\(A=\)\(\frac{6}{7}+\frac{1}{7}\times\frac{2}{7}+\frac{1}{7}\times\frac{5}{7}=\frac{6}{7}+\frac{1}{7}\left(\frac{2}{7}+\frac{5}{7}\right)\)

                                                                                                            \(=\frac{6}{7}+\frac{1}{7}\times1\)

                                                                                                            \(=\frac{6}{7}+\frac{1}{7}\)

                                                                                                             \(=1\)

 Cách 1:                                                      \(B=\frac{4}{9}\times\frac{13}{3}-\frac{4}{3}\times\frac{40}{9}\)

                                                                      \(=\frac{52}{27}-\frac{160}{27}\)

                                                                      \(=-4\)

K cho mk nhé thks bn nhiều

~CHÚC BẠN HỌC TỐT~

A=6/7+1/7(2/7+5/7)

 =6/7+1/7.1

 =6/7+1/7

 =1

A=6/7+1/7.2/7+1/7.5/7

A=6/7+1/7.(2/7+5/7)

A=6/7+1/7.7/7

A=6/7+1/7.1

A=7/7=1

B=4/9.13/3-4/3.4/9

B=4/9.(13/3-4/3)

B=4/9.9/3

B=4/9.3

B=12/9

a)

 \(\begin{array}{l}\left( {0,25 - \frac{5}{6}} \right).1,6 + \frac{{ - 1}}{3}\\ =(\frac{25}{100}-\frac{5}{6}).\frac{16}{10}+\frac{-1}{3}\\= \left( {\frac{1}{4} - \frac{5}{6}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \left( {\frac{6}{{24}} - \frac{{20}}{{24}}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{24}}.\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 5}}{{15}}\\ = \frac{{ - 19}}{{15}}\end{array}\)

b)

\(\begin{array}{l}3 - 2.\left[ {0,5 + \left( {0,25 - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left[ {\frac{1}{2} + \left( {\frac{1}{4} - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left( {\frac{1}{2} + \frac{1}{{12}}} \right)\\ =3-2.(\frac{6}{12}+\frac{1}{12})\\= 3 - 2.\frac{7}{{12}}\\ = 3 - \frac{7}{6}\\=\frac{18}{6}-\frac{7}{6}\\ = \frac{{11}}{6}\end{array}\)

a: \(log_{\dfrac{1}{4}}8=log_{2^{-2}}2^3=\dfrac{-3}{2}\cdot log_22=-\dfrac{3}{2}\)

b: \(log_45\cdot log_56\cdot log_68\)

\(=log_45\cdot\dfrac{log_46}{log_45}\cdot\dfrac{log_48}{log_46}\)

\(=log_48=log_{2^2}2^3=\dfrac{3}{2}\)

\(\left(\frac{1}{4\cdot5-1}-\frac{1}{7\cdot5-2}\right)\text{ : }\frac{40}{77}\)

\(=\left(\frac{1}{19}-\frac{1}{32}\right)\text{ : }\frac{40}{77}\)

\(=\frac{13}{608}\text{ : }\frac{40}{77}\)

\(=\frac{1001}{24320}\)

a) Tại x=-1

\(\Rightarrow x^5-5=\left(-1\right)^5-5=-6\)

\(a,\)Thay \(x=-1\) vào \(x^5-5\)

\(\Rightarrow\left(-1\right)^5-5=-6\)

\(b,\) 

+ TH1:

Thay \(x=1\) vào \(x^2-3x-5\)

\(\Rightarrow1^2-3.1-5=-7\)

+TH2:

Thay \(x=-1\) vào \(x^2-3x-5\)

\(\Rightarrow\left(-1\right)^2-3.\left(-1\right)-5=-1\)

b) Tại x=1

\(\Rightarrow x^2-3x-5=1-3-5=-7\)

Tại x=-1

\(\Rightarrow x^2-3x-5=1+3-5=-1\)

a: Ta có: \(A=\left(x-1\right)\left(x-3\right)+11\)

\(=x^2-4x+3+11\)

\(=x^2-4x+4+8\)

\(=\left(x-2\right)^2+8\ge8\forall x\)

Dấu '=' xảy ra khi x=2

b: Ta có: \(B=-4x^2+4x+5\)

\(=-\left(4x^2-4x+1-6\right)\)

\(=-\left(2x-1\right)^2+6\le6\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

\(a.3x-5y+1=3.\dfrac{1}{3}-5.\left(-\dfrac{1}{5}\right)+1=1+1+1=3\)

b.x=1

\(\Rightarrow3.1^2-2.1-5=-4\)

x=-1

\(\Rightarrow3.\left(-1\right)^2-2.\left(-1\right)-5=3+2-5=0\)

https://khoahoc.vietjack.com/question/661489/bieu-thuc-a-3-45-9-6-7-4-3-12-5-11-3-co-gia-tri-3-11