Thực Hiện Phép Tính Sau :
S=1+3^1+3^2+3^3+....+3^30
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\(3S=3.\left(1+3^1+3^2+...+3^{30}\right)\)
\(3S=3+3^2+...+3^{31}\)
\(3S-S=3+3^2+...+3^{31}-\left(1+3^1+...+3^{30}\right)\)
\(2S=3^{31}-1\)
\(S=\frac{3^{31}-1}{2}\)
\(1,=\left(x+3\right)\left(x-2\right):\left(x+3\right)=x-2\\ 2,=\left(x-5\right)\left(x+6\right):\left(x+6\right)=x-5\\ 3,=\left[3x\left(2x-1\right)-5\right]:\left(2x-1\right)=3x.dư.\left(-5\right)\)
1)\(\left(x+x^2-6\right):\left(x+3\right)=\left[x\left(x+3\right)-2\left(x+3\right)\right]:\left(x+3\right)=\left[\left(x+3\right)\left(x-2\right)\right]:\left(x+3\right)=x-2\)
2) \(\left(x+x^2-30\right):\left(x+6\right)=\left[x\left(x+6\right)-5\left(x+6\right)\right]:\left(x+6\right)=\left[\left(x+6\right)\left(x-5\right)\right]:\left(x+6\right)=x-5\)
3) \(\left(5-3x+6x^2\right):\left(2x-1\right)=\left[3x\left(2x-1\right)+5\right]:\left(2x-1\right)=3x+\dfrac{5}{2x-1}\)
\(a.20-\left[30-\left(5-1\right)^2\right]\) \(b.75-\left(3.5^2-4.2^3\right)\)
\(=20-\left[30-4^2\right]\) \(=75-\left(75-4.2^3\right)\)
\(=20-14=6\) \(=75-\left(75-32\right)\)
\(=75-43=32\)
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
\(a.\frac{7}{4}+\frac{5}{6}\div5-\frac{3}{8}\left(-30\right)^2=\frac{7}{4}+\frac{1}{6}-\frac{675}{2}\)
\(=\frac{23}{12}-\frac{675}{2}\)
\(=-\frac{4027}{12}\)
\(b.\frac{4}{7}+\frac{3}{7}\div3-\frac{3}{8}\left(-2\right)^3=\frac{4}{7}+\frac{1}{7}-\left(-3\right)\)
\(=\frac{4}{7}+\frac{1}{7}+3\)
\(=\frac{5}{7}+3\)
\(=\frac{26}{7}\)
K cho mk nhé thks bn nhìu!
CHÚC BẠN HỌC TỐT
\(S=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+.......+\frac{61}{\left(30.31\right)^2}\)
\(=\frac{1}{1^2.2^2}+\frac{1}{2^2.3^2}+....+\frac{1}{30^2.31^2}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{30}-\frac{1}{31}\)
\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-......-\left(\frac{1}{30}-\frac{1}{30}\right)-\frac{1}{31}\)
\(=1-\frac{1}{31}\\ =\frac{31}{31}-\frac{1}{31}=\frac{30}{31}\)
no mình nha
\(=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{13\cdot15}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}=\dfrac{14}{15}\)