H2O}=n_{H2O}\cdot M_{H2O}=0,3\cdot18=5,4\left(g\right)\)

kết luận: a: 3,7185L; b: 5,4g