Đặt \(a=24-x,b=x-25\)

Khi đó pt ban đầu trở thành :

\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\)

\(\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)

\(\Leftrightarrow30a^2+68ab+30b^2=0\)

\(\Leftrightarrow15a^2+34ab+15b^2=0\)

\(\Leftrightarrow\left(3a+5b\right)\left(5a+3b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3a=-5b\\5a=-3b\end{cases}}\)

Đến đây bạn thay vào là dễ rồi nhé ! Chúc bạn học tốt !

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=24\)

\(\Rightarrow\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)=24\)

\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)=24\)

Đặt \(x^2-5x+5=a,\)ta có

\(\left(a-1\right)\left(a+1\right)=24\Rightarrow a^2=25\Rightarrow a=\pm5\)

Theo cánh đặt,ta có

+,\(x^2-5x+5=5\Rightarrow x\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

+\(x^2-5x+5=-5\Rightarrow x^2-2\cdot\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\)

\(\Rightarrow\left(x-\frac{5}{2}\right)^2+\frac{15}{4}=0\)(vô lí)

Vậy 

Bài 1:

\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2\cdot50=100\)

\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2\cdot52=104\)

=>A<B

Bài 2:

\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

=>\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)

=>\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)

=>4x+13=11

=>4x=-2

=>\(x=-\dfrac{1}{2}\)

a/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

a) (x-2)³+6(x+1)²-x³+12=0

⇒ x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

⇒ x³-6x²+12x-8+6x²+12x+6-x³+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

a.

PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)

b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.

c.

PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$

$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý) 

Do đó pt vô nghiệm.

d.

PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$

$\Leftrightarrow 11x+3=-54x-97$

$\Leftrightarrow 65x=-100$

$\Leftrightarrow x=\frac{-20}{13}$

\(=>2^{x-1}-1=24-9\)
\(2^{x-1}-1=15\)
\(2^{x-1}=16\)
\(=>x-1=4\)
\(x=5\)

\(2^{x-1}-1=24-\left[3^2-\left(2021^0-1\right)\right]\\ 2^{x-1}-1=24-\left[9-\left(1-1\right)\right]\\ 2^{x-1}-1=24-\left[9-0\right]\\ 2^{x-1}-1=24-9\\ 2^{x-1}-1=15\\ 2^{x-1}=15+1\\ 2^{x-1}=16\\ 2^{x-1}=2^4\\ x-1=4\\ x=4+1\\ x=5\)

`2^(x-1) -1 = 24 - [3^2 - (2021^0 -1)]`

`=> 2^(x-1) -1 = 24 - [ 9 - (1-1)]`

`=> 2^(x-1) -1 = 24 - 9`

`=> 2^(x-1) -1 = 15`

`=> 2^(x-1) =15+1`

`=> 2^(x-1) = 16`

`=> 2^(x-1) = 2^4`

`=> x-1=4`

`=> x=4+1`

`=> x=5`

a ) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=24\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=24\)

\(\Leftrightarrow2x=-231\Leftrightarrow x=\dfrac{-231}{2}\)

b ) \(\left(x+3\right)^2-\left(x-4\right)\left(x-8\right)=1\)

\(\Leftrightarrow x^2+6x+9-x^2+12x-32=1\)

\(\Leftrightarrow18x=24\Leftrightarrow x=\dfrac{4}{3}\)

Chúc bạn học tốt !!!!!!!!!!!!

b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)

\(\Leftrightarrow x^2+7x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)