34 : 2 = ?
90 : 3 =?
42 x 2 =?
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42 x 2 = 84
34 x 2 = 68
12 x 3 = 36
21 x 3 = 63
k mk,mk k lại ^_-
B=\(\frac{3\sqrt{x}+4}{3\sqrt{x}-2}-\frac{42\sqrt{x}+34}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{(3\sqrt{x}+4)(5\sqrt{x}+7)-42\sqrt{x}-34}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{15x+20\sqrt{x}+21\sqrt{x}+28-42\sqrt{x}-34}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{15x-\sqrt{x}-6}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+3\right)}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{5\sqrt{x}+3}{5\sqrt{x}+7}\)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
Ta có:
1/30+1/42+1/56+1/72+1/90+1/110+1/132 -x = 2/3
=>1/(5.6) + 1/(6.7) + 1/(7.8) + (1/8.9) + 1/(9.10) + 1/(10.11) + 1/(11.12) - x = 2/3
=>1/5-1/6+1/6-1/7+...+1/11-1/12 - x = 2/3
=>1/5-1/12 - x = 2/3
=>7/60 - x = 2/3
=> x = 7/60 - 2/3
=> x = -11/20
Ta có:
1/30+1/42+1/56+1/72+1/90+1/110+1/132 -x = 2/3
=>1/(5.6) + 1/(6.7) + 1/(7.8) + (1/8.9) + 1/(9.10) + 1/(10.11) + 1/(11.12) - x = 2/3
=>1/5-1/6+1/6-1/7+...+1/11-1/12 - x = 2/3
=>1/5-1/12 - x = 2/3
=>7/60 - x = 2/3
=> x = 7/60 - 2/3
=> x = -11/20
34 : 2 =17
90 : 3 = 30
42 x 2 = 84
k mik nhé mik bị trừ oan nhiều điểm lắm
34 : 2 = 17
90 : 3 = 30
42 x 2 = 84