`30 xx 25 xx 7 xx8`

`=(30 xx7) xx (25xx8)`

`=210 xx200`

`=42000`

`----`

`75 xx 8 xx 12 xx 14`

`=(75xx8) xx(12xx14)`

`=600xx168`

`=100800`

\(a.30\times25\times7\times8=\left(30\times7\right)\times\left(25\times8\right)=210\times200=42000\)

\(b.75\times8\times12\times14=\left(75\times12\right)\times\left(8\times14\right)=900\times112=100800\)

số số hạng là :`(26 - 2) : 2 + 1 = 13(số hạng)`

`13 : 2 = 12 dư 1 `

`=>A = (2 - 4) + (6 - 8) + (10 - 12) + ... + (22 - 24) + 26`

`=>A = (-2) + (-2) + (-2) + ... + (-2) + 26`

`=>A = (-2) xx 6 + 26`

`=> A = -12 + 26`

`=> A = 14`

A=[(26-2):4+1].(2-4)

A=[24:4+1].(-2)

A=7.(-2)

A=-14

a: =16,71*(91+97)-188(6,2+0,67)

=16,71*188-188*(6,2+0,67)

=188(16,71-6,2-0,67)=1849,92

b: 56^2+31^2-13^2+56*62

=(56+31)^2-13^2

=87^2-13^2

=74*100=7400

B1:

\(a.301^2=\left(300+1\right)^2=300^2+2.300.1+1^2\\ =90000+600+1=90601\\ b.88^2+2.88.12+12^2=\left(88+12\right)^2=100^2=10000\\ c.99.100=100^2-100=10000-100=9900\\ d,153^2+94.153+47^2=153^2+2.153.47+47^2=\left(153+47\right)^2=200^2=40000\)

B2:

\(A=x^2-20x+101\\ =x^2-2.x.10+10^2+1\\ =\left(x-10\right)^2+1\ge1\forall x\in R\left(Vì:\left(x-10\right)^2\ge0\forall x\in R\right)\\ \Rightarrow min_A=1\Leftrightarrow x-10=0\Leftrightarrow x=10\)

Mình cần gấp . Giải chi tiết nha ! 

TÍNH NHANH : 

12 x ( 22 - 15 ) - 22 x ( 15 - 12 )

12 x       7      -  22  x 3

84          -   66

18

\(=12\times22-12\times15-22\times15-22\times12\)

\(=\left[\left(12\times22\right)-\left(22\times12\right)\right]-\left[\left(12\times15\right)-\left(22\times15\right)\right]\)

\(=0-15\times\left(12-22\right)\)

\(=0-15\times\left(-10\right)\)

\(=0-\left(-150\right)\)

\(=150\)

=> A = 1.1 + 2.2 + 3.3 + .... + 50.50

=> A = 1.( 2 - 1 ) + 2.( 3 - 1 ) + 3.( 4 - 1 ) + ... + 50.( 51 - 1 )

=> A = 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + .... + 50.51 - 50

=> A = ( 1.2 + 2.3 + 3.4 + ..... + 50.51 ) - ( 1 + 2 + 3 + .... + 50 )

=> A = \(\frac{50\cdot51\cdot52}{3}-\frac{\left(50+1\right)\cdot50}{2}\)

=> A = 44200 - 1275

=> A = 42925

1. Áp dụng công thức tổng cấp số nhân:

\(S_n=u_1.\dfrac{q^n-1}{q-1}=2.\dfrac{2^n-1}{2-1}=2.\left(2^n-1\right)=2^{n+1}-2\)

2. \(\left\{{}\begin{matrix}u_2+u_5=12\\u_4+u_8=22\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(u_1+d\right)+\left(u_1+4d\right)=12\\\left(u_1+3d\right)+\left(u_1+7d\right)=22\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2u_1+5d=12\\2u_1+10d=22\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u_1=1\\d=2\end{matrix}\right.\)

\(\Rightarrow u_n=u_1+\left(n-1\right)d=1+\left(n-1\right)2=2n-1\)

\(\Rightarrow S_n=\dfrac{n\left(u_1+u_n\right)}{2}=\dfrac{n\left(1+2n-1\right)}{2}=n^2\)

3. \(\left\{{}\begin{matrix}u_1+u_2=4\\u_4+u_1=28\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u_1+u_1q=4\\u_1q^3+u_1=28\end{matrix}\right.\)

\(\Rightarrow\dfrac{q^3+1}{q+1}=\dfrac{28}{4}\Rightarrow q^2-q+1=7\)

\(\Rightarrow q^2-q-6=0\Rightarrow\left[{}\begin{matrix}q=3\\q=-2\end{matrix}\right.\)