2(x-5)+3=-7

=>2x-10=-7+3=-4

=>2x=-4+10=6

=>x=6:2=3

4x+(27+2^3)=1

=>4x+(27+8)=1

=>4x+35=1

=>4x=1-35=-34

=>x=-34:4=8.5

lx-3l+5=-27

=>lx-3l=-27-5=-32

mà lx-3l>0 với mọi x

=>k co gt x t/m

-3lx-1l=-27

=>lx-1l=-27:(-3)=9

=>x-1=+9

+)x-1=9=>x=10

+)x-1=-9=>x=-8

\(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^2\left(x+8\right)=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-\left(x^3+8x^2\right)-27=0\)

\(\Leftrightarrow17x-17=0\\ \Leftrightarrow x=1\)

a) \(27\cdot75+25\cdot27-150\cdot2023^0\)

\(=27\cdot\left(75+25\right)-150\cdot1\)

\(=27\cdot100-150\)
\(=2700-150\)

\(=2550\)

b) \(3^x+3^{x+1}+16\cdot4=2^2\cdot5^2\)

\(\Rightarrow3^x+3^x\cdot3=100-64\)

\(\Rightarrow3^x\cdot\left(1+3\right)=36\)

\(\Rightarrow3^x\cdot4=36\)

\(\Rightarrow3^x=36:4\)

\(\Rightarrow3^x=9\)

\(\Rightarrow3^x=3^2\)

\(\Rightarrow x=2\)

a) 27.75 + 25.27 - 150.2023⁰

= 27.(75 + 25) - 150.1

= 27.100 - 150

= 2700 - 150

= 2550

b) 3ˣ + 3ˣ⁺¹ + 16.4 = 2².5²

3ˣ.(1 + 3) + 64 = 4.25

3ˣ.4 + 64 = 100

3ˣ.4 = 100 - 64

3ˣ.4 = 36

3ˣ = 36 : 4

3ˣ = 9

3ˣ = 3²

x = 2

a, 5¹⁴ \(\times\) \(x\) = 5¹⁷

              \(x\) = 5¹⁷ : 5¹⁴

              \(x\) = 5³

b, 5\(^x\) + 5\(^{x+2}\) + 5\(^{x+4}\) = 1948

   5\(^x\) .( 1 + 5² + 5⁴) = 1948

   5\(^x\). ( 1 + 25 + 625) = 1948

  5\(^x\) . 651 = 1948

  5\(^x\)           = 1948 : 651 ( xem lại đề bài )

c, 27 \(\times\) \(x\)  - 100 = 1948

   27 \(\times\) \(x\)            = 1948 + 100

    27 \(\times\) \(x\)           =  2048

             \(x\)           = 2048 : 27

              \(x\)        = \(\dfrac{2048}{27}\)

a. \(\dfrac{x}{5}=\dfrac{8}{10}\)

\(\Rightarrow10x=8.5\)

\(\Rightarrow10x=40\)

\(\Rightarrow x=\dfrac{40}{10}=4\)

b. \(\dfrac{9}{x}=\dfrac{90}{100}\)

\(\Rightarrow9.100=90x\)

\(\Rightarrow900=90x\)

\(\Rightarrow x=\dfrac{900}{90}=10\)

c. \(17+\dfrac{x}{35}=\dfrac{5}{7}\)

\(\Rightarrow\dfrac{595+x}{35}=\dfrac{25}{35}\)

\(\Rightarrow595+x=25\)

\(\Rightarrow x=25-595=-570\)

d. \(x:\dfrac{5}{27}=\dfrac{2}{9}\)

\(\Rightarrow\dfrac{27x:5}{27}=\dfrac{6}{27}\)

\(\Rightarrow27x:5=6\)

\(\Rightarrow27x=5.6\)

\(\Rightarrow27x=30\)

\(\Rightarrow x=\dfrac{30}{27}=\dfrac{10}{9}\approx1,11\)

Ta có: \(\left\{{}\begin{matrix}x\left(x+2y+3z\right)=-5\\y\left(x+2y+3z\right)=27\\z\left(x+2y+3z\right)=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-5}=x+2y+3z\\\dfrac{y}{27}=x+2y+3z\\\dfrac{z}{5}=x+2y+3z\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{-5}=\dfrac{y}{27}=\dfrac{z}{5}\Rightarrow\left\{{}\begin{matrix}y=\dfrac{-27}{5}x\\z=-x\end{matrix}\right.\)

Ta có: \(x\left(x+2y+3z\right)=-5\Rightarrow x\left(x+2.\dfrac{-27}{5}x-3x\right)=-5\)

\(\Rightarrow\dfrac{-64}{5}x^2=-5\Rightarrow x^2=\dfrac{25}{64}\Rightarrow x=\dfrac{5}{8}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=-\dfrac{27}{5}x=-\dfrac{27}{8}\\z=-x=-\dfrac{5}{8}\end{matrix}\right.\)

a. 3 × x = 27

x = 27 : 3

x = 9

b. x : 5 = 4

x = 4 × 5

x = 20

\(2^{27}-x=8^5\\ \Rightarrow x=2^{27}-8^5=2^{27}-2^{15}\\ \Rightarrow x=2^{15}\left(2^{12}-1\right)=32768\cdot4095\\ \Rightarrow x=134184960\)

\(2^{27-x}=8^5=2^{15}\\ \Rightarrow27-x=15\Rightarrow x=12\)