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

\(=\left(2-x\right)^3\)

\(125x^3-75x^2+15x-1\)

\(=\left(5x-1\right)^3\)

\(x^2-xz-9y^2+3yz\)

\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x+3y-z\right)\)

\(x^3-x^2-5x+125\)

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

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

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

\(x^3+2x^2-6x-27\)

\(=x^3+5x^2+9x-3x^2-15x-27\)

\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

\(12x^3+4x^2-27x-9\)

\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(4x^2-9\right)\)

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

\(4x^4+4x^3-x^2-x\)

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

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

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

a

\(5\frac{4}{7}:x+=13\)

\(\frac{39}{7}:x=13\)

\(x=\frac{39}{7}:13\)

\(x=\frac{3}{7}\)

\(\frac{4}{7}x=\frac{9}{8}-0,125\)

\(\frac{4}{7}x=1\)

\(x=1:\frac{4}{7}\)

\(x=\frac{7}{4}=1\frac{3}{4}\)

b1 dễ quá 

vậy thì bạn giải đi

\(a.\)

\(x+5=-10\)

\(\Rightarrow x=-10-5=-15\)

a ) x +5 = -10

x = -10 -5

x = - 15

b) x - ( - 10 ) = 5

x = 5+(-10)

x = -5

c) \(\left|x\right|\) -5 = 3

\(\left|x\right|=8\)

x ϵ { -8 ; 8 }

d) 15 - ( - x ) = 20

Không có số tự nhiên x nào mà 15 ( - x ) = 20

e ) \(\left|x-4\right|=3-\left(-7\right)\\ \left|x-4\right|=10\\ \left|x\right|=14\\ x\in\left\{\pm14\right\}\)

f ) \(\left|x+5\right|=10-\left(-20\right)\\ \left|x+5\right|=30\\ \left|x\right|=25\\ x\in\left\{\pm25\right\}\)

Bài 2:

a) \(\left(x-3\right)^3+27=0\)

\(\Leftrightarrow\left(x-3\right)^3=0-27\)

\(\Leftrightarrow\left(x-3\right)^3=-27\)

\(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\)

\(\Leftrightarrow x-3=-3\)

\(\Leftrightarrow x=\left(-3\right)+3\)

\(\Leftrightarrow x=0\)

b) \(-125-\left(x+1\right)^3=0\)

\(\Leftrightarrow\left(x+1\right)^3=-125-0\)

\(\Leftrightarrow\left(x+1\right)^3=-125\)

\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)

\(\Leftrightarrow x+1=-5\)

\(\Leftrightarrow x=\left(-5\right)-1\)

\(\Leftrightarrow x=-6\)

c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)

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

\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)

\(\Leftrightarrow2x=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{2}:2\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

d) \(2^x+2^{x+1}=24\)

\(\Leftrightarrow2^x+2^x.2=24\)

\(\Leftrightarrow2^x\left(1+2\right)=24\)

\(\Leftrightarrow2^x.3=24\)

\(\Leftrightarrow2^x=24:3\)

\(\Leftrightarrow2^x=8\)

\(\Leftrightarrow2^x=2^3\)

\(\Rightarrow x=3\)

e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)

\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)

g) \(\left|x-3\right|+2x=10\)

\(\Leftrightarrow\left|x-3\right|=10-2x\)

\(\Leftrightarrow\left|x-3\right|=2.5-2x\)

\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)

(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)

Bài 1:

a) \(2^7+2^9⋮10\)

Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)

\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)

\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)

\(\Leftrightarrow\overline{C8}+\overline{D2}\)

\(\Leftrightarrow\overline{E0}\)

Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)

b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)

Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)

\(\Leftrightarrow2^{72}.5^{20}\)

Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)

c) \(3^{10}+3^{12}⋮30\)

Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)

\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)

\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)

\(\Leftrightarrow\overline{C9}+\overline{B1}\)

\(\Leftrightarrow\overline{D0}⋮10\)

(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)

\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)

=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)

=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)

Vậy \(x\in\left\{\frac{9}{20}\right\}\)

\(b,x+\frac{1}{4}=\frac{4}{3}\)

=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)

Vậy \(x\in\left\{\frac{13}{12}\right\}\)

\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)

=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)

Vậy \(x\in\left\{\frac{25}{42}\right\}\)

\(d,\left|x+5\right|-6=9\)

=> \(\left|x+5\right|=9+6=15\)

=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)

Vậy \(x\in\left\{10;-20\right\}\)

\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)

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

=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)

\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{6}\)

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

Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)

\(g,x^2=16\)

=> \(\left|x\right|=\sqrt{16}=4\)

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

vậy \(x\in\left\{4;-4\right\}\)

\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)

=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)

Vậy \(x\in\left\{\frac{5}{6}\right\}\)

\(i,3^3.x=3^6\)

\(x=3^6:3^3=3^3=27\)

Vậy \(x\in\left\{27\right\}\)

\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)

=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)

Vậy \(x\in\left\{\frac{5}{27}\right\}\)

\(k,1\frac{2}{3}:x=6:0,3\)

=> \(\frac{5}{3}:x=20\)

=> \(x=\frac{5}{3}:20=\frac{1}{12}\)

Vậy \(x\in\left\{\frac{1}{12}\right\}\)

a: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5

=>x=3/5:2/3=3/5x3/2=9/10

b: \(\Leftrightarrow x\cdot2.8-50=34\)

=>2,8x=84

=>x=30

c: \(\Leftrightarrow\dfrac{1}{6}x=\dfrac{5}{12}\)

hay x=5/2

d: \(\Leftrightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}+\dfrac{7}{4}=\dfrac{41}{4}\)

=>2x-3/4=41/4 hoặc 2x-3/4=-41/4

=>2x=44/4=11 hoặc 2x=-19/2

=>x=11/2 hoặc x=-19/4

Bạn đăng ít một thôi!

mk lỡ đăng rồi bạn ạ 

Bài 2: a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-3\right).7=\left(x+5\right).5\)

\(\Leftrightarrow7x-21=5x+25\)

\(\Leftrightarrow7x-5x=21+25\)

\(\Leftrightarrow2x=46\)

\(\Rightarrow x=46:2=23\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=63\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

\(\Rightarrow x^2=\left(\pm8\right)^2\)

\(\Rightarrow x=8\) hoặc \(x=-8\)

2)a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)

\(7x-21=5x+25\)

\(7x-5x+25=21\)

\(2x+25=21\)

\(2x=-4\Rightarrow x=-2\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(7.9=\left(x+1\right)\left(x-1\right)\)

\(63=x\left(x-1\right)+1\left(x-1\right)\)

\(63=x^2-x+x-1\)

\(x^2=63+1=64\)

\(x=\left\{\pm8\right\}\)

c) \(\dfrac{x+4}{20}=\dfrac{2}{x+4}\)

\(\Leftrightarrow\left(x+4\right)\left(x+4\right)=2.20=40\)

\(x\left(x+4\right)+4\left(x+4\right)=40\)

\(x^2+4x+4x+16=40\)

\(x^2+8x=40-16=24\)

\(x\left(x+8\right)=24\)

\(x\in\left\{\varnothing\right\}\)

d) \(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)

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

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

\(x^2-2x+2x-4=x^2+3x-x-3\)

\(\)\(x^2-4=x^2+2x-3\)

\(\Leftrightarrow x^2-x^2-2x+3=4\)

\(-2x+3=4\)

\(-2x=1\)

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