$log_{3}(x-2)=log_{4}(x^{2}-4x+3)$
dk : $x\geq 3$
dat $log_{3}(x-2)=t$(1)
va      $log_{4}(x^{2}-4x+3)=t$ (2)
tu (1) =>$x=3^{t} +2$
thay vao (2) ta co :$(3^{t}+2)^{2}-4(3^{t}+2)+3=4^{t}$
      <=>  $3^{2t}-2^{2t}=1$          (day la ham f(t))
xet $f(t)=3^{2t}-2^{2t}-1$
=> dao ham $f(t)=2.3^{2t}ln(3)-2.2^{2t}.ln(2) \geq 0$
day la ham dong bien
nhan thay $f(\frac{1}{2})$=0
=>$t=\frac{1}{2}$ la mot nghiem cua phuong trinh
voi $t>\frac{1}{2}$ va $t<\frac{1}{2}$ khong phai la nghiem cua phuong trinh

vay$t=\frac{1}{2}$ la nghiem cua phuong trinh
voi $t=\frac{1}{2}$ =>$x=\sqrt{3}+2$
vay phuong trinh co nghiem la: $x=\sqrt{3}+2$









Đk: $x>3$

$log_{3}(x-2)=log_{4}(x^2-4x+3)=t \Rightarrow \begin{cases}x-2=3^t \\x^2-4x+3=4^t \end{cases} $$\Leftrightarrow (2+3^t)^2-4(2+3^t)+3=4^t \Leftrightarrow 4+4.3^t+9^t-8-4.3^t+3=4^t \Leftrightarrow 5^t=1 \Leftrightarrow t=0$ 
Khi đó t a có:
                                \begin{cases}x-2=1 \\ x^2-4x+2=0 \end{cases} 
Giải hệ này ra ta th ấy không tìm được x thỏa mãn.
Vậy phtrinh vô nghiệm.
mihn da lam va dang len roi do cach cua pan chi dung phan dau con phan sau thi sai –  lehue2141995 13-03-13 05:12 PM
Ban thu lam theo cach cua ban minh xem voi. Vi theo de bai la log_3(x-2)=log_4(x^2-4x 3) nen minh dat log_3(x-2)=log_4(x^2-4x 3) roi suy ra he tren ban a –  quachthailang2706 13-03-13 04:57 PM
cach lam sai roi –  lehue2141995 13-03-13 04:44 PM

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