Ta có: \(\left(5x+2,5\right)^4-\left(5x-1,5\right)^4=80\)

=>\(\left\lbrack\left(5x+2,5\right)^2-\left(5x-1,5\right)^2\right\rbrack\left\lbrack\left(5x+2,5\right)^2+\left(5x-1,5\right)^2\right\rbrack=80\)

=>\(\left(5x+2,5-5x+1,5\right)\left(5x+2,5+5x-1,5\right)\left\lbrack25x_{}^2+25x+6,25+25x^2-15x+2,25\right\rbrack=80\)

=>\(4\cdot\left(10x+1\right)\left(50x^2+10x+8,5\right)=80\)

=>\(\left(10x+1\right)\left(50x^2+10x+8,5\right)=20\)

=>\(500x^3+100x^2+85x+50x^2+10x+8,5=20\)

=>\(500x^3+150x^2+95x-11,5=0\)

=>\(500x^3-50x^2+200x^2-20x+115x-11,5=0\)

=>\(50x^2\left(10x-1\right)+20x\left(10x-1\right)+11,5\left(10x-1\right)=0\)

=>\(\left(10x-1\right)\left(50x^2+20x+11,5\right)=0\)

mà \(50x^2+20x+11,5=50\left(x^2+\frac25x+\frac{23}{100}\right)=5\left(x^2+\frac25x+\frac{1}{25}+\frac{19}{100}\right)=5\left(x+\frac15\right)^2+\frac{19}{20}\ge\frac{19}{20}>0\forall x\)

nên 10x-1=0

=>10x=1

=>\(x=\frac{1}{10}\)

Giải phương trình:

\(\left(\right. 5 x + 2 , 5 \left.\right)^{4} - \left(\right. 5 x - 1 , 5 \left.\right)^{4} = 80\)

Đặt \(A = 5 x + 2 , 5 , \textrm{ }\textrm{ } B = 5 x - 1 , 5\).

Khi đó:

\(A^{4} - B^{4} = \left(\right. A - B \left.\right) \left(\right. A + B \left.\right) \left(\right. A^{2} + B^{2} \left.\right)\)

Ta có:

\(A - B = 4 , A + B = 10 x + 1\) \(A^{2} + B^{2} = \left(\right. 5 x + 2 , 5 \left.\right)^{2} + \left(\right. 5 x - 1 , 5 \left.\right)^{2} = 50 x^{2} + 10 x + 8 , 5\)

Vậy phương trình trở thành:

\(4 \left(\right. 10 x + 1 \left.\right) \left(\right. 50 x^{2} + 10 x + 8 , 5 \left.\right) = 80\) \(\left(\right. 10 x + 1 \left.\right) \left(\right. 50 x^{2} + 10 x + 8 , 5 \left.\right) = 20\)

Khai triển:

\(500 x^{3} + 150 x^{2} + 95 x + 8 , 5 = 20\) \(500 x^{3} + 150 x^{2} + 95 x - 11 , 5 = 0\)

Nhân cả phương trình với 2:

\(1000 x^{3} + 300 x^{2} + 190 x - 23 = 0\)

Thử nghiệm \(x = 0 , 1\):

\(1000 \left(\right. 0 , 1 \left.\right)^{3} + 300 \left(\right. 0 , 1 \left.\right)^{2} + 190 \left(\right. 0 , 1 \left.\right) - 23 = 0\)

→ \(x = 0 , 1\) là nghiệm.

Chia bậc ba cho \(\left(\right. x - 0 , 1 \left.\right)\), ta được:

\(1000 x^{2} + 400 x + 230 = 0\)

\(\Delta < 0\) nên vô nghiệm thực.

Đáp số:

\(x = 0 , 1\)

tick cho em nha

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

\(\Leftrightarrow x^2-4x+4-5x+4=0\)

\(\Leftrightarrow x^2-9x+8=0\)

\(\Leftrightarrow x^2-8x-x+8=0\)

\(\Leftrightarrow\left(x^2-x\right)-\left(8x-8\right)=0\)

\(\Leftrightarrow x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

Vậy ..........................

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

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

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

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

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)

a) ta có : \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)

\(\Leftrightarrow36x^2-12x-36x^2+27x=30\Leftrightarrow15x=30\Leftrightarrow x=2\)

b) điều kiện : \(x\ne\dfrac{1}{5};x\ne1;x\ne\dfrac{3}{5}\)

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

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

\(\Leftrightarrow\dfrac{x+7}{3-3x}=\dfrac{4}{3-5x}\Leftrightarrow\left(x+7\right)\left(3-5x\right)=4\left(3-3x\right)\)

\(\Leftrightarrow-5x^2-20+9=0\)

ta có : \(\Delta'=\left(10\right)^2+5\left(9\right)=145>0\) \(\Rightarrow\) phương trình có 2 nghiệm phân biệt

\(x=\dfrac{10+\sqrt{145}}{-5};x=\dfrac{10-\sqrt{145}}{-5}\)

\(2x^4+5x^2-7=0\left(1\right)\)

Đặt \(t=x^2\left(t\ge0\right)\)

\(\left(1\right):2t^2+5t-7=0\\ \Leftrightarrow2t^2+7t-2t-7=0\\ \Leftrightarrow t\left(2t+7\right)-\left(2t+7\right)=0\\ \Leftrightarrow\left(2t+7\right)\left(t-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2t+7=0\\t-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{7}{2}\left(KTM\right)\\t=1\left(TM\right)\end{matrix}\right.\)

Với \(t=1\Leftrightarrow x^2=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy nghiệm phương trình là \(S=\left\{1;-1\right\}\)

\(4\left(5x-3\right)-3\left(2x+1\right)=9\)

\(\Leftrightarrow20x-12-6x-3=9\)

\(\Leftrightarrow14x-15=9\)

\(\Leftrightarrow14x=24\)

\(\Leftrightarrow x=\dfrac{12}{7}\)

Vậy phương trình có nghiệm là: \(x=\dfrac{12}{7}\)

cac bn giúp mk vs

\(5x-4\left(6x+18-x^2-3x\right)=\left(12-8x-6x+4x^2\right)+2\)

\(\Leftrightarrow5x-4\left(-x^2+3x+18\right)=\left(4x^2-14x+12\right)+2\)

\(\Leftrightarrow4x^2-7x-72=4x^2-14x+14\Leftrightarrow7x=86\Leftrightarrow x=\dfrac{86}{7}\)