\(5,\left(x\cdot0,5-\frac{3}{7}\right):\frac{1}{2}=1\frac{1}{7}\)

\(\Leftrightarrow x\cdot0,5:\frac{1}{2}-\frac{3}{7}:\frac{1}{2}=1\frac{1}{7}\)

\(\Leftrightarrow x-\frac{6}{7}=\frac{8}{7}\)

\(\Leftrightarrow x=2\)

\(6,x\cdot1,75=1\frac{3}{10}+45\%\)

\(\Leftrightarrow x\cdot\frac{7}{4}=\frac{13}{10}+\frac{9}{20}\)

\(\Leftrightarrow x\cdot\frac{7}{4}=\frac{7}{4}\)

\(\Leftrightarrow x=1\)

\(7,\frac{5-x}{15}+\frac{5}{12}-\frac{1}{8}=\frac{3}{8}\)

\(\Leftrightarrow\frac{5-x}{15}=\frac{3}{8}-\frac{5}{12}+\frac{1}{8}\)

\(\Leftrightarrow\frac{5-x}{15}=\frac{1}{12}\)

\(\Leftrightarrow60-12x=15\)

\(\Leftrightarrow12x=45\)

\(\Leftrightarrow x=\frac{15}{4}\)

\(8,\left|x-\frac{25}{33}\right|-\frac{3}{11}=\frac{2}{3}\)

\(\Leftrightarrow\left|\frac{x-25}{33}\right|=\frac{31}{33}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{25}{33}=\frac{31}{33}\\x-\frac{25}{33}=-\frac{31}{33}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{56}{33}\\x=-\frac{2}{11}\end{cases}}\)

\(9,-\frac{9}{8}+\frac{-3}{8}\cdot x=-\frac{1}{8}\)

\(\Leftrightarrow\frac{-9}{8}+\frac{-3}{8}\cdot x+\frac{1}{8}=0\)

\(\Leftrightarrow-1-\frac{3}{8}x=0\)

\(\Leftrightarrow\frac{3}{8}x=-1\)

\(\Rightarrow x=-\frac{8}{3}\)

bạn có thể đừng làm tắt bước được không

A. \(\frac{3}{4}\) x \(\frac{8}{9}\)x \(\frac{15}{16}\)x .... x \(\frac{899}{900}\)

= \(\frac{1.3}{2^2}\) x \(\frac{2.4}{3^3}\)x \(\frac{3.5}{4^2}\)x ... x \(\frac{29.31}{30^2}\)

= \(\left(\frac{1.2.3...29}{2.3.4...30}\right).\left(\frac{3.4.5...31}{2.3.4...30}\right)\)

= \(\frac{1}{30}.\frac{31}{2}\)= \(\frac{31}{60}\)

B. 

\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}=\frac{8}{24}+\frac{9}{24}-\frac{14}{24}=\frac{8+9-14}{24}=\frac{3}{24}=\frac{1}{8}\)

Bài 2:

a; -8/9 và -7/9

Vì 8/9 > 7/9

- 8/9 < - 7/9 (nhân cả hai vế của bất đẳng thức với một số âm thì dấu của bất đẳng thức đổi chiều.

bài khó quá giải cũng dài luôn

\(Ai\)\(giúp\)\(mình\)\(bài\)\(kia\)\(đi\)

* x/7 = -6/21

=> x = -6.7 : 21

=> x = -2

* x/-2 = -8/x

=> x.x = -8.(-2) = 16

=> x^2 = 16

=> x = 4 hoặc x = -4

\(\frac{-2}{7}\)= \(\frac{-6}{21}\)

\(\frac{4}{-2}\)= \(\frac{-8}{4}\)

Học tốt ^-^

Câu a:

33 ⋮ (x+ 1)

(x+ 1) ∈ Ư(33) = {-33; -11; -3; -1; 1; 3; 11}

x ∈ {-34; -12; -4; -2; 0; 2; 10}

Vậy: {-34; -12; -4; -2; 0; 2; 10}

Câu b:

x ∈ ƯC(250; 48)

250 = 2.5^3; 48 = 2^4.3

ƯCLN(250; 48) = 2

x ∈ ƯC(2) = {-2; -1}

Vậy x ∈ {-2; -1}

a) \(\frac{x}{3}-\frac{1}{4}=\frac{-5}{6}\)

\(\Rightarrow\frac{x}{3}=\frac{-5}{6}+\frac{1}{4}\)

\(\Rightarrow\frac{x}{3}=\frac{-7}{12}\)

\(\Rightarrow x=\frac{\left(-7\right).3}{12}\)

\(\Rightarrow x=\frac{-7}{4}\)

Vậy x = \(\frac{-7}{4}\)

b) \(\frac{x+3}{15}=\frac{1}{3}\)

\(\Rightarrow\left(x+3\right).3=15\)

\(\Rightarrow x+3=15:3\)

\(\Rightarrow x+3=5\)

\(\Rightarrow x=5-3\)

\(\Rightarrow x=2\)

Vậy x = 2

c) \(\frac{x-12}{4}=\frac{1}{2}\)

\(\Rightarrow\left(x-12\right).2=4\)

\(\Rightarrow x-12=4:2\)

\(\Rightarrow x-12=2\)

\(\Rightarrow x=2+12\)

\(\Rightarrow x=14\)

Vậy x = 14

d) \(\frac{25}{9}-\frac{12}{13}x=\frac{7}{9}\)

\(\Rightarrow\frac{12}{13}x=\frac{25}{9}-\frac{7}{9}\)

\(\Rightarrow\frac{12}{13}x=2\)

\(\Rightarrow x=2:\frac{12}{13}\)

\(\Rightarrow x=\frac{13}{6}\)

Vậy x = \(\frac{13}{6}\)

_Chúc bạn học tốt_

a)\(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\)

\(\frac{x}{3}=-\frac{5}{6}+\frac{1}{4}\)

\(\frac{x}{3}=-\frac{7}{12}\)

\(x=-\frac{7}{12}\times3\)

\(\Rightarrow x=-\frac{7}{4}\)

b) \(\frac{x+3}{15}=\frac{1}{3}\)

\(\frac{x+3}{15\div5}=\frac{1}{3}\)

\(\Rightarrow x=5-3\)

\(\Rightarrow x=2\)

c) \(\frac{x-12}{4}=\frac{1}{2}\)

\(\frac{x-12}{4\div2}=\frac{1}{2}\)

\(x=12+2\)

\(\Rightarrow x=14\)

d) tự làm nhé cũng dễ mà

A = 1/3^2 + 1/4^2 + 1/5^2 + ... + 1/50^2

1/3^2 = 1/9

1/4^2 < 1/3.4 = 1/3 - 1/4

1/5^2 < 1/4.5 = 1/4 - 1/5

.............................................

1/50^2 < 1/49.50 = 1/49 - 1/50

Cộng vế với vế ta có:

A = 1/3^2+1/4^2+..+1/50^2 = 1/9 + 1/3 - 1/50

A = 4/9 - 1/50 < 4/9

1/3^2 = 1/9

1/4^2 > 1/4.5 = 1/4 - 1/5

1/5^2 > 1/5.6 = 1/5 - 1/6

............................................

1/50^2 > 1/49.50 = 1/49 - 1/50

Cộng vế với vế ta có:

A = 1/3^2+1/4^2+ ...+ 1/50^2 > 1/9+1/4-1/50

A > 1/4 + (1/9 - 1/50)

1/9 > 1/50

1/9 - 1/50 > 0

A > 1/4 + 1/9 - 1/50 > 1/4

Vậy 1/4 < A < 4/9 (đpcm)

a) \(\frac{13}{26}-\frac{1}{3}-\frac{1}{2}+\frac{7}{21}\)
\(=\frac{1}{2}-\frac{1}{3}-\frac{1}{2}+\frac{1}{3}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\)
\(=0+0\)
\(=0\)

b) \(\left(\frac{-5}{12}+\frac{6}{11}\right)+\left(\frac{7}{17}+\frac{5}{17}+\frac{5}{12}\right)\)
\(=\frac{-5}{12}+\frac{6}{11}+\frac{7}{17}+\frac{5}{17}+\frac{5}{12}\)
\(=\left(\frac{-5}{12}+\frac{5}{12}\right)+\left(\frac{7}{17}+\frac{5}{17}\right)+\frac{6}{11}\)
\(=0+\frac{12}{17}+\frac{6}{11}\)
\(=\frac{132}{187}+\frac{102}{187}\)
\(=\frac{234}{187}\)

c) \(\left(\frac{13}{5}+\frac{7}{16}\right)-\left(\frac{11}{16}-\frac{12}{10}\right)\)
\(=\left(\frac{13}{5}+\frac{7}{16}\right)-\left(\frac{11}{16}-\frac{6}{5}\right)\)
\(=\frac{13}{5}+\frac{7}{16}-\frac{11}{16}+\frac{6}{5}\)
\(=\left(\frac{13}{5}+\frac{6}{5}\right)+\left(\frac{7}{16}-\frac{11}{16}\right)\)
\(=\frac{19}{5}+\left(\frac{-4}{16}\right)\)
\(=\frac{19}{5}-\frac{1}{4}\)
\(=\frac{76}{20}-\frac{5}{20}\)
\(=\frac{71}{20}\)

d) \(-\left(\frac{3}{10}-\frac{6}{11}\right)-\left(\frac{21}{30}-\frac{5}{11}\right)\)
\(=-\left(\frac{3}{10}-\frac{6}{11}\right)-\left(\frac{7}{10}-\frac{5}{11}\right)\)
\(=-\frac{3}{10}+\frac{6}{11}-\frac{7}{10}+\frac{5}{11}\)
\(= \left(-\frac{3}{10}-\frac{7}{10}\right)+\left(\frac{6}{11}+\frac{5}{11}\right)\)
\(=\frac{-10}{10}+\frac{11}{11}\)
\(=-1+1\)
\(=0\)