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#)Giải :
\(A=\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\)
\(A< \frac{1}{100.101}+\frac{1}{101.102}+\frac{1}{102.103}+\frac{1}{103.104}+\frac{1}{104.105}\)
\(A< \frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}\)
\(A< \frac{1}{100}-\frac{1}{105}\)
\(A< \frac{1}{2100}=\frac{1}{2^2.3.5^2.7}\)
\(\Rightarrow A< B\)
P/s : Hình như viết sai đề ở chỗ 32 thì phải ??? Bài tui làm là đã sửa lại đề rùi nhé !
Tìm x:
\(\dfrac{4-x}{6-x}\)=\(\dfrac{x-3}{x-8}\)\(\Rightarrow\)(4-x)(x-8)=(6-x)(x-3)
\(\Rightarrow\)12x-x2-32=9x-x2-18
\(\Rightarrow\)3x=14\(\Rightarrow\)x=\(\dfrac{14}{3}\).
\(\dfrac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)
=\(\dfrac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)
=\(\dfrac{7^{48}.5^{30}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{48}}\)
=5.(1-7.22) = 5.(1-28) = 5.(-27) = -135
\(\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)
\(=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}.\left(2^4\right)^2.7^{48}}\)
\(=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)
\(=\frac{7^{48}.5^{30}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{48}}\)
\(=5.\left(1-7.4\right)\)
\(=5.\left(1-28\right)\)
\(=5.\left(-27\right)=-135\)
a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)
\(\Rightarrow x+3=-3\)
\(\Rightarrow x=-6\)
b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)
\(\Rightarrow2x+2=-2\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)
\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)
\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}+\dfrac{5}{48}=\dfrac{6}{48}=\dfrac{1}{8}\)
\(\Leftrightarrow-\dfrac{1}{12}< x< \dfrac{1}{8}\)
=>x=0
c: \(\Leftrightarrow x=\dfrac{-1}{2}\cdot\dfrac{1}{4}=\dfrac{-1}{8}\)
d: \(\Leftrightarrow x^8=x^7\)
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
e: \(\Leftrightarrow3^x=\dfrac{3^{10}}{3^9}=3\)
hay x=1
f: =>x-1=20
hay x=21
Cũng khuya rồi , mình làm câu 1 thôi nhé !
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{2.5^{22}-9.5^{21}}{\left(5^2\right)^{10}}\)
\(\frac{5^{21}.\left(2.5-9\right)}{5^{20}}=5.\left(10-9\right)=5\)
\(H=\frac{8}{1^2\cdot3^2}+\frac{16}{3^2\cdot5^2}+...+\frac{48}{11^2\cdot13^2}\)
\(H=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{5^2}+...+\frac{1}{11^2}-\frac{1}{13^2}\)
\(H=1-\frac{1}{13^2}\)
\(H=\frac{168}{169}\)
Phương thiếu bước nhé
\(H=\frac{8}{1^2.3^2}+\frac{16}{3^2.5^2}+\frac{24}{5^2.7^2}+...+\frac{48}{11^2.13^2}\)
\(H=\frac{3^2-1^2}{1^2.3^2}+\frac{5^2-3^2}{3^2.5^2}+\frac{7^2-5^2}{5^2.7^2}+...+\frac{13^2-11^2}{11^2.13^2}\)
\(H=\frac{3^2}{1^2.3^2}-\frac{1^2}{1^2.3^2}+\frac{5^2}{3^2.5^2}-\frac{3^2}{3^2.5^2}+\frac{7^2}{5^2.7^2}-\frac{5^2}{5^2.7^2}+...+\frac{13^2}{11^2.13^2}-\frac{11^2}{11^2.13^2}\)
\(H=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{5^2}+\frac{1}{5^2}-\frac{1}{7^2}+...+\frac{1}{11^2}-\frac{1}{13^2}\)
\(H=1-\frac{1}{13^2}=1-\frac{1}{169}=\frac{168}{169}\)
Chúc bạn học tốt ~