38/45 phần 76/45 

đó dễ mà

\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}=\frac{1}{2}\)

\(8-12x+6x^2-x^3\)

\(=\left(2-x\right)^3\)

\(125x^3-75x^2+15x-1\)

\(=\left(5x-1\right)^3\)

\(x^2-xz-9y^2+3yz\)

\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x+3y-z\right)\)

\(x^3-x^2-5x+125\)

\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

\(x^3+2x^2-6x-27\)

\(=x^3+5x^2+9x-3x^2-15x-27\)

\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

\(12x^3+4x^2-27x-9\)

\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(4x^2-9\right)\)

\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)

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

\(=4x^3\left(x+1\right)-x\left(x+1\right)\)

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

\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)

Gọi các phân tử của A là a.

Ta có:

\(B\left(12\right)=\left\{0;12;24;...\right\}\)

\(\Rightarrow A=\left\{0;12;24;...\right\}\)

\(\Rightarrow a\in\left\{0;12;24;...\right\}\)

Mà a nhỏ hơn 12 \(\Rightarrow a=0\)

Vậy tập hợp A có 1 phần tử.

Phần I :

bài này cậu hỏi và có người giải đúng rồi mà

Ta có 5\(\dfrac{1}{5}\)=5\(\dfrac{3}{15}\)

Từ đó ta tìm đc các phần nguyên của x là :5;6;7;8

Bài 1: Tính ( hợp lý nếu có thể )

\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)

\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)

\(=-1+1+\dfrac{2}{-5}\)

\(=0+\dfrac{2}{-5}\)

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

\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)

\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)

\(=0+\dfrac{2}{3}\)

\(=\dfrac{2}{3}\)

\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)

\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)

\(=-1+1\)

\(=0\)

\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)

\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)

\(=\dfrac{1}{6}+\dfrac{-1}{6}\)

\(=0\)

Bài 2: Tìm x,biết:

a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)

\(x=\dfrac{4}{5}-\dfrac{2}{3}\)

\(x=\dfrac{2}{15}\)

Vậy \(x=\dfrac{2}{15}\)

b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)

\(\Rightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}+\dfrac{2}{3}\)

\(x=\dfrac{3}{3}=1\)

Vậy \(x=1\)

c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!

\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)

\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)

\(x=\dfrac{1}{44}\)

Vậy \(x=\dfrac{1}{44}\)

d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(x=-\dfrac{3}{20}\)

Vậy \(x=-\dfrac{3}{20}\)

a) \(\frac{2^{12}x3^5-4^6.9^2}{\left(2^2x3\right)^6+8^4x3^5}=\frac{2^{12}x3^5+\left(2^2\right)^6x\left(3^2\right)^2}{2^{12}x3^6+\left(2^3\right)^4x3^5}\)

\(=\frac{2^{12}x3^5-2^{12}x3^4}{2^{12}x3^6+2^{12}x3^5}=\frac{2^{12}x3^4x\left(3-1\right)}{2^{12}x3^5x\left(3+1\right)}\)

\(=\frac{2}{3.4}=\frac{1}{3.2}=\frac{1}{6}\)

b) \(\frac{1}{9x10}-\frac{1}{8x9}-\frac{1}{7x8}-\frac{1}{6x7}-\frac{1}{5x6}-\frac{1}{4x5}-\frac{1}{3x4}-\frac{1}{2x3}-\frac{1}{1x2}\)

\(=-\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}\right)\)

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

\(=-\left(1-\frac{1}{10}\right)\)

\(=\frac{-9}{10}\)

sorry bn nha! mk ko bk lm phần c

trả lời :

a) 1/6

b)-9/10

Bài làm

a) \(-\frac{3}{7}+\frac{3}{4}:\frac{3}{14}\)

= \(-\frac{3}{7}+\frac{3}{4}.\frac{14}{3}\)

= \(-\frac{3}{7}+\frac{7}{2}\)

\(=-\frac{7}{14}+\frac{49}{14}\)

\(=\frac{42}{14}=3\)

b) \(5-\frac{7}{39}:\frac{7}{13}+\frac{8}{9}:4\)

\(=5=\frac{7}{39}.\frac{13}{7}+\frac{8}{9}.\frac{1}{4}\)

\(=5-\frac{1}{3}+\frac{2}{9}\)

\(=\frac{45}{9}-\frac{3}{9}+\frac{2}{9}\)

\(=\frac{44}{9}\)

c) \(\left(\frac{5}{12}:\frac{11}{6}+\frac{5}{12}:\frac{11}{5}\right)-\frac{-7}{12}\)

\(=\left(\frac{5}{12}.\frac{6}{11}+\frac{5}{12}.\frac{5}{11}\right)+\frac{7}{12}\)

\(=\left[\frac{5}{12}\left(\frac{6}{11}+\frac{5}{11}\right)\right]+\frac{7}{12}\)

\(=\frac{5}{12}+\frac{7}{12}\)

\(=\frac{12}{12}=1\)

d) \(-\frac{5}{9}+\frac{14}{9}\left(\frac{3}{4}-\frac{2}{5}\right):49\)

\(=-\frac{5}{9}+\frac{14}{9}\left(\frac{15}{20}-\frac{8}{20}\right):49\)

\(=-\frac{5}{9}+\frac{14}{9}.\frac{7}{20}.\frac{1}{49}\)

\(=-\frac{5}{9}+\frac{7}{9}.\frac{7}{10}.\frac{1}{7.7}\)

\(=-\frac{5}{9}+\frac{1}{90}\)

\(=-\frac{50}{90}+\frac{1}{90}=-\frac{49}{90}\)