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Đổi \(1dm=10cm\)

Diện tích hình thoi là:

\(8\times10:2=40\left(cm^2\right)\)

Chọn B

Gọi tập hợp tất cả các số nguyên \(x\) thỏa mãn đề bài trên là \(A.\)

\(\Rightarrow A=\left\{-1;0;1\right\}\)

\(\Leftrightarrow B.\left\{-1;0;1\right\}\)

\(320-\left[32-2.\left(5-2\right)^2\right]\)

\(=320-\left[32-2.3^2\right]\)

\(=320-\left[32-2.9\right]\)

\(=320-\left[32-18\right]\)

\(=320-14\)

\(=306\)

\(\Rightarrow B.306\)

Đáp án C.660

Loại B,D vì không chia hết cho 5

Ta có: 4 + 2 + 5 = 11 không chia hết cho 3

6 + 6 + 0 = 12 chia hết cho 3

Vậy 425 không chia hết cho 3,  660 chia hết cho 3

Nên chọn C

\(5^5.5^2\)

\(=5^{5+2}\)

\(=5^7\)

\(\Rightarrow B.5^7\)

B,5⁷  nha

a=7

b=0

tích cho mình nhé

tặng coin cx đc

Bài 4:

\(A=2^1+2^2+...+2^{100}\\ =2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{99}\left(1+2\right)\\ =2.3+2^3.3+...+2^{99}.3\\ =3.\left(2+2^3+...+2^{99}\right)⋮3\)

Bài 2:

\(a,x+3x=4^4:4-4^2\\ 4x=4^{4-1}-4^2\\ 4x=4^3-4^2\\ 4x=64-16\\ 4x=48\\ x=\dfrac{48}{4}=12\\ ---\\ b,\left|5-x\right|-1=4\\ \left|5-x\right|=4+1\\ \left|5-x\right|=5\\ \left[{}\begin{matrix}5-x=5\\5-x=-5\end{matrix}\right.\\ \left[{}\begin{matrix}x=5-5=0\\x=5-\left(-5\right)=10\end{matrix}\right.\)

Em cần giúp gì thế em?

bạn muốn giúp gì vậy ạ

Bài 2:

\(a,\dfrac{-1}{3}.\dfrac{5}{7}=\dfrac{-5}{21}\\ b,\dfrac{1}{2}.\dfrac{-3}{4}=\dfrac{-3}{8}\\ c,\dfrac{19}{7}.\dfrac{7}{15}=\dfrac{19.1}{1.15}=\dfrac{19}{15}\\ d,\dfrac{5}{11}:\left(-9\right)=\dfrac{5}{11}.\dfrac{-1}{9}=\dfrac{-5}{99}\\ e,-1:\dfrac{3}{5}=-1.\dfrac{5}{3}=-\dfrac{5}{3}\\ f,\dfrac{-2}{9}:\dfrac{2}{9}:\dfrac{-9}{14}=-\left(\dfrac{2}{9}:\dfrac{2}{9}\right).\dfrac{-14}{9}=-1.\dfrac{-14}{9}=\dfrac{14}{9}\\ k,\left(-\dfrac{2}{7}\right)^2=\left(-1\right)^2.\left(\dfrac{2}{7}\right)^2=1.\dfrac{2^2}{7^2}=\dfrac{4}{49}\\ l,\dfrac{1}{3}.\left(\dfrac{1}{3}\right)^3=\left(\dfrac{1}{3}\right)^4=\dfrac{1^4}{3^4}=\dfrac{1}{243}\\ m,\left(-\dfrac{1}{2}\right)^2.\left(\dfrac{1}{2}\right)^3=\left(-1\right)^2.\left(\dfrac{1}{2}\right)^{2+3}=1.\left(\dfrac{1}{2}\right)^5=1.\dfrac{1^5}{2^5}=\dfrac{1}{32}\)

Bài 1:

\(a,\dfrac{1}{-8}+\dfrac{-5}{8}=\dfrac{-1}{8}+\dfrac{-5}{8}=\dfrac{-\left(1+5\right)}{8}=-\dfrac{6}{8}=\dfrac{-6:2}{8:2}=-\dfrac{3}{4}\\ b,\dfrac{1}{7}+\dfrac{-3}{7}=\dfrac{1-3}{7}=-\dfrac{2}{7}\\ c,\dfrac{-12}{35}+\dfrac{-7}{35}=\dfrac{-\left(12+7\right)}{35}=\dfrac{-19}{35}\\ d,\dfrac{1}{6}+\dfrac{2}{5}=\dfrac{1.5+2.6}{6.5}=\dfrac{5+12}{30}=\dfrac{17}{30}\\ e,\dfrac{3}{5}+\dfrac{-7}{4}=\dfrac{3.4-7.5}{5.4}=\dfrac{12-35}{20}=\dfrac{-23}{20}\\ g,-2-\left(-\dfrac{1}{5}\right)=-2+\dfrac{1}{5}=\dfrac{-2.5+1}{5}=\dfrac{-9}{5}\\ h,\dfrac{2}{3}-\left(-1\right)=\dfrac{2}{3}+1=\dfrac{5}{3}\\ i,4-\dfrac{2}{3}=\dfrac{4.3-2}{3}=\dfrac{12-2}{3}=\dfrac{10}{3}\\ j,\dfrac{3}{4}-2=\dfrac{3-2.4}{4}=\dfrac{-5}{4}\\ k,-1-\left(-\dfrac{2}{3}\right)=-1+\dfrac{2}{3}=\dfrac{-1.3+2}{3}=\dfrac{-3+2}{3}=-\dfrac{1}{3}\)