a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)

\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)

\(\Leftrightarrow-12x=12\)

hay x=-1

|2(x-5)|-|5-x|-7|x-5|=-10

|x-5|(2-1-7)=-10

|x-5|.-6=-10

|x-5\=5/3

tu lam roi h cho tao

b,|5-x|=|3-x| roi tu lam nha

look at my avatar

`#3107`

b)

`2.3^x = 162`

`\Rightarrow 3^x = 162 \div 2`

`\Rightarrow 3^x = 81`

`\Rightarrow 3^x = 3^4`

`\Rightarrow x = 4`

Vậy, `x = 4`

c)

`(2x - 15)^5 = (2 - 15)^3`

\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`

\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)

`d)`

\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!

`e)`

\(7\cdot4^{x-1}+4^{x-1}=23?\)

\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)

Bạn xem lại đề!

`f)`

\(2\cdot2^{2x}+4^3\cdot4^x=1056\)

\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)

Vậy, `x = 2`

_____

\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)

\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)

\(\Rightarrow\left(x\div3+17\right)\div10=2\)

\(\Rightarrow x\div3+17=20\)

\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)

Vậy, `x = 9.`

a) x = 2

b) x = 7.

c) x = 12

d) x = 45

e) x = 18.

f) x  = 10

a) x = 2.               

b) x = 7.

c) x= 12.

d)  x= 45.

e) x = 18.

f) x  = 10.

a) \(\Rightarrow\left(2x-3\right)^2=49\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b) \(\Rightarrow\left(x-5\right)\left(2x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)

c) \(\Rightarrow x\left(x-5\right)+2\left(x-5\right)=0\Rightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a, ⇒ (2x - 3)² = 49

    ⇒  (2x - 3)² = \(\left(\pm7\right)^2\)

    ⇒ \(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b, ⇒ 2x.(x - 5) + 7.(x - 5) = 0

    ⇒ (x - 5).(2x + 7)  = 0

    ⇒ \(\left[{}\begin{matrix}x-5=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\2x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)

c, ⇒ x² - 5x + 2x - 10 = 0

    ⇒ (x² - 5x) + (2x - 10) = 0

    ⇒ x.(x - 5) +2.(x - 5)    = 0

    ⇒ (x - 5).(x + 2)=0

    \(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)

tách ra

\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)

\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)

1)10-x=5-7

x=10-(-2)

x=12

2)x+5=0+32

x=32-5

x=27

3) -12+2x-9+x=0

-12+(2x+x-9)=0

3x-9=0+12

3x=12+9

x=21:3

x=7

4) 15-x=1-11

x=15-(-10)

x=25

5) 4-(27+3)=x-9

4-30=x-9

-26+9=x

x=-17