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1:
\(\Leftrightarrow4\cdot3^x\cdot\dfrac{1}{9}+2\cdot3^x\cdot3=4\cdot3^4+2\cdot3^7\)
\(\Leftrightarrow3^x\cdot\left(\dfrac{4}{9}+6\right)=3^4\cdot\left(4+2\cdot3^3\right)\)
\(\Leftrightarrow3^x=729\)
hay x=6
2: \(\Leftrightarrow3^x\cdot4\cdot\dfrac{1}{3}+3^x\cdot2\cdot9=4\cdot3^6+2\cdot3^9\)
\(\Leftrightarrow3^x\cdot\dfrac{58}{3}=42282\)
=>3x=2187
hay x=7
A) \(2.3^{x+2}+4.3^{x+1}=10.3^6\)
=> \(2.3.3^{x+1}+4.3^{x+1}=10.3^6\)
=> \(6.3^{x+1}+4.3^{x+1}=10.3^6\)
=> \(\left(6+4\right).3^{x+1}=10.3^6\)
=> \(10.3^{x+1}=10.3^6\)
=> \(3^{x+1}=3^6\)
=> \(x+1=6\)
=> \(x=6-1\)
=> \(x=5\)
Vậy \(x=5.\)
B) \(6.8^{x-1}+8^{x+1}=6.8^{19}+8^{21}\)
=> \(6.8^{x-1}+8^{x-1}.8^2=6.8^{19}+8^{19}.8^2\)
=> \(8^{x-1}.\left(6+8^2\right)=8^{19}.\left(6+8^2\right)\)
=> \(8^{x-1}=8^{19}\)
=> \(x-1=19\)
=> \(x=19+1\)
=> \(x=20\)
Vậy \(x=20.\)
Còn câu c) thì mình đang nghĩ nhé.
Chúc bạn học tốt!
2.3x+2+4.3x+1=10.36
=>2.3.3x+1+4.3x+1=10.36
=>(6+4).3x+1=10.36
=>10.3x+1=10.36
=>3x+1=36
=>x+1=6
=>x=5
3S= 3+2.32+3.33+...+101.3101
<=> 2S= 101.3101-(3100+399+398+....+3)-1 (1)
Ta có
A=3100+399+...+3
<=> 3A=3101+3100+...+32
<=> A=\(\frac{3^{101^{ }}-3}{2}\)(2)
Thay (2) vào (1) ta có
S= \(\frac{101.3^{101}-\frac{3^{101}-3}{2}-1}{2}\)
<=> S=\(\frac{3^{101}.201-1}{2}.\frac{1}{2}\)=\(\frac{3^{101}.201-1}{4}\)
Mik nghĩ vậy k bt đúng k
a) \(TH1:x-3\ge0\Rightarrow x\ge3\)
\(\Rightarrow|x-3|=x-3\)
Thay vào biểu thức ban đầu , ta được :
\(12-\left(x-3\right)=5x-8\)
\(\Rightarrow12-x+3=5x-8\)
\(\Rightarrow15-x=5x-8\)
\(\Rightarrow5x+x=15+8\)
\(\Rightarrow6x=23\)\(\Rightarrow x=\frac{23}{6}\ge3\)( thoả mãn )
\(TH2:x-3< 3\Rightarrow x< 3\)
\(\Rightarrow|x-3|=-\left(x-3\right)=-x+3=3-x\)
Thay vào biểu thức ban đầu , ta được
\(12-\left(3-x\right)=5x-8\)
\(\Rightarrow12-3+x=5x-8\)
\(\Rightarrow9+x=5x-8\)
\(\Rightarrow5x-x=9+8\)
\(\Rightarrow4x=17\)
\(\Rightarrow x=\frac{17}{4}\)( loại )
Vậy \(x=\frac{17}{4}\)
b) \(\Rightarrow4.3^{2x}-2.3^{2x}=54\)
\(\Rightarrow3^{2x}.\left(4-2\right)=54\)
\(\Rightarrow3^{2x}.2=54\)
\(\Rightarrow3^{2x}=27=3^3\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\frac{3}{2}\)
Vậy \(x=\frac{3}{2}\)
Câu 1 :
\(\text{a) }B=\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\\ B=\dfrac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot\left(2^3\cdot3\cdot5\right)}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-\left(2\cdot3\right)^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\\ B=\dfrac{2\cdot6}{3\cdot5}\\ B=\dfrac{4}{5}\\ \)
\(\text{b) }C=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\\ C=\dfrac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{2^{29}\cdot3^{18}\left(10-9\right)}{2^{28}\cdot3^{18}\left(15-14\right)}\\ C=\dfrac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}\\ C=2\\ \)
\(\text{c) }D=\dfrac{49^{24}\cdot125^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot4^5}{5^{29}\cdot16^2\cdot7^{48}}\\ D=\dfrac{\left(7^2\right)^{24}\cdot\left(5^3\right)^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot\left(2^2\right)^5}{5^{29}\cdot\left(2^4\right)^2\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-28\right)}{5^{29}\cdot2^8\cdot7^{48}}\\ D=5\cdot\left(-27\right)\\ D=-135\)
Câu 2 :
\(\text{a) }9^{x+1}-5\cdot3^{2x}=324\\ \Leftrightarrow9^x\cdot9-5\cdot9^x=81\cdot4\\ \Leftrightarrow9^x\left(9-5\right)=9^2\cdot4\\ \Leftrightarrow9^x\cdot4=9^2\cdot4\\ \Leftrightarrow9^x=9^2\\ \Leftrightarrow x=2\\ \text{Vậy }x=2\\ \)
Sorry . Mình chỉ biết đến đây thôi
a)\(x^2\left(x+2\right)+4\left(x+2\right)=0\)
\(\Rightarrow\left(x^2+4\right)\left(x+2\right)=0\)
\(x^2+4>0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
b) \(3^{x+2}+4.3^{x+1}+3^{x-1}=6^6\)
\(\Rightarrow3^x.9+3^x.12+3^x.\dfrac{1}{3}=46656\)
\(\Rightarrow3^x\left(9+12+\dfrac{1}{3}\right)=46656\Leftrightarrow3^x.\dfrac{64}{3}=46656\Leftrightarrow3^x=2187\Leftrightarrow x=7\)
Giải:
a) \(x^2\left(x+2\right)+4\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+4\right)=0\)
Vì \(x^2+4>0;\forall x\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy ..
b) \(3^{x+2}+4.3^{x+1}+3^{x-1}=6^6\)
\(\Leftrightarrow3^{x-1+3}+4.3^{x-1+2}+3^{x-1}=6^6\)
\(\Leftrightarrow3^{x-1}\left(3^3+4.3^2+3\right)=6^6\)
\(\Leftrightarrow3^{x-1}.66=6^6\)
\(\Leftrightarrow3^{x-1}=\dfrac{6^6}{66}\)
\(\Leftrightarrow3^x-3=\dfrac{7779}{11}\)
\(\Leftrightarrow3^x=\dfrac{7809}{11}\)
Tìm x rồi kết luận
đề gì mà ghở vậy
lop 7 do