\(5x^2+25x-750=0\)

\(\Leftrightarrow5\left(x^2+5x-150\right)=0\)

\(\Leftrightarrow5\left(x^2+15x-10x-150\right)=0\)

\(\Leftrightarrow5\left[\left(x^2+15x\right)-\left(10x+150\right)\right]=0\)

\(\Leftrightarrow5\left[x\left(x+15\right)-10\left(x+15\right)\right]=0\)

\(\Leftrightarrow5\left(x-10\right)\left(x+15\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-10=0\\x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-15\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{10;-15\right\}\)

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

  A = (5\(x\) + 1)² + (5\(x\) - 1)² - 2.( 5\(x\) +1).(5\(x\) - 1) tại \(x\) = 1

Thay \(x\) = 1 vào A ta có:

A = (5.1 + 1)² + (5.1 - 1)² - 2.(5.1 + 1).(5.1 - 1)

A = 6² + 4² - 2.6.4 

A = 36 + 16 - 48

A = 52 - 48

A = 4

\(\left(5x-1\right)^2+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)

\(=\left(5x-1\right)^2-2\left(5x-1\right)\left(5x+4\right)+\left(5x+4\right)^2\)

\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2\)

\(=\left(5x-1-5x-4\right)^2\)

\(=\left(-5\right)^2\)

\(=25\)

Ai giúp mình với. Mình tích cho ạ

\(\dfrac{3}{{5x - 1}} + \dfrac{2}{{3 - 5x}} = \dfrac{4}{{\left( {1 - 5x} \right)\left( {x - 3} \right)}}\)

ĐKXĐ: \(x \ne \dfrac{1}{5};x\ne \dfrac{3}{5};x \ne 3\)

\( \Leftrightarrow 3\left( {3 - 5x} \right)\left( {x - 3} \right) + 2\left( {5x - 1} \right)\left( {x - 3} \right) + 4\left( {3 - 5x} \right) = 0\\ \Leftrightarrow 9x - 27 - 15{x^2} + 45x + 10{x^2} - 30x - 2x + 6 + 12 - 20x = 0\\ \Leftrightarrow - 5{x^2} + 2x - 9 = 0 \)

\(\Rightarrow\) Phương trình vô nghiệm.

\(5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2\)

=>\(5^x+5^x\cdot5+5^x\cdot25+5^x\cdot125=88\cdot\dfrac{\left(88+1\right)}{2}-16\)

=>\(156\cdot5^x=44\cdot89-16=3900\)

=>\(5^x=\dfrac{3900}{156}=25\)

=>x=2

   P = (5x − 1) + 2(1 − 5x)(4 + 5x) +  5 x + 4 2

      = 5x – 1 + (2 – 10x).( 4+ 5x) +  5 x + 4 2

      = 5x – 1 + 8 + 10x – 40x – 50 x 2  + 25 x 2  + 40x + 16

      = (- 50 x 2  + 25 x 2 )+ ( 5x + 10x – 40x + 40x) + (- 1+ 8 + 16)

      = -25 x 2  + 15x + 23

Lời giải:

Tại $x=4$ thì:

\(A=5(x^5-x^4+x^3-x^2+x-1)-1\)

\(=(x+1)(x^5-x^4+x^3-x^2+x-1)-1=x^6+1-1=x^6\)

\(=4^6=4096\)