cứu emm vs ạaa
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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:
\(a,\dfrac{1}{-8}+\dfrac{-5}{8}=\dfrac{-1}{8}+\dfrac{-5}{8}=\dfrac{-\left(1+5\right)}{8}=-\dfrac{6}{8}=\dfrac{-6:2}{8:2}=-\dfrac{3}{4}\\ b,\dfrac{1}{7}+\dfrac{-3}{7}=\dfrac{1-3}{7}=-\dfrac{2}{7}\\ c,\dfrac{-12}{35}+\dfrac{-7}{35}=\dfrac{-\left(12+7\right)}{35}=\dfrac{-19}{35}\\ d,\dfrac{1}{6}+\dfrac{2}{5}=\dfrac{1.5+2.6}{6.5}=\dfrac{5+12}{30}=\dfrac{17}{30}\\ e,\dfrac{3}{5}+\dfrac{-7}{4}=\dfrac{3.4-7.5}{5.4}=\dfrac{12-35}{20}=\dfrac{-23}{20}\\ g,-2-\left(-\dfrac{1}{5}\right)=-2+\dfrac{1}{5}=\dfrac{-2.5+1}{5}=\dfrac{-9}{5}\\ h,\dfrac{2}{3}-\left(-1\right)=\dfrac{2}{3}+1=\dfrac{5}{3}\\ i,4-\dfrac{2}{3}=\dfrac{4.3-2}{3}=\dfrac{12-2}{3}=\dfrac{10}{3}\\ j,\dfrac{3}{4}-2=\dfrac{3-2.4}{4}=\dfrac{-5}{4}\\ k,-1-\left(-\dfrac{2}{3}\right)=-1+\dfrac{2}{3}=\dfrac{-1.3+2}{3}=\dfrac{-3+2}{3}=-\dfrac{1}{3}\)
Đề mờ quá. Bạn nên gõ đề ra để mọi người hỗ trợ tốt hơn nhé.
Bài 1:
\(A=3+3^2+...+3^{100}\)
=>\(3\cdot A=3^2+3^3+...+3^{101}\)
=>\(3A-A=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)
=>\(2A=3^{101}-3\)
=>\(2A+3=3^{101}\)
mà \(2A+3=3^n\)
nên n=101
Bài 2:
a: \(M=3+3^2+3^3+3^4+...+3^{100}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)
\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)\)
\(=12\left(1+3^2+...+3^{98}\right)⋮12\)
=>\(M=4\cdot3\cdot\left(1+3^2+...+3^{98}\right)⋮4\)
b: \(M=3+3^2+...+3^{100}\)
=>\(3M=3^2+3^3+...+3^{101}\)
=>\(3M-M=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)
=>\(2M=3^{101}-3\)
=>\(2M+3=3^{101}\)
=>n=101
a: Xét (O) có
ΔABC nội tiếp
AB là đường kính
Do đó: ΔABC vuông tại C
Bài 1:
a: \(k=\dfrac{y}{x}=\dfrac{y_1}{x_1}=\dfrac{2}{3}\)
=>\(y=\dfrac{2}{3}x\)
b: Khi \(x_2=9\) thì \(y_2=\dfrac{2}{3}\cdot x_2=\dfrac{2}{3}\cdot9=6\)
c: Khi \(y_3=8\) thì \(\dfrac{2}{3}\cdot x_3=8\)
=>\(x_3=8:\dfrac{2}{3}=12\)
Bài 2:
a: x,y tỉ lệ thuận
nên \(\dfrac{x_1}{x_2}=\dfrac{y_1}{y_2}\)
=>\(\dfrac{6}{y_2}=\dfrac{4}{9}\)
=>\(y_2=6\cdot\dfrac{9}{4}=\dfrac{3}{2}\cdot9=\dfrac{27}{2}\)
b: \(\dfrac{x_1}{x_2}=\dfrac{y_1}{y_2}\)
=>\(\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}=\dfrac{x_1+x_2}{y_1+y_2}=\dfrac{2}{-10}=-\dfrac{1}{5}\)
=>\(\dfrac{x}{y}=-\dfrac{1}{5}\)
=>y=-5x