A

a: \(\dfrac{11}{4}+\left(-\dfrac{3}{5}+\dfrac{1}{3}\right):\dfrac{-4}{9}\)

\(=\dfrac{11}{4}+\dfrac{-9+5}{15}\cdot\dfrac{9}{-4}\)

\(=\dfrac{11}{4}+\dfrac{-4}{15}\cdot\dfrac{9}{-4}\)

\(=\dfrac{11}{4}+\dfrac{3}{5}=\dfrac{55+12}{20}=\dfrac{67}{20}\)

b: \(\dfrac{2}{6}-\left(-\dfrac{3}{3}+\dfrac{4}{7}\right):\dfrac{-9}{5}\)

\(=\dfrac{2}{6}-\left(-1+\dfrac{4}{7}\right)\cdot\dfrac{-5}{9}\)

\(=\dfrac{2}{6}-\dfrac{-3}{7}\cdot\dfrac{-5}{9}\)

\(=\dfrac{1}{3}-\dfrac{15}{63}\)

\(=\dfrac{6}{63}=\dfrac{2}{21}\)

c: \(21-\dfrac{15}{4}:\left(\dfrac{3}{8}-\dfrac{1}{6}\right)\)

\(=21-\dfrac{15}{4}:\dfrac{3\cdot3-4}{24}\)

\(=21-\dfrac{15}{4}\cdot\dfrac{24}{5}\)

\(=21-3\cdot6=3\)

\(a.\dfrac{11}{4}+\left(-\dfrac{3}{5}+\dfrac{1}{3}\right):-\dfrac{4}{9}\)

\(=\dfrac{11}{4}+\left(-\dfrac{4}{15}\right):-\dfrac{4}{9}\)

\(=\dfrac{11}{4}+\dfrac{3}{5}=\dfrac{67}{20}\)

\(b.\dfrac{2}{6}-\left(-\dfrac{3}{3}+\dfrac{4}{7}\right):-\dfrac{9}{5}\)

\(=\dfrac{2}{6}-\left(-\dfrac{3}{7}\right):-\dfrac{9}{5}\)

\(=\dfrac{2}{6}-\dfrac{5}{21}=\dfrac{2}{21}\)

\(c.21-\dfrac{15}{4}:\left(\dfrac{3}{8}-\dfrac{1}{6}\right)\)

\(=21-\dfrac{15}{4}:\dfrac{5}{24}\)

\(=21-18=3\)

Bài 2: 

a) Ta có: \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)

\(=3^{n+1}\cdot10+2^{n+3}\cdot3⋮6\)

b) Ta có: \(4^{13}+32^5-8^8\)

\(=2^{26}+2^{25}-2^{24}\)

\(=2^{24}\left(2^2+2-1\right)\)

\(=2^{24}\cdot5⋮5\)

c) Ta có: \(2014^{100}+2014^{99}\)

\(=2014^{99}\left(2014+1\right)\)

\(=2014^{99}\cdot2015⋮2015\)

\(\dfrac{11}{10}< \dfrac{9}{30}\)

\(\dfrac{6}{7}>\dfrac{3}{5}\)

\(\dfrac{25}{100}< \dfrac{3}{4}\)

a \(\dfrac{11}{10}>\dfrac{3}{10}\)

b \(\dfrac{30}{35}>\dfrac{21}{35}\)

c \(\dfrac{1}{4}< \dfrac{3}{4}\)

`a) 5 / 13 xx 4 / 15 xx 13 = [ 5 xx 4 xx 13 ] / [ 13 xx 5 xx 3 ] = 4 / 3`

`b) ( 3 / 7 + 5 / 2 ) xx 7 / 5 = 3 / 7 xx 7 / 5 + 5 / 2 xx 7 / 5 = 3 / 5 + 7 / 2 = 6 / 10 + 35 / 10 = 41 / 10`

`c) 1 / 5 xx 11 / 18 + 11 / 18 xx 3 / 5 = 11 / 18 xx ( 1 / 5 + 3 / 5 ) = 11 / 18 xx 4 / 5 = [ 11 xx 2 xx 2 ] / [ 2 xx 9 xx 5 ] = 22 / 45`

\(a,\dfrac{5}{13}\times\dfrac{4}{15}\times13=\dfrac{4}{3}\)

\(b,\left(\dfrac{3}{7}+\dfrac{5}{2}\right)\times\dfrac{7}{5}=\left(\dfrac{6}{14}+\dfrac{35}{14}\right)\times\dfrac{7}{5}=\dfrac{41}{14}\times\dfrac{7}{15}=\dfrac{41}{30}\)

\(c,\dfrac{1}{5}\times\dfrac{11}{18}+\dfrac{11}{18}\times\dfrac{3}{5}=\dfrac{11}{18}\times\left(\dfrac{1}{5}+\dfrac{3}{5}\right)=\dfrac{11}{18}\times\dfrac{4}{5}=\dfrac{22}{45}\)

Lời giải:

a. $7-4(x+1)=3x-5$
$7-4x-4=3x-5$

$3-4x=3x-5$

$3+5=3x+4x$

$8=7x$

$x=\frac{8}{7}$

------------------------

c.

$\frac{5x+8}{3}=\frac{4-2x}{4}$

$\Rightarrow 4(5x+8)=3(4-2x)$

$\Rightarrow 20x+32=12-6x$

$\Rightarrow 20x+6x=12-32$

$\Rightarrow 26x=-20$

$\Rightarrow x=\frac{-20}{26}=\frac{-10}{13}$

-------------------

d. 

$(9x-1)^2=5=(\sqrt{5})^2=(-\sqrt{5})^2$

$\Rightarrow 9x-1=\sqrt{5}$ hoặc $9x-1=-\sqrt{5}$

$\Rightarrow x=\frac{\sqrt{5}+1}{9}$ hoặc $x=\frac{1-\sqrt{5}}{9}$

D

D

Câu hỏi của Thị Kim Vĩnh Bùi - Toán lớp 8 - Học toán với OnlineMath

Thya các giá trị của a, b, c., d vào M . Tính đc M = 0

Câu hỏi của Thị Kim Vĩnh Bùi - Toán lớp 8 - Học toán với OnlineMath

Ở link trên đã tìm đc các giá trị của a, b, c, d thay vào tìm đc M = 0.

Mai tui giải cho hem

Ta có:

\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)

\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)

\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)

\(B=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)

\(B=\frac{9}{5}\)