Đặt a=4x-19; b=4x-20

=>a^4+b^4=(a+b)^4

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

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

=>(4x-19)(4x-20)=0

=>x=19/4 hoặc x=20/4=5

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

\(x^4-4x^2+8x+4=0\)

\(\Leftrightarrow x^2\left(x^2-4\right)+8\left(x+2\right)=0\)

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

\(\Leftrightarrow\left(x+2\right)\left(x^3-2x^2+8\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x^3-2x^2+8=0\end{array}\right.\)

Tới đây tự giải nhé :)

Đầu tiên ta phân tích : \(x^4+4=\left(x^4+4x^2+4\right)-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2\)

\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

Suy ra pt : \(\left(x^2-2x+2\right)\left(x^2+2x+2\right)-4x\left(x-2\right)=0\)

Nhận thấy x = 0 không là nghiệm của pt, do đó chia cả hai vế của pt cho \(x^4\ne0\) được : 

\(\left(1-\frac{2}{x}+\frac{2}{x^2}\right)\left(1+\frac{2}{x}+\frac{2}{x^2}\right)-4\left(\frac{1}{x^2}-\frac{2}{x^3}\right)=0\)

Đặt \(t=\frac{2}{x}\) , pt trở thành : \(\left(1-2t+2t^2\right)\left(1+2t+2t^2\right)-4\left(t^2-2t^3\right)=0\)

Tới đây thử giải pt với ẩn t xem có đc k

\(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)

\(\Leftrightarrow\dfrac{8x^2}{3\left(1-2x\right)\left(1+2x\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{1+8x}{4\left(1+2x\right)}\)

\(\Leftrightarrow\dfrac{-32x^2}{12\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x.4\left(1+2x\right)-\left(1+8x\right).3\left(2x-1\right)}{12\left(2x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow8x\left(1+2x\right)-\left(1+8x\right).3.\left(2x-1\right)=-32x^2\)

\(\Leftrightarrow8x+16x^2-6x+3-48x^2+24x+32x^2=0\)

\(\Leftrightarrow26x+3=0\)

\(\Leftrightarrow x=-\dfrac{3}{26}\)

Vậy:......

5x-16=40+x

=> 5x-16-x = 40

=> 5x-x -16=40

4x-16=40

4x= 40+16

4x=56

x= 56:4

x=14

Vậy...

4x-10=15-x

=> 4x-10+x= 15

4x+x -10=15

5x= 15+10

5x= 25

x= 25:5

x=5

Vậy....

\(1.\sqrt{16-8x+x^2}=4-x\)

\(\sqrt{\left(4-x\right)^2}=4-x\)

\(4-x-4+x=0\)

= 0 phương trình vô nghiệm.

\(2.\sqrt{4x^2-12x+9}=2x-3\)

\(\)\(\sqrt{\left(2x-3\right)^2}=2x-3\)

\(2x-3-2x+3=0\)

= 0 phương trình vô nghiệm.

a: Ta có: \(\sqrt{16-8x+x^2}=4-x\)

\(\Leftrightarrow\left|4-x\right|=4-x\)

hay \(x\le4\)

b: Ta có: \(\sqrt{4x^2-12x+9}=2x-3\)

\(\Leftrightarrow\left|2x-3\right|=2x-3\)

hay \(x\ge\dfrac{3}{2}\)

(4x - 3)² - (2x + 1)² = 0

\(\Leftrightarrow\) (4x - 3 - 2x - 1)(4x - 3 + 2x + 1) = 0

\(\Leftrightarrow\) (2x - 4)(6x - 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy ...

3x - 12 - 5x(x - 4) = 0

\(\Leftrightarrow\) 3x - 12 - 5x² + 20x = 0

\(\Leftrightarrow\) -5x² + 23x - 12 = 0

\(\Leftrightarrow\) 5x² - 23x + 12 = 0

\(\Leftrightarrow\) 5x² - 20x - 3x + 12 = 0

\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0

\(\Leftrightarrow\) (x - 4)(5x - 3) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-4=0\\5x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy ...

(8x + 2)(x² + 5)(x² - 4) = 0

\(\Leftrightarrow\) (8x + 2)(x² + 5)(x - 2)(x + 2) = 0

Vì x² \(\ge\) 0 \(\forall\) x nên x² + 5 > 0 \(\forall\) x

\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}8x+2=0\\x-2=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=2\\x=-2\end{matrix}\right.\)

Vậy ...

Chúc bn học tốt!

a) Ta có: \(\left(4x-3\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left(4x-3-2x-1\right)\left(4x-3+2x+1\right)=0\)

\(\Leftrightarrow\left(2x-4\right)\left(6x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)

b) Ta có: \(3x-12-5x\left(x-4\right)=0\)

\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)

c) Ta có: \(\left(8x+2\right)\left(x^2+5\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)

mà \(2>0\)

và \(x^2+5>0\forall x\)

nên \(\left(4x+1\right)\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-1\\x=2\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=2\\x=-2\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)