B= 121-[16-(16+889)]

B= 121-[16-905]

B= 121+889

B= 1010

Giải:

\(B=121-\left[2^4-\left(4^2+889\right)\right]\)

\(\Leftrightarrow B=121-\left[16-\left(16+889\right)\right]\)

\(\Leftrightarrow B=121-16+16+889\)

\(\Leftrightarrow B=121+889\)

\(\Leftrightarrow B=1010\)

câu 1 bằng 1756

câu 2 bằng a 26 b 27 c có

câu 3 bằng 20

câu 4 bằng 50

100% chắc

a, 25*5*13*4=6500

87*125*8*4=348000

b,50*76*2=7600

19*4*25*8=15200

c,303*20*6*5=181800

4*889*25=88900

a)6500,348000.

b)7600,15200.

c)181800,88900.

a,= ( 25x4 ) x ( 13x5 )

   = 100 x 65

    = 6500

     = ( 125x8 ) x ( 87x4 )

      = 1000 x 348

       = 348000

b,    = ( 50x2 ) x 76

        = 100 x 76

        = 7600

         = ( 4x25 ) x ( 19x8 )

          =100 x 152

           = 15200

          c, = ( 20x5 ) x ( 303x6 )

               = 100 x 1818

                = 181800

               = ( 4x25 ) x 889

               = 100 x 889

                = 88900

889x2=1778

5689 x 4=22756

2873 x 8=22984

4638 x 7=32466

889 x 2 = 1778

5689 x 4 = 22756

2873 x 8 = 22984

4638 x 7 = 32466

Hãy tính :

889 x 2 = 1778

5689 x 4 = 22756

2873 x 8 = 22984

4638 x 7 = 37104

K mk nha

Năn nỉ bạn đó

\(\frac{\frac{2}{7}+\frac{2}{9}-\frac{2}{121}}{\frac{4}{7}+\frac{4}{9}-\frac{4}{121}}=\frac{2.\hept{ }\frac{1}{7}+\frac{1}{9}-\frac{1}{121}}{4.\hept{ }\frac{1}{7}+\frac{1}{9}-\frac{1}{121}}=\frac{2}{4}=\frac{1}{2}\)

bạn nhớ thay ngoặc nhọn thành ngoặc tròn nhé và nhớ đóng mở ngoặc

dễ thế mà ko bk làm hả em

dễ thế mà ko biết

a) \(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)^2=\dfrac{4}{25}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{15}\\x=-\dfrac{2}{15}\end{matrix}\right.\)

b) \(\Rightarrow\left(1-\dfrac{1}{4}x\right)^2=\dfrac{121}{49}\)

\(\Rightarrow\left[{}\begin{matrix}1-\dfrac{1}{4}x=\dfrac{11}{7}\\1-\dfrac{1}{4}x=-\dfrac{11}{7}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{16}{7}\\x=\dfrac{72}{7}\end{matrix}\right.\)

\(B=1+2+4+5+7+8+10+...+119+121+122\)

Ta có : 

\(A=1+2+3+...+121+122\)

\(A=\left[\left(122-1\right):1+1\right]\left(1+122\right):2\)

\(A=122.\left(123\right):2=7503\)

Ta lại có :

\(C=3+6+9+...+120\)

\(C=\left[\left(120-3\right):3+1\right]\left(3+120\right):2\)

\(C=40.123:2=2460\)

Ta thấy : \(B=A-C\)

\(B=7503-2460=5043\)

b: =3/4x(-1)=-3/4

Tổng quát:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)\(=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow S=\dfrac{10}{11}\)

Ta có công thức tổng quát như sau:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left[\left(n+1\right)\sqrt{n}+n\sqrt{n+1}\right]\left[\left(n+1\right)\sqrt{n}-n\sqrt{n+1}\right]}\)

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}\)

\(=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\)

\(=\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}\)

Áp dụng vào tổng S ta có:

\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{121\sqrt{120}+120\sqrt{121}}\)

\(S=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{120}}+\dfrac{1}{\sqrt{121}}\)

\(S=1-\dfrac{1}{\sqrt{121}}=1-\dfrac{1}{11}=\dfrac{10}{11}\)