52*1,8-13,3*1,8=(52-13,3)*1,8=38,7*1,8=69,66

Do tất cả số hạng đều lớn hơn hoặc bằng 0 với mọi x

=> 16x lớn hơn hoặc bằng 0

=> x lớn hơn hoặc = 0

=> bỏ tất cả dấu trị tuyệt đối

=> 11x + 1 + 2 + 3 + ... + 10 = 16x

=> 5x = 55

=> x = 11

(13.3^x-2-3^x):2=162

<=>3^x-2(13-3²)=162.2

<=>3^x-2.4=324

<=>3^x-2=324:4

<=>3^x-2=81=3⁴

<=>x-2=4

<=>x=6

Vậy...

Ta có: \(\left(13.3^{x-2}-3^x\right)\div2=162\)

\(\Leftrightarrow13.3^{x-2}-3^{x-2+2}=162.2\)

\(\Leftrightarrow13.3^{x-2}-3^2.3^{x-2}=324\)

\(\Leftrightarrow\left(13-9\right).3^{x-2}=324\)

\(\Leftrightarrow4.3^{x-2}=324\)

\(\Leftrightarrow3^{x-2}=\dfrac{324}{4}=81=3^4\)

\(\Leftrightarrow3^{x-2}=3^4\Leftrightarrow x-2=4\Leftrightarrow x=6\)

Vậy \(x=6\)

Ta có :

\(\left(13.3^{x-2}-3x\right):2=162\)

\(\Rightarrow3^{x-2}\left(13-3^2\right)=162.2\)

\(\Rightarrow3^{x-2}.4=324\)

\(\Rightarrow3^{x-2}=324:4\)

\(3^{x-2}=81\)

\(3^{x-2}=3^4\)

\(\Rightarrow x-2=4\)

\(x=4+2\)

\(x=6\)

Vậy \(x=6\) là giá trị cần tìm

~ Chúc bn học tốt ~

x⁴+5x³-x²-16x+10=0

<=>x⁴+3x³-5x²+2x³+6x²-10x-5x²-15x+25=0

<=>x²(x²+3x-5)+2x(x²+3x-5)-5(x²+3x-5)=0

<=>(x²+2x-2)(x²+3x-5)=0

<=>x²+2x-2=0 hoặc x²+3x-5=0

\(\Leftrightarrow x=\pm\sqrt{3}-1\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{\sqrt{29}+3}{2}\\x=\frac{\sqrt{29}-3}{2}\end{cases}}\)

a) x(x-5)+2(x-5) = (x-5)(x+2)
b) (x-7)(x-2)

c) (x+2)(x^2+2x+4)+5y(x+2) = (x+2)(x^2+2x+4+5y) 

d) (x^2+8)^2 -16x^2 = (x^2+8-4x)(x^2+8+4x)

Ta có: \(16x+40=10\cdot3^2+5\cdot\left(1+2+3\right)\)

\(\Leftrightarrow16x+40=90+5\cdot6=120\)

hay x=5

A=\(\frac{16x^2-40xy}{8x^2-24xy}=\frac{8x(2x-5y)}{ 8x(x-3y)} =\frac{2x-5y}{x-3y} \)

\(\frac{x}{y}=\frac{10}{3}<=>10y=3x <=>y=\frac{3}{10}x \)

=>A=(\(2x-\frac{3}{2}x):(x-\frac{9}{10}x) \)

=\(\frac{1}{2}x:\frac{1}{10}x=\frac{1}{2}x.\frac{10}{x}=5 \)