`3.`

`b)`

\(\left(-\dfrac{1}{3}\right)^3\cdot x=\dfrac{1}{81}?\\ \Rightarrow\left(-\dfrac{1}{3}\right)^3\cdot x=\left(-\dfrac{1}{3}\right)^4\\ \Rightarrow x=\left(-\dfrac{1}{3}\right)^4\div\left(-\dfrac{1}{3}\right)^3\\ \Rightarrow x=-\dfrac{1}{3}\)

Vậy, `x = -1/3`

`c)`

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{3}+\dfrac{1}{2}\\ \Rightarrow x=\dfrac{5}{6}\)

Vậy, `x = 5/6`

`d)`

\(\left(x+\dfrac{1}{2}\right)^4=\dfrac{16}{81}\\ \Rightarrow\left(x+\dfrac{1}{2}\right)^4=\left(\pm\dfrac{2}{3}\right)^4\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{3}\\x+\dfrac{1}{2}=-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

Vậy, `x \in {1/6; -7/6}`

`e)`

\(\left(2x-1\right)^3=8\\ \Rightarrow\left(2x-1\right)^3=2^3\\ \Rightarrow2x-1=2\\ \Rightarrow2x=2+1\\ \Rightarrow2x=3\\ \Rightarrow x=\dfrac{3}{2}\)

Vậy, `x = 3/2`

`f)`

\(x^3=-27\\ \Rightarrow x^3=\left(-3\right)^3\\ \Rightarrow x=-3\)

Vậy, `x = -3.`

`#3107.101107`

`2.`

`a)`

\(-6\cdot\left(-\dfrac{2}{3}\right)\cdot0,25\\ =4\cdot0,25\\ =1\)

`b)`

\(-\dfrac{15}{4}\cdot\left(-\dfrac{7}{15}\right)\cdot\left(-2\dfrac{2}{5}\right)\\ =\dfrac{15}{4}\cdot\dfrac{7}{15}\cdot\left(-\dfrac{12}{5}\right)\\ =\dfrac{7}{4}\cdot\left(-\dfrac{12}{5}\right)\\ =-\dfrac{21}{5}\)

`c)`

\(\left(-2\dfrac{1}{5}\right)\cdot\left(-\dfrac{9}{11}\right)\cdot\left(-1\dfrac{1}{14}\right)\cdot\dfrac{2}{5}\\ =2\dfrac{1}{5}\cdot\dfrac{9}{11}\cdot\left(-\dfrac{15}{14}\right)\cdot\dfrac{2}{5}\\ =\dfrac{11}{5}\cdot\dfrac{9}{11}\cdot\left(-\dfrac{3}{7}\right)\\ =\dfrac{9}{5}\cdot\left(-\dfrac{3}{7}\right)\\ =-\dfrac{27}{35}\)

`d)`

\(\left(-5\dfrac{1}{2}\right)\cdot\left(-\dfrac{1}{2}\right)-\dfrac{2}{3}\cdot\left(-\dfrac{2}{3}\right)\\ =5\dfrac{1}{2}\cdot\dfrac{1}{2}+\dfrac{2}{3}\cdot\dfrac{2}{3}\\ =\dfrac{11}{4}+\dfrac{4}{9}\\ =\dfrac{115}{36}\)

`e)`

\(\left(1\dfrac{1}{4}\right)\cdot\left(-\dfrac{8}{15}\right)-\dfrac{3}{5}+\dfrac{2}{5}\cdot\left(-\dfrac{3}{4}\right)\\ =-\dfrac{2}{3}-\dfrac{3}{5}-\dfrac{3}{10}\\ =-\dfrac{19}{15}-\dfrac{3}{10}\\ =-\dfrac{47}{30}\)

`f)`

\(\left(-0,125\right)\cdot\left(-16\right)\cdot\left(-\dfrac{8}{9}\right)\cdot\left(-0,25\right)\\ =0,125\cdot16\cdot\dfrac{8}{9}\cdot0,25\\ =\left(0,125\cdot\dfrac{8}{9}\right)\cdot\left(16\cdot0,25\right)\\ =\dfrac{1}{9}\cdot4\\ =\dfrac{4}{9}\)

HẠN CHẾ đăng bài spam nếu không có Câu hỏi muốn được giải đáp, làm trôi CH của người khác và làm phiền đến diễn đàn!

(n - 1) + 5 vì n - 1 + 5 = n + 4 Bạn nhé.

`#3107.101107`

`a)`

`A = {12; 24; 36; 48}`

`b)`

`B = {40; 51; 62; 73; 84; 95}`

`c)`

`C = {12 - 19; 23 - 29; 34 - 39; 45 - 49; 56 - 59; 67 - 69; 78; 79; 89}.`

`#3107.101107`

`1)`

`(2x - 1)^2 - (4x^2 - 1) = 0`

`=> (2x - 1)^2 - [(2x)^2 - 1^2] = 0`

`=> (2x - 1)^2 - (2x - 1)(2x + 1) = 0`

`=> (2x - 1)(2x - 1 - 2x - 1) = 0`

`=>`\(\left[{}\begin{matrix}2x-1=0\\2x-1-2x-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=1\\-2=0\left(\text{vô lý}\right)\end{matrix}\right.\)

`=> x = 1/2`

Vậy, `x = 1/2`

`2)`

\(\left(x+2\right)^2-x\left(x-3\right)=2\\ \Rightarrow x^2+4x+4-x^2+3x=2\\ \Rightarrow7x+4=2\\ \Rightarrow7x=2-4\\ \Rightarrow7x=-2\Rightarrow x=-\dfrac{7}{2}\)

Vậy, `x = -7/2`

`3)`

\(\left(x-5\right)^2-x\left(x+2\right)=5\\ \Rightarrow x^2-10x+25-x^2-2x=5\\ \Rightarrow-12x+25=5\\ \Rightarrow-12x=5-25\\ \Rightarrow-12x=-20\\ \Rightarrow12x=20\\ \Rightarrow x=\dfrac{5}{3}\)

Vậy, `x = 5/3`

`4)`

\(\left(x-1\right)^2+x\left(4-x\right)=11\\ \Rightarrow x^2-2x+1+4x-x^2=11\\ \Rightarrow2x+1=11\\ \Rightarrow2x=11-1\\ \Rightarrow2x=10\\ \Rightarrow x=5\)

Vậy, `x = 5.`

`#3107.101107`

`1.`

`a,`

`7,7 + 7,3 \times 7,4 = 7,7 + 54,02 = 61,72`

`b,`

`4,57 \times 5,1 + 2,43 \times 5,1`

`= 5,1 \times (4,57 + 2,43)`

`= 5,1 \times 7`

`= 35,7`

`c,`

\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}\\ =\dfrac{1}{6}+\dfrac{1}{4}\\ =\dfrac{5}{12}\)

`2.`

`a,`

\(x\times15,45=30,2\\ x=30,2\div15,45\\ x\approx1,95\)

Vậy, `x \approx 1,95`

`b,`

\(\dfrac{22}{7}\div\left(11-x\right)+\dfrac{2}{3}=\dfrac{7}{5}\\ \dfrac{22}{7}\div\left(11-x\right)=\dfrac{7}{5}-\dfrac{2}{3}\\ \dfrac{22}{7}\div\left(11-x\right)=\dfrac{11}{15}\\ 11-x=\dfrac{22}{7}\div\dfrac{11}{15}\\ 11-x=\dfrac{30}{7}\\ x=11-\dfrac{30}{7}\\ x=\dfrac{47}{7}\)

Vậy, \(x=\dfrac{47}{7}.\)

The teacher demonstrates how to use the equipment.

He kept his notes for the speech in a folder.

`#3107.101107`

`a)`

Gọi CT chung: \(\text{Fe}_x\text{O}_y\)

`%O = 100% - 70% = 30%`

Ta có:

\(x\div y=\dfrac{\%\text{Fe}}{56}\div\dfrac{\%\text{O}}{16}\\ =\dfrac{70\%}{56}\div\dfrac{30\%}{16}\\ =\dfrac{1}{80}\div\dfrac{3}{160}\\ =\dfrac{2}{3}\)

Vậy, `x = 2; y = 3`

`\Rightarrow CTHH: Fe_2O_3`

Tương tự như 2 bài sau

`b)`

Gọi CT chung: \(\text{C}_x\text{H}_y\)

`%H = 100% - 75% = 25%`

Ta có:

\(x\div y\\ =\dfrac{\%\text{C}}{12}\div\dfrac{\%\text{H}}{1}\\ =\dfrac{75\%}{12}\div\dfrac{25\%}{1}\\ =\dfrac{1}{16}\div\dfrac{1}{4}\\ =\dfrac{1}{4}\)

CTHH: `CH_4`

`c)`

Gọi CT chung: `Fe_xS_yO_z`

`%O = 100% - 28% - 24% = 48%`

Ta có:

\(x\div y\div z\\ =\dfrac{\%\text{Fe}}{56}\div\dfrac{\%\text{S}}{32}\div\dfrac{\%\text{O}}{16}\\ =\dfrac{28\%}{56}\div\dfrac{24\%}{32}\div\dfrac{\%48}{16}\\ =\dfrac{1}{200}\div\dfrac{3}{400}\div\dfrac{3}{100}\\ =0,005\div0,0075\div0,03\\ =2\div3\div12\)

`\Rightarrow CTHH: Fe_2S_3O_12`

Hay `Fe_2(SO_4)_3`

`#3107.101107`

`a)`

\(\left(\dfrac{2}{3}\right)^3\div\left(\dfrac{8}{27}\right)^3\\ =\left(\dfrac{2}{3}\right)^3\div\left(\dfrac{2^3}{3^3}\right)^3\\ =\left(\dfrac{2}{3}\right)^3\div\left(\dfrac{2}{3}\right)^9\\ =\left(\dfrac{2}{3}\right)^{3-9}\\ =\left(\dfrac{2}{3}\right)^{-6}\\ =\left(\dfrac{3}{2}\right)^6\\ =\dfrac{729}{64}\)

`b)`

\(\left(-\dfrac{7}{5}\right)^5\div\left(-\dfrac{14}{18}\right)^5\\ =\left(-\dfrac{7}{5}\div-\dfrac{14}{18}\right)^5\\ =\left(\dfrac{7}{5}\cdot\dfrac{18}{14}\right)^5\\ =\left(\dfrac{9}{5}\right)^5\\ =\dfrac{59049}{3125}=18,89568\)

`c)`

\(\left(-\dfrac{1}{7}\right)^{2018}\div\left(\dfrac{1}{7}\right)^{2018}\\ =\left(\dfrac{1}{7}\right)^{2018}\div\left(\dfrac{1}{7}\right)^{2018}\\ =\left(\dfrac{1}{7}\right)^{2018-2018}\\ =\left(\dfrac{1}{7}\right)^0\\ =1\)