rút gọn 3^4*2^3-3^4*4 phần 3^5*3^2-3^5*5
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\(\frac{1}{4}+\frac{3}{5}=\frac{17}{20}\)
\(\frac{5}{2}+\frac{7}{9}=\frac{59}{18}\)
\(\frac{3}{2}+\frac{2}{3}=\frac{13}{6}\)
\(\frac{4}{5}+\frac{3}{2}=\frac{23}{10}\)
\(\frac{4}{5}+\frac{3}{15}=\frac{4}{5}+\frac{1}{5}=1\)
\(\frac{2}{3}+\frac{32}{24}=\frac{2}{3}+\frac{4}{3}=2\)
\(\frac{5}{6}+\frac{15}{18}=\frac{5}{6}+\frac{5}{6}=\frac{10}{6}=\frac{5}{3}\)
\(\frac{8}{15}+\frac{2}{3}=\frac{8}{15}+\frac{10}{15}=\frac{18}{15}=\frac{6}{5}\)
\(\frac{3}{7}+\frac{4}{8}=\frac{24}{56}+\frac{28}{56}=\frac{52}{56}=\frac{13}{14}\)
_HT_
\(\dfrac{-2^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot5^3\cdot2^4\cdot42}\)
\(=\dfrac{-2^6\cdot3^3\cdot5^3\cdot7}{3\cdot5^3\cdot2^4\cdot2\cdot3\cdot7}\)
\(=\dfrac{-2^6\cdot3^3\cdot5^3}{2^5\cdot3^2\cdot5^3}=-2\cdot3=-6\)
bài 4:so sánh
5/2 lớn hơn 3/7
4/3 lớn hơn,3/2 lớn hơn
bài 6:rút gọn các phân số sau:
3/9=1/3 9/12=3/4 8/18=4/9 60/36=10/6 17/34=1/2 17/51=1/3 35/100=7/20 25/100=1/4 8/1000=1/125 24/30=4/5 18/54=1/3 72/42=12/7
đay nhé mk chưa làm hết đc bn viết liền quá mk nhìn khó mà mk hỏi bài 7 là nhân hay cộng vậy?
\(A=\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.2^4.5^3.14}=\frac{\left(-8\right).3^2.3.5^3.7.8}{3.2^3.2.5^3.2.7}=\frac{\left(-8\right).9}{2.2}=\frac{-72}{4}=-18\)
\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}=\frac{8}{25-3}=\frac{8}{22}=\frac{4}{11}\)
\(\frac{3^4.5-3^6}{3^4.13+3^4}=\frac{3^4.\left(5-3^2\right)}{3^4.\left(13+1\right)}=\frac{5-9}{14}=\frac{-4}{14}=\frac{-2}{7}\)
\(\frac{-2}{7}=\frac{-22}{77}\)
\(\frac{4}{11}=\frac{28}{77}\)
1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)
\(=x^2+4x+4+x^2-6x+9\)
\(=2x^2-2x+13\)
2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)
\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)
\(=-2x+7\)
3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=x^2-25-x^2-10x-25\)
=-10x-50
4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16\)
=-6x+25
5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)
\(=y^2-6y+9-y^2+6y-9\)
=0
6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)
\(=4x^2+12x+9-4x^2+9\)
=12x+18
Ta có: \(\dfrac{3^4\cdot2^3-3^4\cdot4}{3^5\cdot3^2-3^5\cdot5}\)
\(=\dfrac{3^4\cdot\left(2^3-4\right)}{3^5\cdot\left(3^2-5\right)}\)
\(=\dfrac{3^4\cdot4}{3\cdot3^4\cdot4}=\dfrac{1}{3}\)
\(\dfrac{3^4\cdot2^3-3^{\text{4}}\cdot4}{3^5\cdot3^2-3^5\cdot5}\)=\(\dfrac{3^4\left(8-4\right)}{3^5\left(9-5\right)}\)=\(\dfrac{4}{3.4}\)=\(\dfrac{1}{4}\)