[ 20 x 2⁴ + 12 x 2⁴ - 48 x 2² ] : 8²

= [ 20 x 16 + 12 x 16 - 48 x 4 ] : 64

= [ 320 + 192 - 192 ] : 64

= [ 512 - 192 ] : 64

= 320 : 64

= 5

1000 mình làm violympic roi

ko hiểu

\(A=2+2^2+2^3+...+2^{10}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)

\(A=2.3+2^3.3+...+2^9.3\)

\(A=3\left(2+2^3+...+2^9\right)\)

Ta có : \(3⋮3\Rightarrow A=3\left(2+2^3+...+2^9\right)⋮3\)

không

a) \(x-2=-6\)

\(x=-6+2\)

\(x=-4\)

b) \(15-\left(x-7\right)=-21\)

\(x-7=36\)

\(x=43\)

c) \(4.\left(3x-4\right)-2=18\)

\(4\left(3x-4\right)=20\)

\(3x-4=5\)

\(3x=9\)

\(x=3\)

d) \(\left(3x-6\right)+3=32\)

\(3x-6=29\)

\(3x=29+6\)

\(3x=35\)

\(x=\frac{35}{3}\)

e) \(\left(3x-6\right).3=32\)

\(3x-6=\frac{32}{3}\)

\(3x=\frac{32}{3}+6\)

\(3x=\frac{50}{3}\)

\(x=\frac{50}{9}\)

f) \(\left(3x-6\right):3=32\)

\(3x-6=96\)

\(3x=102\)

\(x=34\)

g) \(\left(3x-6\right)-3=32\)

\(3x-6=35\)

\(3x=41\)

\(x=\frac{41}{3}\)

h) \(\left(3x-2^4\right).7^3=2.7^4\)

\(\left(3x-2^4\right)=2.7=14\)

\(\left(3x-16\right)=14\)

\(3x=14+16=30\)

\(x=10\)

i) \(\left|x\right|=\left|-7\right|\)

\(\left|x\right|=7\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

k) \(\left|x+1\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)

l) \(\left|x-2\right|=3\)

\(\Rightarrow\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

m) \(x+\left|-2\right|=0\)

\(x+2=0\)

\(x=-2\)

o) \(72-3\left|x+1\right|=9\)

\(3\left|x-1\right|=63\)

\(\left|x-1\right|=21\)

\(\Rightarrow\orbr{\begin{cases}x-1=21\\x-1=-21\end{cases}\Rightarrow\orbr{\begin{cases}x=22\\x=-20\end{cases}}}\)

p) Ta có: \(\left|x-1\right|=3\)

\(\Rightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}\)

mà \(x+1< 0\)

\(\Rightarrow x-1=-3\)

\(\Rightarrow x=-2\)

q) \(\left(x-2\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)

hok tốt!!

Giải:

a) \(x\left(x-2\right)-\left(x+3\right).x+7+9x=6\)

\(\Leftrightarrow x^2-2x-\left(x^2+3x\right)+7+9x=6\)

\(\Leftrightarrow x^2-2x-x^2-3x+7+9x=6\)

\(\Leftrightarrow4x=-1\)

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

Vậy ...

b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

\(\Leftrightarrow21x-35-15x^2+25x-\left(10x+2-15x^2+6x\right)=4\)

\(\Leftrightarrow21x-35-15x^2+25x-10x-2+15x^2-6x=4\)

\(\Leftrightarrow30x-37=4\)

\(\Leftrightarrow30x=41\)

\(\Leftrightarrow x=\dfrac{41}{30}\)

Vậy ...

c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14x\) (Sửa đề)

\(\Leftrightarrow x^3+8-x^3-3=14x\)

\(\Leftrightarrow5=14x\)

\(\Leftrightarrow x=\dfrac{5}{14}\)

Vậy ...

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

\(\Leftrightarrow x^3+1-x^3-3x=2\)

\(\Leftrightarrow1-3x=2\)

\(\Leftrightarrow-3x=1\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

Vậy ...

a) \(x\left(x-2\right)-\left(x+3\right)x+7+9x=6\)

=> \(x^2-2x-x-3x+7+9x=6\)

=> \(x^2-2x-x^2-3x+7+9x=6\)

=> \(\left(x^2-x^2\right)+\left(-2x-3x+9x\right)=6-7\)

=> \(4x=-1\)

Vậy \(x=\dfrac{-1}{4}\)

b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

=>\(21x-15x^2-35+25x-10x+15x^2-4+6x=4\)

=> \(\left(21x+25x-10x+6x\right)\)\(+\left(-15x^2+15x^2\right)\)\(=4+35+4\)

=> \(42x=43\)

Vậy \(x=\dfrac{43}{42}\)

c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14\)

=> \(x^3-2x^2+4x+2x^2-4x+8-x^3-3\)\(=14x\)

=>\(\left(x^3-x^3\right)+\left(-2x^2+2x^x\right)+\left(4x-4x\right)+\left(8-3\right)\)\(=14x\)

=> \(5=14x\)

Vậy \(x=\dfrac{5}{14}\)

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

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

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

=> \(-3x=1\)

Vậy \(x=\dfrac{-1}{3}\)

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

\(< =>\left\{{}\begin{matrix}x^3-1=0\\x^2+1=0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x^3=1\\x^2=-1\left(kxđ\right)\end{matrix}\right.\)

<=>x=1

vậy ...

\(2.\left(2x+6\right)\left(3x^2-12\right)=0\)

\(< =>\left\{{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x=-3\\x^2=4\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

vậy ...

Trong Th này bn nên dùng dấu ''hoặc''

a,\(\left(x^3-1\right)\left(x^2+1\right)=0\)

\(\left[{}\begin{matrix}x^3-1=0\\x^2+1=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x^3=1\\x^2=-1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\x=\pm1\end{matrix}\right.\)

b, \(\left(2x+6\right)\left(3x^2-12\right)=0\)

\(\left[{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-3\\x^2=4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-3\\x=\pm2\end{matrix}\right.\)

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

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}}\)