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Đặt A=2.22+3.23+4.24+...+n.2nA=2.22+3.23+4.24+...+n.2n
Ta có:
A=2.22+3.23+4.24+...+n.2nA=2.22+3.23+4.24+...+n.2n
⇒2A=2(2.22+3.23+4.24+...+n.2n)⇒2A=2(2.22+3.23+4.24+...+n.2n)
⇒2A=2.23+3.24+4.25+...+n.2n+1⇒2A=2.23+3.24+4.25+...+n.2n+1
⇒2A−A=2.22+(3.23−2.23)+...+(n−n+1).2n−n.2n+1⇒2A−A=2.22+(3.23−2.23)+...+(n−n+1).2n−n.2n+1
⇒A=2.22+23+24+...+2n−n.2n+1⇒A=2.22+23+24+...+2n−n.2n+1
⇒A=22+(22+23+...+2n+1)−(n+1).2n+1⇒A=22+(22+23+...+2n+1)−(n+1).2n+1
⇒A=−22−(22+23+...+2n+1)+(n+1).2n+1⇒A=−22−(22+23+...+2n+1)+(n+1).2n+1
Đặt B=22+23+...+2n+1B=22+23+...+2n+1
⇒2B=23+24+...+2n+2⇒2B=23+24+...+2n+2
⇒2B−B=2n+2−22⇒B=2n+2−22⇒2B−B=2n+2−22⇒B=2n+2−22
⇒A=22−2n+2+22+(n+1).2n+1⇒A=22−2n+2+22+(n+1).2n+1
⇒A=(n+1).2n+1−2n+2⇒A=(n+1).2n+1−2n+2
⇒A=2n+1(n+1−2)⇒A=2n+1(n+1−2)
⇒A=(n−1).2n+1=2(n−1).2n⇒A=(n−1).2n+1=2(n−1).2n
Mà A=2(n−1).2n=2n+10A=2(n−1).2n=2n+10
⇒2(n+1)=210⇒n−1=29⇒2(n+1)=210⇒n−1=29
⇒n−1=512⇒n=513⇒n−1=512⇒n=513
Vậy n=513
Đặt A = 2.2 2 + 3.2 3 + 4.2 4 + . . . + m .2 m Ta có: A = 2.2 2 + 3.2 3 + 4.2 4 + . . . + m .2 m ⇒ 2 A = 2 ( 2.2 2 + 3.2 3 + 4.2 4 + . . . + n .2 n ) ⇒ 2 A = 2.2 3 + 3.2 4 + 4.2 5 + . . . + m .2 m + 1 ⇒ 2 A − A = 2.2 2 + ( 3.2 3 − 2.2 3 ) + . . . + ( m − m + 1 ) .2 m − m .2 m + 1 ⇒ A = 2.2 2 + 2 3 + 2 4 + . . . + 2 n − n .2 n + 1 ⇒ A = 2 2 + ( 2 2 + 2 3 + . . . + 2 m + 1 ) − ( m + 1 ) .2 m + 1 ⇒ A = − 2 2 − ( 2 2 + 2 3 + . . . + 2 m + 1 ) + ( m + 1 ) .2 m + 1 Đặt B = 2 2 + 2 3 + . . . + 2 m + 1 ⇒ 2 B = 2 3 + 2 4 + . . . + 2 m + 2 ⇒ 2 B − B = 2 m + 2 − 2 2 ⇒ B = 2 m + 2 − 2 2 ⇒ A = 2 2 − 2 m + 2 + 2 2 + ( m + 1 ) .2 m + 1 ⇒ A = ( m + 1 ) .2 m + 1 − 2 m + 2 ⇒ A = 2 m + 1 ( m + 1 − 2 ) ⇒ A = ( m − 1 ) .2 m + 1 = 2 ( m − 1 ) .2 n Mà A = 2 ( m − 1 ) .2 m = 2 m + 10 ⇒ 2 ( m + 1 ) = 2 10 ⇒ m − 1 = 2 9 ⇒ m − 1 = 512 ⇒ m = 513 Vậy m = 513
Đặt \(A=2.2^2+3.2^3+4.2^4+...+n.2^n\)
Ta có:
\(A=2.2^2+3.2^3+4.2^4+...+n.2^n\)
\(\Rightarrow2A=2\left(2.2^2+3.2^3+4.2^4+...+n.2^n\right)\)
\(\Rightarrow2A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}\)
\(\Rightarrow2A-A=2.2^2+\left(3.2^3-2.2^3\right)+...+\left(n-n+1\right).2^n-n.2^{n+1}\)
\(\Rightarrow A=2.2^2+2^3+2^4+...+2^n-n.2^{n+1}\)
\(\Rightarrow A=2^2+\left(2^2+2^3+...+2^{n+1}\right)-\left(n+1\right).2^{n+1}\)
\(\Rightarrow A=-2^2-\left(2^2+2^3+...+2^{n+1}\right)+\left(n+1\right).2^{n+1}\)
Đặt \(B=2^2+2^3+...+2^{n+1}\)
\(\Rightarrow2B=2^3+2^4+...+2^{n+2}\)
\(\Rightarrow2B-B=2^{n+2}-2^2\Rightarrow B=2^{n+2}-2^2\)
\(\Rightarrow A=2^2-2^{n+2}+2^2+\left(n+1\right).2^{n+1}\)
\(\Rightarrow A=\left(n+1\right).2^{n+1}-2^{n+2}\)
\(\Rightarrow A=2^{n+1}\left(n+1-2\right)\)
\(\Rightarrow A=\left(n-1\right).2^{n+1}=2\left(n-1\right).2^n\)
Mà \(A=2\left(n-1\right).2^n=2^{n+10}\)
\(\Rightarrow2\left(n+1\right)=2^{10}\Rightarrow n-1=2^9\)
\(\Rightarrow n-1=512\Rightarrow n=513\)
Vậy \(n=513\)
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`@` `\text {Ans}`
`\downarrow`
\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=\dfrac{2^3}{1}=2^3=8\)
_____
\(\dfrac{4^8\cdot9^4}{6^6\cdot8^3}\)
`=`\(\dfrac{\left(2^2\right)^8\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}\)
`=`\(\dfrac{2^{16}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)
`=`\(\dfrac{2^{16}\cdot3^8}{2^{15}\cdot3^6}\)
`=`\(\dfrac{3^2}{2}\) `=`\(\dfrac{9}{2}\)
______
\(\dfrac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}\)
`=`\(\dfrac{\left(3^3\right)^4\cdot2^3-3^{10}\cdot\left(2^2\right)^3}{2^4\cdot3^4\cdot\left(3^2\right)^3}\)
`=`\(\dfrac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^4\cdot3^4\cdot3^6}\)
`=`\(\dfrac{3^{10}\cdot\left(3^2\cdot2^3-2^6\right)}{3^{10}\cdot2^4}\)
`=`\(\dfrac{72-2^6}{2^4}=\dfrac{8}{16}=\dfrac{1}{2}\)
\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=2^3=8\)
\(\dfrac{4^8.9^4}{6^6.8^3}=\dfrac{2^{16}.3^8}{2^6.3^6.2^9}=2.3^2=18\)
\(\dfrac{27^4.2^3-3^{10}.4^3}{6^4.9^3}=\dfrac{3^{12}.2^3-3^{10}.2^6}{2^4.3^4.3^6}=\dfrac{2^3.3^{10}.\left(3^2-2^3\right)}{2^4.3^{10}}=\dfrac{9-8}{2}=\dfrac{1}{2}\)
-3/4. 2/11 + -3/4.9/11+11/4
= -3/4.(2/11+ 9/11)+11/4
= 3/4.1+11/4
= 7/2
\(-\frac{3}{4}.\frac{2}{11}+\left(\frac{-3}{4}\right).\frac{9}{11}+\frac{11}{4}\)
=\(\frac{-3}{4}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{11}{4}\)
=\(\frac{-3}{4}.1+\frac{11}{4}\)
=\(\frac{-3}{4}+\frac{11}{4}\) = \(\frac{8}{11}\)
Hok tốt !