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

175-34+125+234-150

=(175+125)+(234-34)-150

=300+200-150

=500-150=350

câu a:

\(\begin{cases}x-2y=1\\ 2x=y+4\end{cases}\Leftrightarrow\begin{cases}2x-4y=2\left(1\right)\\ 2x-y=4\left(2\right)\end{cases}\)

lấy (1) - (2) ta được:

-3y=-2⇒ y=\(\frac{-2}{-3}=\frac23\) (3)

thay (3) vào (1) ta được:

\(2x-4\cdot\frac23=2\)

\(2x-\frac83=2\Rightarrow2x=2+\frac83=\frac{14}{3}\)

\(\Rightarrow x=\frac{14}{3}:2=\frac73\)

vậy \(\left(x;y\right)=\left(\frac73;\frac23\right)\)

câu b:

\(\begin{cases}\frac12x+y=1\\ 2y=10-3x\end{cases}\Leftrightarrow\begin{cases}3x+6y=6\left(1\right)\\ 3x+2y=10\left(2\right)\end{cases}\)

lấy (1) - (2) ta được:

4y=-4 ⇒ y = -1 (3)

thay (3) vào (1) ta được:

\(3x+6\cdot\left(-1\right)=6\)

\(3x-6=6\)

\(3x=6+6=12\)

\(x=12:3=4\)

vậy \(\left(x;y\right)=\left(4;-1\right)\)

câu c:

\(\begin{cases}\frac{x}{2}=\frac{y}{3}\\ \frac{x+8}{y+4}=\frac94\end{cases}\Rightarrow\begin{cases}3x-2y=0\\ 4x-9y=4\end{cases}\Rightarrow\begin{cases}12x-8y=0\left(1\right)\\ 12x-27y=12\left(2\right)\end{cases}\)

lấy (1)-(2) ta được:

19y=-12 ⇒ y= \(-\frac{12}{19}\) (3)

thay (3) vào (1) ta được

\(12x-8\cdot\left(-\frac{12}{19}\right)=0\)

\(12x+\frac{96}{19}=0\)

\(12x=-\frac{96}{19}\Rightarrow x=-\frac{8}{19}\)

kết luận: \(\left(x;y\right)=\left(-\frac{8}{19};-\frac{12}{19}\right)\)

\(B=\frac{7}{1\cdot9}+\frac{7}{9\cdot17}+\cdots+\frac{7}{2011\cdot2019}\)

\(=\frac78\cdot\left(\frac11-\frac19+\frac19-\frac{1}{17}+.\ldots+\frac{1}{2011}-\frac{1}{2019}\right)\)

\(=\frac78\cdot\left(\frac11-\frac{1}{2019}\right)=\frac78\cdot\frac{2018}{2019}=\frac{7063}{8076}\)

\(C=\frac{3^2}{1\cdot4}+\frac{3^2}{4\cdot7}+\cdots+\frac{3^2}{2017\cdot2020}\)

\(=3\cdot\left(\frac11-\frac14+\frac14-\frac17+\ldots+\frac{1}{2017}-\frac{1}{2020}\right)\)

\(=3\cdot\left(\frac11-\frac{1}{2020}\right)=3\cdot\frac{2019}{2020}=\frac{6057}{2020}\)

\(D=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdots+\frac{1}{18\cdot19\cdot20}\)

\(=\frac12\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\cdots+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)

\(=\frac12\cdot\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)=\frac12\cdot\left(\frac12-\frac{1}{380}\right)\)

\(=\frac12\cdot\frac{189}{380}=\frac{189}{760}\)

\(\frac{1^2}{2^2}\cdot\frac{2^2}{3^2}\cdot\frac{3^2}{4^2}\cdot\ldots\cdot\frac{99^2}{100^2}\)

\(=\frac{1^2}{100^2}=\frac{1}{10000}\)