a: =>x-3=9

=>x=12

b: =>10-x=-26

=>x=36

c: =>x:4-1=2

=>x:4=3

=>x=12

d: =>x^2=4

=>x=2 hoặc x=-2

e: =>(x-2)^2=100

=>x-2=10 hoặc x-2=-10

=>x=12 hoặc x=-8

A. x = 2

B. \(\dfrac{3}{8}=\dfrac{6}{x}\)\(\Leftrightarrow x=\dfrac{6.8}{3}=16\)

C. x = 3

D. \(x=\dfrac{4.6}{8}=3\)

E. \(x=\dfrac{7}{3}\)

G.\(\dfrac{14}{13}=\dfrac{28}{10-x}\)

<=>\(14\left(10-x\right)=364\)

<=> 10 - x = 26 

<=> x = -16 

H. \(3\left(x+2\right)=4\left(x-5\right)\)

<=> 3x + 6  = 4x - 20 

<=> -x = -26

<=> x = 26

K. \(\dfrac{x}{2}=\dfrac{8}{x}\)

<=> \(x^2=16\)

<=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

M. \(\left(x-2\right)^2=100\)

<=> \(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

a=2

b=16

c=3

d=3

mik chỉ biết thế này thôi(ko chắc đúng=3)

A, \(\dfrac{x}{5}=\dfrac{2}{5}\Rightarrow x=2\)

B,\(\dfrac{3}{8}=\dfrac{6}{x};\dfrac{3}{8}=\dfrac{3\cdot2}{8\cdot2}=\dfrac{6}{16}\Rightarrow x=16\)

C,\(\dfrac{1}{9}=\dfrac{x}{27};\dfrac{1}{9}=\dfrac{1\cdot3}{9\cdot3}=\dfrac{3}{27}\Rightarrow x=3\)

D,\(\dfrac{4}{x}=\dfrac{8}{6};\dfrac{4}{x}=\dfrac{4\cdot2}{x\cdot2}=\dfrac{8}{2x}\)

   \(\Rightarrow2x=6;6=x:2;x=3\)

\(a.13.\left(78+22\right)=13.100=1300\)

\(b.25.70+28.25+25.2=25.\left(70+28+2\right)=25.100=2500\)

\(c.\dfrac{15.3.14.2.19}{19.14.15}=6\)

\(d.\dfrac{132.70.60}{12.35.20}=\dfrac{11.12.35.2.20.3}{12.35.20}=66\)

Mọi người ơi, giúp em với ạ

a) \(\left(9^4.8+9^4.5\right):\left(9^2.\left(10-1\right)\right)\)

=\(9^4.13:9^3=13.9=117\)

b) 100-(75-25)=100-50=50

\(a,\left(2^3\cdot9^4+9^3\cdot45\right):\left(9^2\cdot10-9^2\right)\)

\(< =>\left(8\cdot6561+729\cdot45\right):\left(81\cdot10-81\right)\)

\(< =>\left(52488+32805\right):\left(810-81\right)\)

\(< =>85293:729\)

\(< =>117\)

\(b,100-\left[75-\left(7-2\right)^2\right]\)

\(< =>100-\left[75-\left(5\right)^2\right]\)

\(< =>100-\left[75-25\right]\)

\(< =>100-50\)

\(< =>50\)

\(c,\left(20\cdot2^4+12\cdot2^4\right)-\left(48\cdot2^2\right):8^2\)

\(< =>\left[2^4\cdot\left(20+12\right)\right]-\cdot\left(48\cdot4\right):8^2\)

\(< =>\left[2^4\cdot32\right]-192:8^2\)

\(< =>\left[16\cdot32\right]-192:64\)

\(< =>512-3\)

\(< =>509\)

a) 3x4-3x-2x+2=x+20   ​​​b) 10+13-(x²- 6x+9)=0

    12-5x=x+20                 -x²+ 6x+14=0... 

    6x=-8

    x=-3/2

Bài 2

\(a,\left(x-5\right)^4=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)

\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)

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

\(\Rightarrow\left(x-5\right)^4\left(x-4\right)\left(x-6\right)=0\)

\(\Rightarrow x\in\left\{4;5;6\right\}\)

\(b,\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\left[\left(2xs-15\right)^2-1\right]=0\)

\(\Rightarrow\left(2x-15\right)^3\left(2x-15+1\right)\left(2x-15-1\right)=0\)

\(\Rightarrow\left(2x-15\right)^3\left(2x-14\right)\left(2x-16\right)\)

\(\Rightarrow x\in\left\{\frac{15}{2};7;8\right\}\)

Mà \(\frac{15}{2}\notin n\)

\(\Rightarrow x\in\left\{7;8\right\}\)

#)Giải :

Bài 1 :

a)\(A=\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)

b)\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{11.3^{29}-3^{29}.3}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.2^3}{2^2.3^{28}}=6\)

Bài 2 : 

a) \(\left(x-5\right)^2=\left(x-5\right)^6\)

\(\Leftrightarrow x^4-625=x^6-15625\)

\(\Leftrightarrow x^6-x^4=15000\)

\(\Leftrightarrow x^6-x^4=5^6-5^4\)

\(\Leftrightarrow x=5\)

b)\(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Leftrightarrow2x-15=1\)

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=8\)

Bài 1 :

\(A=\frac{2^{13}+2^5}{2^{10}+2^2}\)\(=\frac{2^{10}.2^3+2^2.2^3}{2^{10}+2^2}\) \(=\frac{2^3.\left(2^{10}+2^2\right)}{2^{10}+2^2}=2^3=8\)

Vậy A = 8

\(a)\frac{-3}{x}=\frac{2^{10}}{2^{13}}\)

\(\Rightarrow\frac{-3}{x}=\frac{1}{2^3}=\frac{1}{8}\)

\(\Rightarrow-3\cdot8=x\)       \(\Rightarrow x=-24\)

\(b)\frac{7}{x}=\frac{y}{36}=\frac{-14}{28}\)

\(\Rightarrow\frac{7}{x}=\frac{y}{36}=-0,5\)

\(\Rightarrow x=7:\left(-0,5\right)=-14\)

     \(y=-0,5\cdot36=-18\)

Vậy x = -14; y = -18

\(c)\frac{10}{x}=\frac{15}{39}\)

\(\Rightarrow10\cdot39=15x\)             \(\Rightarrow390=15x\)

=> x = 390 : 15 = 26

Ah mk nhầm 1 chút!
x,y thuộc N nên phần a và b x và y không có giá trị thỏa mãn yêu cầu đề bài

Sorry!!!!

1. -28 + 32

2. 50 + 21

3. (-45) + (-30)

4. x + (-80)

5. 7 + (-a)

6. -25 + a