a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)

\(\Leftrightarrow49-21x+60x+24=84x+1092\)

\(\Leftrightarrow39x-84x=1092-73\)

=>-45x=1019

hay x=-1019/45

b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)

=>21x+63-14=20x+36-49x+63

=>21x+49=-29x+99

=>50x=50

hay x=1

c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)

=>14x+7-15x-6-21x-63=0

=>-22x-64=0

hay x=-32/11

d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)

=>70x-105-30x-45=84x+63-1785

=>40x-150-84x+1722=0

=>-44x+1572=0

hay x=393/11

a, msc 12.7=84 

Chuyển vế về =0 rồi làm

b,msc 28

c,làm tương tự

a, \(\Rightarrow49-21x+60x+24=84x+1092\)

\(\Leftrightarrow-45x=1019\Leftrightarrow x=-\dfrac{1019}{45}\)

b, \(\Rightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)

\(\Leftrightarrow21x+63-14=20x+36-49x+63\)

\(\Leftrightarrow50x=50\Leftrightarrow x=1\)

c, \(\Rightarrow14x+7-15x-6=21x+63\Leftrightarrow-22x=62\Leftrightarrow x=-\dfrac{31}{11}\)

d, \(\Rightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-105.17\)

\(\Leftrightarrow70x-105-30x-45=84x+63-1785\)

\(\Leftrightarrow-44x=-1572\Leftrightarrow x=\dfrac{393}{11}\)

a)A=\(x^5-\dfrac{1}{2}x+7x^3-2x+\dfrac{1}{5}x^3+3x^4-x^5+\dfrac{2}{5}x^4+15\)

=\(=\dfrac{-5}{2}x+\dfrac{36}{5}x^3+\dfrac{17}{5}x^4+15\)

b)B=\(3x^2-10+\dfrac{2}{5}x^3+7x-x^2+8+7x^2\)

\(=9x^2+\dfrac{2}{5}x^3+7x+2\)

c)C=\(\dfrac{1}{7}x-2x^4+5x+6\)

c)C=\(\dfrac{36}{7}x-2x^4+6\)

\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)

\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)

\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)

\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)

Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:

\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)

Tick plzz

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

\(\Leftrightarrow x^2+x-6-12x+6-x^2+8x-16=0\)

\(\Leftrightarrow-3x=16\)

hay \(x=-\dfrac{16}{3}\)

b: Ta có: \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)

\(\Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\)

\(\Leftrightarrow-14x+7+4x-6=0\)

\(\Leftrightarrow10x=1\)

hay \(x=\dfrac{1}{10}\)

c: Ta có: \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)

\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)

\(\Leftrightarrow23x+70=10x+200\)

\(\Leftrightarrow x=10\)

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

\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)

\(=\dfrac{1}{2}x^3+x^2-15x-18\)

2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)

\(=4x^3+6x^2-6x^2-9x+10x+15\)

\(=4x^3+x+15\)

3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)

\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)

\(=3x^5-x^4+5x^3+10x^2+26x-5\)

4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)

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

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

\(1\text{)}\left(x-5\right)^2+3\left(x-5\right)=0\)

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

\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

\(2\text{)}\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)

\(\Leftrightarrow\dfrac{7\left(2x-1\right)-3\left(5x+2\right)}{21}=\dfrac{21\left(x+13\right)}{21}\)

\(\Leftrightarrow14x-7-15x-6=21x+273\)

\(\Leftrightarrow-x-13=21x+273\)

\(\Leftrightarrow-22x=286\)

\(\Rightarrow x=-\dfrac{286}{22}=-\dfrac{143}{11}\)

\(3\text{)}\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{7x-6}{4-x^2}\left(đk:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{x-1}{2+x}+\dfrac{x}{2-x}=\dfrac{7x-6}{4-x^2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(2-x\right)+x\left(2+x\right)}{4-x^2}=\dfrac{7x-6}{4-x^2}\)

\(\Leftrightarrow2x-x^2-2+x+2x+x^2=7x-6\)

\(\Leftrightarrow x-2=7x-6\)

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

\(\Rightarrow x=\dfrac{2}{3}\)

1.(x−5)²+3(x−5)=0

=>(x-5)(x-5)+3.(x-5)=0

=>(x-5).(x-5+3)=0

=>x-5=0 hoặc x-2=0

=>x=5 hoặc x=2

2)\(\dfrac{2x-1}{3}\)-\(\dfrac{5x+2}{7}\)=x+13

=>\(\dfrac{7.\left(2x-1\right)}{7.3}\)-\(\dfrac{3.\left(5x+2\right)}{3.7}\)=\(\dfrac{21.\left(x+13\right)}{21}\)

=>khử mẫu:

=>7.(2x-1)-3.(5x+2)=21.(x+13)

=>14x-7-15x-6=21x+273

=>14-7-15x-6-21x-273=0

=>-36x-272=0

=>-36x=272

=>x=...

a: =>10x-4=15-9x

=>19x=19

hay x=1

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>30x-32x=60-9

=>-2x=51

hay x=-51/2

c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)

=>3x=6/5

hay x=2/5

d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)

\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)

=>21x-120x+1080=80x+60

=>-179x=-1020

hay x=1020/179

e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

=>35x-5+60x=96-6x

=>95x+6x=96+5

=>x=1

f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)

=>6x+24-30x+120=10x-15x+30

=>-24x+96=-5x+30

=>-19x=-66

hay x=66/19

a: \(2^{x^2-2x+1}=1\)

=>\(2^{\left(x-1\right)^2}=2^0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1

b: \(7^{x^2+7x}=5764801\)

=>\(7^{x^2+7x}=7^8\)

=>\(x^2+7x=8\)

=>\(x^2+7x-8=0\)

=>(x+8)(x-1)=0

=>\(\left[{}\begin{matrix}x+8=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\)

c: \(6^{x^2+12x}=6^{7x}\)

=>\(x^2+12x=7x\)

=>\(x^2+5x=0\)

=>x(x+5)=0

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

d: \(\left(\dfrac{1}{3}\right)^{x-1}=3^{2x-5}\)

=>\(3^{-x+1}=3^{2x-5}\)

=>-x+1=2x-5

=>-x-2x=-5-1

=>-3x=-6

=>x=2

e: \(\left(\dfrac{1}{5}\right)^{3x+5}=5^{2x+1}\)

=>\(5^{-3x-5}=5^{2x+1}\)

=>-3x-5=2x+1

=>-5x=6

=>\(x=-\dfrac{6}{5}\)

Bài 3 :

\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)

\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)

\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)

\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)

.....

\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)

\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)

Bạn xem lại đề 2, phần mẫu của N

@Nguyễn Đức Trí: Đề bài nó như vậy mà