a) \(x⋮15;x⋮35;x⋮42\&250< x< 850\) (sửa dấu chia thành chia hết)

\(BCNN\left(15;35;42\right)=210\)

\(\Rightarrow x\in BC\left(15;35;42\right)=\left\{0;210;420;630;840;...\right\}\)

mà \(250< x< 850\)

\(\Rightarrow x\in\left\{420;630;840\right\}\)

b) x nhỏ nhất khác 0 thỏa mãn \(x⋮15;x⋮115\)

\(BCNN\left(15;115\right)=345\)

Vậy \(x\in\left\{345\right\}\) thỏa mãn đề bài

Bài c bạn xem lại đề

phần c lỗi nha ạ, mình sửa lại 

c) (x−1) chia hết 52,(x−1) chia hết 35 và 1000 < x < 2000

\(2x+20=80-60\)

\(2x+20=20\)

\(2x=0\)

\(x=0\)

b,100x=1000:10

     100x=100

         x=100:100

         x=1

c,250x+250=1750

    250(x+1)=1750

           x+1=1750:250

           x+1=7

               x=6

\(\left(\frac{1}{4}\right)^3\cdot4^3=\left(\frac{1}{4}\cdot4\right)^3=1^3=1\)

\(\frac{1000^4}{250^4}=4^4=256\)

\(2^2\cdot9\cdot\frac{1}{54}\cdot\left(\frac{4}{9}\right)^2=2^2\cdot3^2\cdot2\cdot3^3\cdot\left(\frac{4}{9}\right)^2=\left[\left(2\cdot3\cdot\frac{4}{9}\right)^2\right]\cdot2\cdot3^3=\frac{64}{9}\cdot2\cdot27=384\)

2. a) 2^x = 9 => x không thỏa mãn

b) x² = 9 => x = \(\pm\)3

c) (x + 1)² = 4 => (x + 1)² = \(\pm\)2²

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

Bài 1 :

\(a,\left(\frac{1}{4}\right)^3.4^3\)

\(=\frac{1}{4^3}.4^3\)

\(=1\)

\(b,\frac{1000^4}{250^4}=\frac{\left(250.4\right)^4}{250^4}=\frac{250^4.4^4}{250^4}=4^4=256\)

\(d,2^2.9.\frac{1}{54}.\left(\frac{4}{9}\right)^2\)

\(=36.\frac{1}{54}.\frac{4^2}{9^2}\)

\(=\frac{18.2.16}{18.3.81}\)

\(=\frac{32}{243}\)

Bài 2 :

\(a,2^x=9\)

\(\Rightarrow\)x không thỏa mãn

\(b,x^2=9\)

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

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

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

\(\Rightarrow\left(x+1\right)^2=2^2\)

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

Học tốt

Làm lại câu d , xin lỗi bạn vì sự nhầm lẫn

\(2^2\cdot9\cdot\frac{1}{54}\cdot\left(\frac{4}{9}\right)^2=4\cdot9\cdot\frac{1}{54}\cdot\frac{16}{81}\)

\(=36\cdot\frac{1}{54}\cdot\frac{16}{81}=2\cdot\frac{1}{3}\cdot\frac{16}{81}=\frac{32}{81}\cdot\frac{1}{3}=\frac{32}{243}\)

x+1000-100+100=250+750

x+1000-100+100=1000

x+1000-100=1000-100

x+1000-100=900

x+1000=900+100

x+1000=1000

x=1000-1000

x=0

x + 1000 - 100 +100 = 250 +750

x + 1000 - 100 +100 = 1000

x+1000                    = 1000

x                              =0

vậy x=0.

ai tk mk, mk tk lại.

x = 0

k mk nha

Ta có:\(1001=1000+1=x+1\)

\(x^8-1001x^7+1001x^6+...+1001x^2-1001x+250\\ =x^8-\left(x+1\right)x^7+\left(x+1\right)x^6+...+\left(x+1\right)x^2-\left(x+1\right)x\\ =x^8-x^8-x^7+x^7+x^6+...+x^3+x^2-x^2-x+250\\ =-x+250=-1000+250\\ =-750\)

P(x)=x^8-1001x^7+...-1001x+250

=x^8-x^7(x+1)+x^6(x+1)-...-x(x+1)+250

=x^8-x^8-x^7+x^7+...-x^2-x+250

=250-x

=-750