`#3107.101107`

\(1-\left(x+\dfrac{1}{3}\right)^2=\dfrac{3}{4}\\ \left(x+\dfrac{1}{3}\right)^2=1-\dfrac{3}{4}\\ \left(x+\dfrac{1}{3}\right)^2=\dfrac{1}{4}\\ \left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{1}{2}\right)^2\)

TH1: 

\(x+\dfrac{1}{3}=\dfrac{1}{2}\\ x=\dfrac{1}{2}-\dfrac{1}{3}\\ x=\dfrac{1}{6}\)

TH2: 

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

Vậy, \(\in\left\{\dfrac{1}{6};-\dfrac{5}{6}\right\}.\)

1. His brother goes to the theater once a month.

2. I usually store onions basket in the kitchen shelf.

1. You should not eat a large meal before bedtime.

2. It takes him two hours a day to read a book about adventure.

3. Would you like to buy tickets for the concert?

Xem lại bài nhé.Nếu bạn làm theo cách thông thường thì vế trái khai triển như thế là sai hoàn toàn.

`#3107.101107`

\(\left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{2}{3}\right)^6\\ \left(x+\dfrac{1}{3}\right)^2=\left[\pm\left(\dfrac{2}{3}\right)^3\right]^2\\ \left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{8}{27}\right)^2\)

TH1: \(x+\dfrac{1}{3}=\dfrac{8}{27}\\ x=\dfrac{8}{27}-\dfrac{1}{3}\\ x=-\dfrac{1}{27}\)

TH2: 

\(x+\dfrac{1}{3}=-\dfrac{8}{27}\\ x=-\dfrac{8}{27}-\dfrac{1}{3}\\ x=-\dfrac{17}{27}\)

Vậy, `x \in {-1/27; -17/27}.`

`#3107.101107`

`2.`

`a)`

\(\dfrac{2}{3}-\dfrac{5}{2}x=-\dfrac{13}{3}\\ \dfrac{5}{2}x=\dfrac{2}{3}-\left(-\dfrac{13}{3}\right)\\ \dfrac{5}{2}x=\dfrac{2}{3}+\dfrac{13}{3}\\ \dfrac{5}{2}x=\dfrac{15}{3}\\ x=5\div\dfrac{5}{2}\\ x=2\)

`b)`

\(2\cdot\left|3-2x\right|+\dfrac{1}{2}=\dfrac{5}{2}\\ 2\left|3-2x\right|=\dfrac{5}{2}-\dfrac{1}{2}\\ 2\left|3-2x\right|=\dfrac{4}{2}\\ 2\left|3-2x\right|=2\\ \left|3-2x\right|=1\)

TH1: `3 - 2x = 1`

`2x = 3 - 1`

`2x = 2`

`x = 1`

TH2: `3 - 2x = -1`

`2x = 3 - (-1)`

`2x = 4`

`x = 2`

Vậy, `x \in {1; 2}`

`c)`

\(x^2\cdot\left(2^x-6\right)-2x^2=0\\ x^2\left(2^x-6\right)=2x^2\\ 2^x-6=2\\ 2^x=8\\ 2^x=2^3\\ x=3\)

Vậy, `x = 3.`

`#3107.101107`

`a)`

\(\dfrac{23}{7}+\dfrac{4}{3}-\dfrac{9}{7}+\dfrac{10}{6}\\ =\left(\dfrac{23}{7}-\dfrac{9}{7}\right)+\left(\dfrac{4}{3}+\dfrac{10}{6}\right)\\ =\dfrac{14}{7}+\left(\dfrac{4}{3}+\dfrac{5}{3}\right)\\ =\dfrac{14}{7}+\dfrac{9}{3}\\ =2+2\\ =4\)

`b)`

\(\left(\dfrac{5}{8}-\dfrac{\sqrt{9}}{12}\right)\div\dfrac{3}{4}+\dfrac{11}{8}\div\dfrac{3}{4}\)

\(=\left(\dfrac{5}{8}-\dfrac{3}{12}\right)\cdot\dfrac{4}{3}+\dfrac{11}{8}\cdot\dfrac{3}{4}\)

\(=\left(\dfrac{5}{8}-\dfrac{1}{4}\right)\cdot\dfrac{4}{3}+\dfrac{11}{8}\cdot\dfrac{4}{3}\)

\(=\dfrac{4}{3}\cdot\left(\dfrac{5}{8}-\dfrac{1}{4}+\dfrac{11}{8}\right)\\ =\dfrac{4}{3}\cdot\left(\dfrac{16}{8}-\dfrac{1}{4}\right)\\ =\dfrac{4}{3}\cdot\left(2-\dfrac{1}{4}\right)\\ =\dfrac{4}{3}\cdot\dfrac{7}{4}\\ =\dfrac{7}{3}\)

`c)`

\(\left(0,\left(3\right)+\dfrac{\left|-2\right|}{3}\right)\div\dfrac{\sqrt{25}}{4}-\left(2^3+3^2\right)^0\\ =\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\div\dfrac{5}{4}-\left(8+9\right)^0\\ =\dfrac{3}{3}\div\dfrac{5}{4}-17^0\\ =1\div\dfrac{5}{4}-1\\ =\dfrac{5}{4}-1\\ =\dfrac{1}{4}\)

Có tham khảo thì ghi cre vào bạn nhé.

Xem lại bài b

`#3107.101107`

`1.`

`a)`

\(\left\{{}\begin{matrix}x+2y=12\left(1\right)\\3x-y=1\left(2\right)\end{matrix}\right.\)

Từ `(1)`: `x + 2y = 12`

`x = 12 - 2y`

Thay vào pt `(2)`:

`3*(12 - 2y) - y = 1`

`36 - 6y - y = 1`

`36 - 7y = 1`

`7y = 35`

`y = 5`

Ta có: `x = 12 - 2y`

`x = 12 - 2*5`

`x = 12 - 10`

`x = 2`

Vậy, hệ pt có nghiệm `(2; 5)`

`b)`

\(\left\{{}\begin{matrix}2x+y=7\left(1\right)\\5x-4y=-2\left(2\right)\end{matrix}\right.\)

Từ `(1)`: `2x + y = 7`

`y = 2x - 7`

Thay vào `(2)`:

`5x - 4(2x + 7) = -2`

`5x - 8x - 14 = -2`

`-3x = 12`

`x = -4`

Ta có: `y = 2x - 7`

`y = 2*(-4) - 7`

`y = -8-7`

`y = -15`

Vậy, hệ pt có nghiệm (`-4; -15).`