8x⁵+4x³+(x²-2).(-8x³)=20

8x⁵⁺4x³-8x⁵+16x³=20

20x³                     =20

x³                         =1

x                           =1

vậy x=1

 ( 2x - 1 )² - ( 2x + 5 )( 2x - 5 ) = 20

\(=4x^2+1-4x-\left(4x^2-25\right)=20\)

\(=4x^2+1-4x-4x^2+25=20\)

\(=26-4x=20\)

\(4x=26-20=6\)

\(x=\frac{6}{4}=1,5\)

Bài 1: 

a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)

\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

b) Ta có: \(\left(2x-3\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)

\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)

\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Bài 2: 

a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)

b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)

c) \(3+3^2+3^3+...+3^{2007}\)

\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{2005}\right)⋮13\)

\(\left(\dfrac{1}{20}-\dfrac{1}{5}\right)\cdot2^{x+1}+2^{x+2}=\dfrac{-148}{5}\)

\(\Rightarrow-\dfrac{3}{20}\cdot2^{x+1}+2^{x+1}\cdot2=-\dfrac{148}{5}\)

\(\Rightarrow2^{x+1}\cdot\left(-\dfrac{3}{20}+2\right)=\dfrac{-148}{5}\)

\(\Rightarrow2^{x+1}\cdot\dfrac{37}{20}=\dfrac{-148}{5}\)

\(\Rightarrow2^{x+1}=\dfrac{-148}{5}:\dfrac{37}{20}\)

\(\Rightarrow2^{x+1}=-16\)

Xem lại bài 

\(\left(\dfrac{1}{20}-\dfrac{1}{5}\right)\cdot2^{x+1}+2^{x+2}=-\dfrac{148}{5}\)

\(\Leftrightarrow-\dfrac{3}{20}\cdot2^{x+1}+2^{x+1}\cdot2=\dfrac{-148}{5}\)

\(\Leftrightarrow2^{x+1}\left(-\dfrac{3}{20}+2\right)=\dfrac{-148}{5}\)

\(\Leftrightarrow2^{x+1}\cdot\dfrac{37}{20}=\dfrac{-148}{5}\)

\(\Leftrightarrow2^{x+1}=\dfrac{-148}{5}:\dfrac{37}{20}\)

\(\Leftrightarrow2^{x+1}=-16\)

\(\Leftrightarrow2^{x+1}=-2^4\)(vô lí)

Xem lại đề bài!

Ta có :           [( 2^2x +10 ) . 2 + 20 ] -18 = 20

                           ( 2^2x  +10 ) . 2 + 20 = 20 + 18 = 38 

                                ( 2^2x +10 ) . 2 = 38 -20 = 18

                                     2^2x +10 = 18 :2 = 9

                   Vì 10 > 9 nên x = rỗng.

Đề hỏi gì bạn 

tìm gtln

Áp dụng tính chất dễ thôi bạn 

C=/2x-20/-/2x-3/\(\le\)/2x-20-2x-3/ (Do tính chất /a/-/b/ bé hơn hoặc bằng /a-b/

hay C\(\le\)-23

a. 5x.(12x+7)-3x.(20x-5)=-150

x=-3

b. ( 2x-1).(3-x)+(x+4).(2x-5)=20

x=43/10

c. 9x²-1+(3x-1)²=0

x=1/3

d. 3x.(x-2)-(3x+2).(x-1)=7

x=-5/2

e. (2x-1)²-(2x+5).(2x-5)=20

x=3/2

f. 4x²-5=4

x=3/2

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

Lời giải:

Đặt \(\sqrt{2x^2+4x+8}=t\Rightarrow x^2+2x+4=\frac{t^2}{2}\)

Khi đó, PT đã cho trở thành:

\(\frac{t^2}{2}=t+24\Leftrightarrow t^2-2t-48=0\)

\(\Leftrightarrow (t+6)(t-8)=0\Rightarrow t=8\) (do \(t\geq 0\))

Khi đó \(x^2+2x+4=\frac{t^2}{2}=32\Leftrightarrow x^2+2x-28=0\)

\(\Leftrightarrow \) \(\left[{}\begin{matrix}x=-1+\sqrt{29}\\x=-1-\sqrt{29}\end{matrix}\right.\) (đều thỏa mãn)