Giải phương trình
C1 : $3.25^{x-2} + (3x-10)5^{x-2} + 3-x =0$
C2: $(5-\sqrt{21})^{x}+7(5+\sqrt{21})^{x}=2^{x+3}$
C3: $(2-\sqrt{3})^{x} + (2+\sqrt{3})^{x} = 4^{x}$
C4: $2^{x} + 3^{x}=6^{x}+1$
Câu 1. Đặt $5^{x-2} = t,\ t>0$ pt đưa về

$3t^2+(3x-10)t+3-x=0$

$\Delta = (3x-10)^2 -4.3.(3-x)=(3x-8)^2$

$t= 3-x;\ t= \dfrac{1}{3}$

Tôi giải cái khó thôi nha, $5^{x-2}=3-x$ ta thấy $VT$ và $VP$ lần lượt là 2 hàm đồng biến và nghịch biến, do đó pt có nghiệm duy nhất $x=2$
Câu 2) Ta có: $\frac{5-\sqrt{21}}{2}.\frac{5+\sqrt{21}}{2}=1\Rightarrow \frac{5-\sqrt{21}}{2}=\frac{1}{\frac{5+\sqrt{21}}{2}}$
Từ đó pt:  $(5-\sqrt{21})^{x}+7(5+\sqrt{21})^{x}=2^{x+3}\Leftrightarrow 1+7(\frac{5+\sqrt{21}}{2})^{2x}=8(\frac{5+\sqrt{21}}{2})^x$
... dễ ùi nhé!

Câu 4) $2^{x} + 3^{x}=6^{x}+1\Leftrightarrow 2^x+3^x=2^x.3^x+1\Leftrightarrow -2^x(3^x-1)+3^x-1=0\Leftrightarrow (3^x-1)(1-2^x)=0\Leftrightarrow ...$
* các câu còn lại mình sẽ giải sau?
bạn cứ biến đổi từ từ là ra à ? –  HọcTạiNhà 24-11-14 06:15 PM
câu 2 ko hiểu lắm –  HIHI 23-11-14 10:51 PM

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