Đặt 4X - 19 =a; 4X -20 =b => 8X-39 = a + b

Từ đó ta có:
a^4 + b^4 = (a+b)^4 = a^4 + b^4 + 4a^3.b + 6a^2b^2 + 4ab^3
=> 4a^3.b + 6a^2.b^2 + 4a.b^3 = 0
ab(4a^2 + 6ab + 4b^2) =0
=> ab = 0 hoặc 4a^2 + 6ab +4b^2 = 0
TH1: ab = 0 -> 4x -19 =0 hoặc 4x-20 =0 => x =19/4 hoặc x = 20/4 =5
TH2: 4a^2 + 6ab +4b^2 = 0 => 2a^2 + 3ab +2b^2 = 0
Mà a - b = 1 ->a = 1+b
Thế vào ta có: 2(1+b)^2 + 3(1+b)+2b^2
= 2(1+2b+b^2) + 3b +3 + 2b^2
= 4b^2 + 7b +5
detal = 7*7 - 4*4*5 < 0 , phương trình vô nghiệm b

Vậy Phương trình ban đầu có 2 nghiệm X1 = 19/4, X2 =5

Đặt a=4x-19; b=4x-20

=>a+b=4x-19+4x-20=8x-39

=>\(\left(a+b\right)^4=\left(8x-39\right)^4=\left(39-8x\right)^4\)

Phương trình sẽ trở thành:

\(a^4+b^4=\left(a+b\right)^4\)

=>\(a^4+b^4=\left(a^2+2ab+b^2\right)^2\)

=>\(a^4+b^4=a^4+4a^2b^2+b^4+4a^3b+2a^2b^2+4ab^3\)

=>\(4a^3b+6a^2b^2+4ab^3=0\)

=>\(ab\left(4a^2+6ab+4b^2\right)=0\)

=>\(ab\left(a^2+\frac32ab+b^2\right)=0\)

=>\(ab\left(a^2+2\cdot a\cdot\frac34b+\frac{9}{16}b^2+\frac{7}{16}b^2\right)=0\)

=>\(ab\left\lbrack\left(a+\frac34b\right)^2+\frac{7}{16}b^2\right\rbrack=0\)

=>\(\left[\begin{array}{l}a=0\\ b=0\\ \left(a+\frac34b\right)^2+\frac{7}{16}b^2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}a=0\\ b=0\\ a=b=0\end{array}\right.\)

=>\(\left[\begin{array}{l}4x-19=0\\ 4x-20=0\\ 4x-19=4x-20\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{19}{4}\\ x=5\end{array}\right.\)

Giải phương trình:

\(\left(\right. 4 x - 19 \left.\right)^{4} + \left(\right. 4 x - 20 \left.\right)^{4} = \left(\right. 39 - 8 x \left.\right)^{4}\)

Đặt \(a = 4 x - 19\).
Khi đó:

• \(4 x - 20 = a - 1\)
• \(39 - 8 x = 1 - 2 a\)

Phương trình trở thành:

\(a^{4} + \left(\right. a - 1 \left.\right)^{4} = \left(\right. 1 - 2 a \left.\right)^{4}\)

Khai triển:

\(a^{4} + \left(\right. a^{4} - 4 a^{3} + 6 a^{2} - 4 a + 1 \left.\right) = 16 a^{4} - 32 a^{3} + 24 a^{2} - 8 a + 1\)

Rút gọn:

\(2 a^{4} - 4 a^{3} + 6 a^{2} - 4 a + 1 = 16 a^{4} - 32 a^{3} + 24 a^{2} - 8 a + 1\) \(\Rightarrow 14 a^{4} - 28 a^{3} + 18 a^{2} - 4 a = 0\) \(\Rightarrow a \left(\right. 7 a^{3} - 14 a^{2} + 9 a - 2 \left.\right) = 0\)

Thử nghiệm \(a = 1\) đúng, vậy:

\(7 a^{3} - 14 a^{2} + 9 a - 2 = \left(\right. a - 1 \left.\right) \left(\right. 7 a^{2} - 7 a + 2 \left.\right)\)

Phần còn lại vô nghiệm thực.

→ Nghiệm thực: \(a = 0\) hoặc \(a = 1\).

tick cho mik nhé ạ!

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow-2\sqrt{x-1}=-2\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

ĐKXĐ: \(x\ne0\)

Phương trình tương đương:

\(\dfrac{4}{4x-8+\dfrac{7}{x}}+\dfrac{3}{4x-10+\dfrac{7}{x}}=1\)

Đặt \(4x-10+\dfrac{7}{x}=t\)

\(\Rightarrow\dfrac{4}{t+2}+\dfrac{3}{t}=1\)

\(\Rightarrow4t+3\left(t+2\right)=t\left(t+2\right)\)

\(\Leftrightarrow t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x-10+\dfrac{7}{x}=-1\\4x-10+\dfrac{7}{x}=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x^2-9x+7=0\left(vn\right)\\4x^2-16x+7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

pt <=> x^4+8x-4x^3-5 = 0

<=> (x^4-x^3)-(3x^3-3x)+(5x-5) = 0

<=> x^3.(x-1)-3.x.(x-1).(x+1)+5.(x-1) = 0

<=> (x-1).(x^3-3x^2-3x+5) = 0

<=> (x-1).[(x^3-x^2)-(2x^2-2x)-(5x-5)] = 0

<=> (x-1)^2.(x^2-2x-5) = 0

<=> x-1=0 hoặc x^2-2x-5=0

<=> x=1 hoặc x = \(1+-\sqrt{6}\)

Vậy ...............

Tk mk nha

ĐK: x khác 1; - 1

\(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}.\)

<=> \(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}+\frac{12x-1}{4x-4}.\)

<=> \(\frac{6.4}{4\left(x^2-1\right)}+\frac{5\left(x^2-1\right)}{4\left(x^2-1\right)}=\frac{\left(8x-1\right)\left(x-1\right)}{4\left(x^2-1\right)}+\frac{\left(12x-1\right)\left(x+1\right)}{4\left(x^2-1\right)}.\)

<=> \(24+20x^2-20=8x^2-x-8x+1+12x^2-x+12x-1\)

<=> \(2x=4\)

<=> x = 2 thỏa mãn.

Bài 1: 

b: \(\Leftrightarrow x-2=0\)

hay x=2

anh ơi, vậy là sai đề hả anh, chứ đề kêu chứng minh phương trình vô nghiệm mà em thấy anh ghi x=2

\(\frac{x^2-2x+2}{x-1}+\frac{x^2-8x+20}{x-4}=\frac{x^2-4x+6}{x-2}+\frac{x^2-6x+12}{x-3}\)\(ĐKXĐ:x\ne1;2;3;4\)

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

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

\(\Leftrightarrow x-1+\frac{1}{x-1}+x-4+\frac{1}{x-4}=x-2+\frac{1}{x-2}+x-3+\frac{1}{x-3}\)

\(\Leftrightarrow\frac{1}{x-1}+\frac{4}{x-4}=\frac{2}{x-2}+\frac{3}{x-3}\)

\(\Leftrightarrow\frac{x-4+4x-4}{\left(x-1\right)\left(x-4\right)}=\frac{2x-6+3x-6}{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{5x-8}{x^2-5x+4}=\frac{5x-12}{x^2-5x+6}\)

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

Tự giải ra rồi tìm x nhé

dấu suy ra số 4 là 1/(x+1) + 1/(x+4) mà.