\(\left(x+\dfrac{5}{3}\right).\dfrac{9}{13}=\dfrac{2}{3}\\ \Rightarrow\dfrac{9}{13}x+\dfrac{15}{13}=\dfrac{2}{3}\\ \Rightarrow\dfrac{9}{13}x=\dfrac{2}{3}-\dfrac{15}{13}=\dfrac{26}{39}-\dfrac{45}{39}=\dfrac{-9}{39}=\dfrac{-3}{13}\\ \Rightarrow\dfrac{9}{13}x=\dfrac{-3}{13}\\ \Rightarrow x=\dfrac{-3}{13}:\dfrac{9}{13}=\dfrac{-3}{13}.\dfrac{13}{9}=\dfrac{-3}{9}=\dfrac{-1}{3}\)

nhầm ở chỗ \(x=\dfrac{3}{35}:\dfrac{4}{5}=\dfrac{3}{35}.\dfrac{5}{4}=\dfrac{15}{140}=\dfrac{3}{28}\) nha

Thay là \(x=\dfrac{4}{5}:\dfrac{3}{35}=\dfrac{4}{5}.\dfrac{35}{3}=\dfrac{140}{15}=\dfrac{28}{3}\) nha

\(\dfrac{4}{5}:x=\dfrac{2}{7}-\dfrac{1}{5}\\ \dfrac{4}{5}:x=\dfrac{10}{35}-\dfrac{7}{35}=\dfrac{3}{35}\\ \Rightarrow\dfrac{4}{5}:x=\dfrac{3}{35}\\ \Rightarrow x=\dfrac{3}{35}:\dfrac{4}{5}=\dfrac{3}{35}.\dfrac{5}{4}=\dfrac{15}{140}=\dfrac{3}{28}\)

ko có gì

a, \(64:2^3=2^6:2^3=2^3\)

b,\(243:3^4=3^5:3^4=3\)

c,\(625:5^3=5^4:5^3=5\)

d,\(7^5.343=7^5:7^3=7^2\)

e,\(100000:10^3=10^5:10^3=10^2\)

f,\(11^5:121=11^5:11^2=11^3\)

g,\(243:3^3:3=3^5:3^3:3=3^2:3=3\)

h,\(4^8:64:16=4^8:4^3:4^2=4^5:4^2=4^3\)

i,\(4^8.32:256=\left(2^2\right)^8.2^5:2^8=2^{16}.2^5:2^8=2^{21}:2^8=2^{13}\)

Gọi số cần tìm là x ta có

\(x-12+24=87\\ \\\Rightarrow x+12=87\\ \Rightarrow x=87-12=75\)

xy-2x=0

x(y-2)=0

Ta có 2 TH

TH1: x=0

TH2:y-2=0 \(\Rightarrow y=0+2=2\)

Vậy:..............

Em làm rồi đó anh xem đi

ok anh (nếu được)

1