Question

[Minh Ngô]

(5x+2,5)^4-(5x-1,5)^4=80 Giải PT

Answer 1

[𝐋𝐘𝐋𝐘]

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.

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Đáp số:

\(x = 0 , 1\)

tick cho em nha

Answer 2

[Nguyễn Lê Phước Thịnh]

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

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