\(\dfrac{\left(-0.25\right)^{-5}\cdot9^4\left(-2\right)^{-3}-2^{-3}\cdot6^9}{2^9\cdot3^6+6^6\cdot40}\)

\(=\dfrac{2^7\cdot3^8-2^6\cdot3^9}{2^9\cdot3^6+3^6\cdot2^9\cdot5}\)

\(=\dfrac{2^6\cdot3^8\left(2-3\right)}{2^9\cdot3^6\cdot6}\)

\(=\dfrac{1}{2^3}\cdot3^2\cdot\dfrac{-1}{6}\)

\(=\dfrac{-9}{6\cdot2^3}=\dfrac{-3}{2^4}=\dfrac{-3}{16}\)

1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42

2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57

3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165

4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315

5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5

\(\frac{\left(-0,25\right)^{-5}.9^4.\left(-2\right)^{-3}-2^{-2}.6^9}{2^9.3^6+6^6.40}\)

\(=\frac{\left(-4\right)^5.\left(3^2\right)^4.\left(-2\right)^{-3}-2^{-2}.\left(3.2\right)^9}{2^9.3^6+\left(2.3\right)^6.2^3.5}\)

\(=\frac{-\left(2^2\right)^5.3^8.\left(-2\right)^{-3}-2^{-2}.3^9.2^9}{2^9.3^6+2^6.3^6.2^3.5}\)

\(=\frac{-2^{10}3^8.\left(-2\right)^{-3}-2^7.3^9}{2^9.3^6+2^9.3^6.5}\)

\(=\frac{2^73^8.-2^7.3^9}{2^9.3^6+2^9.3^6.5}\)

\(=\frac{2^7.3^8.\left(1-3\right)}{2^9.3^6.\left(1+5\right)}\)

\(=\frac{3^2.\left(-2\right)}{2^2.6}\)

\(=\frac{-3}{4}\)

1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}.3\)

\(\left|4-2x\right|=1\)

=>\(4-2x=\pm1\)

+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)

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

\(2x=3\) \(2x=4+1\)

\(x=3:2\) \(2x=5\)

\(x=1,5\) \(x=5:2\)

Vậy x=1,5 \(x=2,5\)

Vậy x=2,5

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

\(9:\left|x+\left(-1\right)\right|=-3\)

\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)

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

=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )

Vậy x = \(\varnothing\)

3) \(\left|\dfrac{3}{2}-x\right|+\left(-\dfrac{4}{5}\right)=-\dfrac{14}{5}\)

\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)-\left(-\dfrac{4}{5}\right)\)

\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)+\dfrac{4}{5}\)

\(\left|\dfrac{3}{2}-x\right|=-\dfrac{10}{5}\)

\(\left|\dfrac{3}{2}-x\right|=-2\)

=> \(\dfrac{3}{2}-x\) không có giá trị nào thỏa mãn ( Vì giá trị tuyệt đối không thể là số nguyên âm )

Vậy x=\(\varnothing\)

cho xin đáp án chi tiết điiiii

\(=(\frac{3}{4}-\frac{1}{4})-\frac{7}{3}+\frac{9}{2}-\frac{5}{6}=\frac{1}{2}+\frac{9}{2}-\frac{14}{6}-\frac{5}{6}=5-\frac{19}{6}=\frac{11}{6}\)

Kết bạn nha bạn

Bài làm

       \(\frac{3}{4}-0,25-\left(\frac{7}{3}+\frac{-9}{2}\right)-\frac{5}{6}\)

\(=\frac{3}{4}-\frac{1}{4}-\left(\frac{14}{6}+\frac{-27}{6}\right)-\frac{5}{6}\)

\(=\frac{1}{2}+\frac{13}{6}-\frac{5}{6}\)

\(=\frac{3}{6}+\frac{13}{6}-\frac{5}{6}\)

\(=\frac{11}{6}\)

# Học tốt #

3/4-0,25-(7/3+(-9/2) )-5/6

=3/4-1/4-(14/6+(-27/6) )-5/6

=1/2-(-13/6)-5/6

=3/6+13/6-5/6

=(3+13-5)/6

=11/6

mn giúp e vs ạT^T

\(a,=\dfrac{3}{2}-\dfrac{5}{6}:\dfrac{1}{4}+\sqrt{\dfrac{1}{4}-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{10}{3}+\sqrt{\dfrac{1}{2}}=-\dfrac{11}{6}+\dfrac{\sqrt{2}}{2}=\dfrac{-33+3\sqrt{2}}{6}\)

\(b,=-\dfrac{4}{3}\cdot\dfrac{9}{2}+\dfrac{13}{12}\cdot\left(-\dfrac{8}{13}\right)=6-\dfrac{2}{3}=\dfrac{16}{3}\\ c,=\dfrac{1}{4}-\left(-\dfrac{1}{6}:4-8\cdot\dfrac{1}{16}\right)=\dfrac{1}{4}-\left(-\dfrac{1}{24}-\dfrac{1}{2}\right)\\ =\dfrac{1}{4}-\dfrac{13}{24}=-\dfrac{7}{24}\\ d,=\dfrac{3^{11}\cdot5^{11}\cdot5^7\cdot3^4}{5^{18}\cdot3^{18}}=\dfrac{1}{3^3}=\dfrac{1}{27}\)