\(i)4^8:x=4^6\\ x=4^8:4^6\\ x=4^2\\ k)12x-33=3^5\\ 12x-33=243\\ 12x=243+3\\ 12x=276\\ x=276:12\\ x=23\\ l)\left(5x+335\right):2=20^2\\ \left(5x+335\right):2=400\\ 5x+335=400.2\\ 5x+335=800\\ 5x=800-335\\ 5x=465\\ x=465:5\\ x=93\) 

\(m)\left(x^2-10\right):5=3\\ x^2-10=3.5\\ x^2-10=15\\ x^2=15+10\\ x^2=25\\ x^2=5^2\\ 740:\left(x+10\right)=10^2-2.13\\ 740:\left(x+10\right)=100-26\\ 740:\left(x+10\right)=74\\ x+10=740:74\\ x+10=10\\ x=10-10\\ x=0.\)

i) 4⁸:x=4⁶

<=> x = 4⁸ : 4⁶ = 4² = 16

k)12x-33=3⁵

<=> x = (3⁵ + 33) : 12 = 23

l)(5x+335):2=20²

<=> x = (20² x 2 - 335) : 5 = 93

m)(x²-10):5=3

<=> x² = 3 x 5 + 10 = 25

<=> x = 5 hoặc x = -5

740:(x+10)=10²-2.13

<=> x = 740 : (10² - 2.13) - 10 = 0

a,4^8:X=4^6

X=4^8:4^6

X=4^2

b,12x-33=3^5

12x-33=243

12x=243+33

12x=276

12*X=276

X= 276:12

x=23

c,(5x+335):2=20^2

(5x+335):2=400

(5x+335)=400*2

(5x+335)=800

5*x+335=800

5*x=800-335

5*x=465

x=465:5

x=93

d, (x^2-10):5=3

x^2-10=3*5

x^2-10=15

x^2=15+10

x^2=25

x^2=5^2

vậy x=5

e,740:(x+10)=10^2 - 2*13

740:(x+10)=10^2-26

740:(x+10)=100-26

740:(x+10)=74

x+10=740:74

x+10=10

x=10-10

x=0

nhớ tik cho mik nhé

\(\left(x^2-5x+8\right)^2-\left(5x-17\right)^2=0\)

\(\Leftrightarrow\left(x^2-5x+8-5x+17\right)\left(x^2-5x+8+5x-17\right)=0\)

\(\Leftrightarrow\left(x^2-10x+25\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left(x^2-5x-5x+25\right)\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[x\left(x-5\right)-5\left(x-5\right)\right]\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-5\right)^2.\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^2=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\\x=-3\end{matrix}\right.\)

\(A=\dfrac{5x_1-x_2}{x_1}+\dfrac{5x_2-x_1}{x_2}\)

\(=\dfrac{5x_1\cdot x_2-x_2^2+5x_1x_2-x_1^2}{x_1x_2}\)

\(=\dfrac{10x_1x_2-\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}{x_1x_2}\)

\(=\dfrac{10\cdot4-\left[5^2-2\cdot4\right]}{4}=\dfrac{40-25+8}{4}=\dfrac{23}{4}\)

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a)  \(x\left(x+4\right)\left(x-4\right)-\left(x^2-1\right)\left(x^2+1\right)\)

\(=x\left(x^2-16\right)-\left(x^4-1\right)\)

\(=x^3-16x-x^4+1\)

bạn ktra lại đề

b)  \(x^4+2x^3+5x^2+4x-12\)

\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)

\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)

\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

Ủa pạn có thể giải ại cái bước thứ 2 đc ko ạk

\(110-3\times\left(8+x\right)=1\)

\(3\times\left(8+x\right)=110-1\)

\(3\times\left(8+x\right)=109\)

\(8+x=\dfrac{109}{3}\)

\(x=\dfrac{109}{3}-8\)

\(x=\dfrac{85}{3}\)

Chúc bạn học tốt

$110-3\times \left(8+x\right)=1$

$3\times \left(8+x\right)=110-1$

$3\times \left(8+x\right)=109$

8+�=10938+x=3109​

�=1093−8x=3109​−8

�=853x=385​
 

\(\dfrac{1}{x}\)+  2\(\sqrt{x-8}\)

ĐK: \(x\) ≠ 0; \(x\) - 8 ≥ 0; ⇒ \(x\) ≥ 8 vậy \(x\) ≥ 8

2:

a: =>(x-9)(x-1)=0

=>x=9 hoặc x=1

b: =>(x+4)(x^2-4x+16)+(x+4)(x-16)=0

=>(x+4)(x^2-4x+16+x-16)=0

=>(x+4)(x^2-3x)=0

=>x(x-3)(x+4)=0

=>x=0;x=3;x=-4

 bài 2 :

a: =>(x-9)(x-1)=0

=>x=9 hoặc x=1

b: =>(x+4)(x^2-4x+16)+(x+4)(x-16)=0

=>(x+4)(x^2-4x+16+x-16)=0

=>(x+4)(x^2-3x)=0

=>x(x-3)(x+4)=0

=>x=0;x=3;x=-4