`#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).`

`#3107.101107`

`2.`

`a)`

\(\left\{{}\begin{matrix}3x-2y=-12\left(1\right)\\2x+5y=11\left(2\right)\end{matrix}\right.\)

Nhân `(1)` với 2: `6x - 4y = -24`

Nhân `(2)` với 3: `6x + 15y = 33`

Khi đó ta có pt mới: \(\left\{{}\begin{matrix}6x-4y=-24\\6x+15y=33\end{matrix}\right.\)

Trừ 2 vế pt ta được:

`6x - 4y - 6x - 15y = -24 - 33`

`-19y = -57`

`y = 3`

Thay `y = 3` vào `(2):`

`2x + 5 * 3 = 11`

`2x + 15 = 11`

`2x = -4`

`x = -2`

Vậy, hệ pt có nghiệm `(-2; 3).`

`b)`

\(\left\{{}\begin{matrix}4x-3y=6\left(1\right)\\8y-6x=10\left(2\right)\end{matrix}\right.\)

Nhân pt `(1)` với 3: `12x - 9y = 18`

Nhân pt `(2)` với 2: `16y - 12x = 20`

Khi đó ta có pt mới: \(\left\{{}\begin{matrix}12x-9y=18\\16y-12x=20\end{matrix}\right.\)

Cộng 2 vế pt ta được: 

`12x - 9y + 16y - 12x = 18 + 20`

`7y = 38`

`y = 38/7`

Thay vào `(1)`:

`4x - 3 * (38/7) = 6`

`4x = 114/7 + 6`

`4x = 156/7`

`x = 39/7`

Vậy, hệ pt có nghiệm `(39/7; 38/7).`

`#3107.101107`

`a)`

`2^x . 2^9 = 2^{1945}`

`2^x = 2^{1945} \div 2^9`

`2^x = 2^{1945 - 9}`

`2^x = 2^{1936}`

`x = 1936`

Vậy, `x = 1936`

`b)`

\(3^x+3^{x+1}+3^{x+2}=3159\\ 3^x+3^x\cdot3+3^x\cdot3^2=3159\\ 3^x\cdot\left(1+3+3^2\right)=3159\\ 3^x\cdot\left(4+9\right)=3159\\ 3^x\cdot13=3159\\ 3^x=243\\ 3^x=3^5\\ x=5\)

Vậy, `x = 5.`

`#3107.101107`

\(\dfrac{2^{4-x}}{16^5}=32^6\\ \dfrac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\\ \dfrac{2^4\div2^x}{2^{20}}=2^{30}\\ 2^4\div2^x=2^{30}\cdot2^{20}\\ 2^4\div2^x=2^{50}\\ 2^x=2^4\div2^{50}\\ 2^x=2^{-46}\\ x=-46\)

Viết hoa V lên nhé