a) {-2).x = -10 => x = (-10): (- 2) => x = 5.

b)  (-18). x = -36 => x = (-36): (-18) => x = 2.

c) 2.x + 1 = 3=>2x = 2 => x = l.

d) (-4).x + 5 = -15 => (-4)x = (-15) -5 => (-4) x = -20 => x = 5.

a) {-2).x = -10 => x = (-10): (- 2) => x = 5.

b)  (-18). x = -36 => x = (-36): (-18) => x = 2.

c) 2.x + 1 = 3=>2x = 2 => x = l.

d) (-4).x + 5 = -15 => (-4)x = (-15) -5 => (-4) x = -20 => x = 5.

Lời giải:

a. 

$(25-2x)^3:5-3^2=4^2$

$(25-2x)^3:5=4^2+3^2=25$

$(25-2x)^3=25.5=5^3$

$\Rightarrow 25-2x=5$

$\Rightarrow 2x=20$

$\Rightarrow x=10$

b.

$2.3^x=10.3^{12}+8.27^4=10.3^{12}+8.3^{12}=18.3^{12}=2.3^{14}$

$\Rightarrow 3^x=3^{14}$

$\Rightarrow x=14$

`@` `\text {Ans}`

`\downarrow`

`2+(x+3)=7`

`\Rightarrow x+3=7-2`

`\Rightarrow x+3=5`

`\Rightarrow x=5-3`

`\Rightarrow x=2`

`5+(3+x)=10`

`\Rightarrow 3+x=10-5`

`\Rightarrow 3+x=5`

`\Rightarrow x=5-3`

`\Rightarrow x=2`

`(4+x)+1=7`

`\Rightarrow 4+x=7-1`

`\Rightarrow 4+x=6`

`\Rightarrow x=6-4`

`\Rightarrow x=2`

`(x+5)+3=9`

`\Rightarrow x+5=9-3`

`\Rightarrow x+5=6`

`\Rightarrow x=6-5`

`\Rightarrow x=1`

`(x-1)-4=7`

`\Rightarrow x-1=7+4`

`\Rightarrow x-1=11`

`\Rightarrow x=11+1`

`\Rightarrow x=12`

`4-(6-x)=1`

`\Rightarrow 6-x=4-1`

`\Rightarrow 6-x=3`

`\Rightarrow x=6-3`

`\Rightarrow x=3`

\(2+\left(x+3\right)=7\)

\(\Rightarrow2+x+3=7\)

\(\Rightarrow x+5=7\)

\(\Rightarrow x=2\)

\(5+\left(3+x\right)=10\)

\(\Rightarrow5+3+x=10\)

\(\Rightarrow x+8=10\)

\(\Rightarrow x=2\)

\(\left(4+x\right)+1=7\)

\(\Rightarrow4+x+1=7\)

\(\Rightarrow x+5=7\)

\(\Rightarrow x=2\)

\(\left(x+5\right)+3=9\)

\(=x+5+3=9\)

\(\Rightarrow x+8=9\)

\(\Rightarrow x=1\)

\(\left(x-1\right)-4=7\)

\(\Rightarrow x-1-4=7\)

\(\Rightarrow x-5=7\)

\(\Rightarrow x=12\)

\(4-\left(6-x\right)=1\)

\(\Rightarrow4-6-x=1\)

\(\Rightarrow-2-x=1\)

\(\Rightarrow x=-3\)

làm Ny anh nhé :))

Đồ điên

hê cái anh này đi đâu cg làm ny anh nhé chắc phải đến mấy chục người rồi nhỉ

a) \(35-3.\left|x\right|=5.\left(2^3-4\right)\)

    \(35-3.\left|x\right|=5.\left(8-4\right)\)

    \(35-3.\left|x\right|=5-4\)

    \(35-3.\left|x\right|=1\)

                \(3.\left|x\right|=35-1\)

                \(3.\left|x\right|=34\)

                    \(\left|x\right|=34:3\)

                     \(\left|x\right|=\frac{34}{3}\)

\(=>x=\frac{34}{3}\)hoặc \(x=-\frac{34}{3}\)

b)\(10+2.\left|x\right|=2.\left(3^2-1\right)\)

   \(10+2.\left|x\right|=2.\left(9-1\right)\)

   \(10+2.\left|x\right|=2.8\)

   \(10+2.\left|x\right|=16\)

               \(2.\left|x\right|=16-10\)

               \(2.\left|x\right|=6\)

                   \(\left|x\right|=6:2\)

                   \(\left|x\right|=3\)

\(=>x=3\)hoặc \(x=-3\)

a)35-3|x|=5(8-4)

35-3|x|=5.4=20

3|x|=35-20=15

|x|=15:3=5

=>x=5 hoặc x=-5

a)

\(35-3|x|=5\left(2^3-4\right)\)

\(\Rightarrow35-3|x|=5\left(8-4\right)\)

\(\Rightarrow35-3|x|=5.4\)

\(\Rightarrow35-3|x|=20\)

\(\Rightarrow3|x|=35-20\)

\(\Rightarrow3|x|=15\)

\(\Rightarrow|x|=15\div3\)

\(\Rightarrow|x|=5\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

Vậy \(x=5\) hoặc \(x=-5\)

b)

\(10+2|x|=2\left(3^2-1\right)\)

\(\Rightarrow10+2|x|=2\left(9-1\right)\)

\(\Rightarrow10+2|x|=2.8\)

\(\Rightarrow10+2|x|=16\)

\(\Rightarrow2|x|=16-10\)

\(\Rightarrow2|x|=6\)

\(\Rightarrow|x|=6\div2\)

\(\Rightarrow|x|=3\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy \(x=3\) hoặc \(x=-3\)

Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.

\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)