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

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.

Ai giúp mình với. Mình tích cho ạ

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

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

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

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

=>x=2

Lời giải:

$5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2$

$5^x(1+5+5^2+5^3)=88.89:2-16$

$5^x.156=3900$

$5^x=3900:156=25=5^2$

$\Rightarrow x=2$

Lời giải:

Tại $x=4$ thì:

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

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

\(=4^6=4096\)

Hình như bạn ghi sai đề. Đề đúng: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x+10=30\)

\(\Leftrightarrow10x=20\)

\(\Leftrightarrow x=2\)

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

\(\Rightarrow9-15x+10x-2=4\)

\(\Leftrightarrow3-5x=0\)

\(\Leftrightarrow x=\frac{3}{5}\)

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

⇒9 − 15x + 10x − 2 = 4

⇔3 − 5x = 0

⇔x =\(\frac{3}{5}\)