=> 72x - 20,25 + 162 = 0

=> 72x = -162 + 20,25

=> x = -63/32

-4,5x² + 72x + 162 = 0

ó -4,5x² – 9x + 81x + 162 = 0

ó -4,5x( x + 2 ) + 81( x + 2 ) = 0

ó ( x + 2 ).( -4,5x + 81 ) = 0

ó x + 2 = 0 hoặc -4,5x + 81 = 0

ó x = 2 hoặc -4,5x = -81

ó x = 2 hoặc x = 18

Vậy PT đã cho có tập nghiệm S={2;18}

bài này đặt nhân tử chung phải không bạn

cho mình hỏi 4,5x² -72-162 =0 thì làm sao

\(-4,5x^2+72x+162=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-18\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-18=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=18\end{matrix}\right.\)

hỏi bài kiểu sang hả bạn

\(-4,5x^2+72x+162=0\\ \Leftrightarrow x^2-16x-36=0\\ \Leftrightarrow x^2+2x-18x-36=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(18x+36\right)=0\\ \Leftrightarrow x\left(x+2\right)-18\left(x+2\right)=0\\ \Leftrightarrow\left(x-18\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-18=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{18;-2\right\}\)

b) \(\left(x-1\right)^3=\dfrac{1}{8}\)

    \(\left(x-1\right)^3=\left(\dfrac{1}{2}\right)^3\)

     \(x-1=\dfrac{1}{2}\)

     \(x=\dfrac{1}{2}+1\)

     \(x=\dfrac{3}{2}\)

a: =>2^x*4-2^x*3=32

=>2^x=32

=>x=5

b: =>(4x-3)^2-(4x-3)=0

=>(4x-3)(4x-3-1)=0

=>(4x-3)(4x-4)=0

=>x=3/4 hoặc x=1

c: =>7^2x+7^2x*7^3=344

=>7^2x=1

=>2x=0

=>x=0

d: =>(7x-3)^2012-(7x-3)^2010=0

=>(7x-3)^2010*[(7x-3)^2-1]=0

=>(7x-3)^2010*(7x-4)(7x-2)=0

=>x=2/7; x=4/7; x=3/7

e: =>(4x^2-3)^3=-8

=>4x^2-3=-2

=>4x^2=1

=>x^2=1/4

=>x=1/2 hoặc x=-1/2

a) 2^x(2² - 3) = 32

2^x.1=2⁵

=> x = 5

b) (4x - 3)² = 4x -3

=> (4x - 3)² - (4x - 3) = 0

(4x-3)[(4x - 3) - 1] = 0

(4x-3)(4x - 4)=0

\(\Rightarrow\left[{}\begin{matrix}4x-3=0\\4x-4=0\end{matrix}\right.\)         \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=1\end{matrix}\right.\)

c) 7^2x + 7^2x+3 = 344

=> 7^2x(1 + 7³) =344

7^2x . 344 = 344

=> 2x = 0  => x = 0

d) (7x - 3)²⁰¹² = (3 - 7x)²⁰¹⁰

=> (7x - 3)²⁰¹² - (7x - 3)²⁰¹⁰ = 0

(7x - 3)²⁰¹⁰ [(7x - 3)² - 1] = 0

\(\Rightarrow\left[{}\begin{matrix}7x-3=0\\\left(7x-3\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\7x=4\\7x=2\end{matrix}\right.\)                 \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\x=\dfrac{4}{7}\\x=\dfrac{2}{7}\end{matrix}\right.\)

e) (4x² - 3)³ + 8 = 0

(4x² - 3)³ = (-2)³

=> 4x² - 3 = -2

4x² = 1

x² = 1/4

=> \(x=\pm\dfrac{1}{2}\)

\(7^{2x-6}=49\\ \Rightarrow7^{2x-6}=7^2\\ \Rightarrow2x-6=2\\ \Rightarrow2x=8\\ \Rightarrow x=4\)

7^2x-6=7²

2x-6=2

2x=2+6

2x=8

x=4

$72x-6=49$

$=>72x=49+6$

$=>72x=55$

$=>x=\dfrac{72}{55}$

\(7^{2x-6}=49\)

\(=>7^{2x-6}=7^2\) hoặc \(7^{2x-6}=\left(-7\right)^2\)

\(=>2x-6=7\) hoặc \(2x-6=-7\)

\(=>2x=7+6\) hoặc \(2x=\left(-7\right)+6\)

\(=>2x=13\) hoặc \(2x=-1\)

\(=>x=\dfrac{13}{2}\) hoặc \(x=-\dfrac{1}{2}\)

Vậy \(x=\dfrac{13}{2}\) hoặc \(x=\dfrac{-1}{2}\)

\(#NqHahh\)

\(7^{2x-6}=49\)

\(\Rightarrow7^{2x-6}=7^2\)

\(\Rightarrow2x-6=2\)

\(\Rightarrow2x=2+6\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=8:2=4\)

Mình xin sửa lại nha:

\(6^{2x-6}=49\)

\(=>7^{2x-6}=7^2\)

\(=>2x-6=2\)

\(=>2x=2+6\)

\(=>2x=8\)

\(=>x=8:2\)

\(=>x=4\)