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Question

tìm x,biết:a) $4^x+4^{x+3}=4160$b)$2^{x-1}+5.2^{x-2}=\frac{7}{32}$

Phạm Nguyễn Tất Đạt · Accepted Answer

b)$2^{x-1}+5\cdot2^{x-2}=\frac{7}{32}$$2^x:2+5\cdot2^x:2^2=\frac{7}{32}$$2^x:2+2^x:\frac{4}{5}=\frac{7}{32}$$2^x\cdot\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}$$2^x\cdot\frac{7}{4}=\frac{7}{32}$$2^x=\frac{7}{32}:\frac{7}{4}=\frac{1}{8}$$2^x=\frac{2^0}{2^3}=2^{-3}$$\Rightarrow x=-3$Câu trả lời: b)$2^{x-1}+5\cdot2^{x-2}=\frac{7}{32}$$2^x:2+5\cdot2^x:2^2=\frac{7}{32}$$2^x:2+2^x:\frac{4}{5}=\frac{7}{32}$$2^x\cdot\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}$$2^x\cdot\frac{7}{4}=\frac{7}{32}$$2^x=\frac{7}{32}:\frac{7}{4}=\frac{1}{8}$$2^x=\frac{2^0}{2^3}=2^{-3}$$\Rightarrow x=-3$

Nguyễn Huy Tú · Answer

a) $4^x+4^{x+3}=4160$$\Rightarrow4^x+4^x.4^3=4160$$\Rightarrow4^x.\left(1+4^3\right)=4160$$\Rightarrow4^x.65=4160$$\Rightarrow4^x=64$$\Rightarrow4^x=4^4$$\Rightarrow x=4$Vậy $x=4$b) $2^{x-1}+5.2^{x-2}=\frac{7}{32}$$\Rightarrow2^x.\frac{1}{2}+5.2^x.\frac{1}{4}=\frac{7}{32}$$\Rightarrow2^x.\left(\frac{1}{2}+5.\frac{1}{4}\right)=\frac{7}{32}$$\Rightarrow2^x.\frac{7}{4}=\frac{7}{32}$$\Rightarrow2^x=\frac{7}{32}:\frac{7}{4}$$\Rightarrow2^x=\frac{1}{8}$$\Rightarrow2^x=2^{-3}$$\Rightarrow x=-3$Vậy $x=-3$  Câu trả lời: a) $4^x+4^{x+3}=4160$$\Rightarrow4^x+4^x.4^3=4160$$\Rightarrow4^x.\left(1+4^3\right)=4160$$\Rightarrow4^x.65=4160$$\Rightarrow4^x=64$$\Rightarrow4^x=4^4$$\Rightarrow x=4$Vậy $x=4$b) $2^{x-1}+5.2^{x-2}=\frac{7}{32}$$\Rightarrow2^x.\frac{1}{2}+5.2^x.\frac{1}{4}=\frac{7}{32}$$\Rightarrow2^x.\left(\frac{1}{2}+5.\frac{1}{4}\right)=\frac{7}{32}$$\Rightarrow2^x.\frac{7}{4}=\frac{7}{32}$$\Rightarrow2^x=\frac{7}{32}:\frac{7}{4}$$\Rightarrow2^x=\frac{1}{8}$$\Rightarrow2^x=2^{-3}$$\Rightarrow x=-3$Vậy $x=-3$

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