a) \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3x-24}\) \(ĐK:x\ne8\)

\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3\left(x-8\right)}\)

\(\Leftrightarrow\frac{3.3}{6.\left(x-8\right)}+\frac{6.\left(3x-20\right)}{6\left(x-8\right)}-\frac{2\left(3x-102\right)}{6\left(x-8\right)}=\frac{-1}{8}\)

\(\Leftrightarrow\frac{9+18x-120-6x+204}{6\left(x-8\right)}=\frac{-1}{8}\)

\(\Leftrightarrow\frac{12x+93}{6\left(x-8\right)}=\frac{-1}{8}\)

\(\Leftrightarrow8\left(12x+93\right)=-6\left(x-8\right)\)

\(\Leftrightarrow96x+744=-6x+48\)

\(\Leftrightarrow102x=-696\)

\(\Leftrightarrow x=\frac{-116}{17}\) (nhận)

Vậy .....

b) \(\frac{1}{3-x}+\frac{14}{x^2-9}=\frac{x-4}{3+x}+\frac{7}{3+x}\) \(ĐK:x\ne\pm3\)

\(\Leftrightarrow\frac{1}{3-x}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{x-4}{3+x}+\frac{7}{3+x}\)

\(\Leftrightarrow-\frac{3+x}{\left(x-3\right)\left(3+x\right)}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{-3-x+14}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)

\(\Leftrightarrow-3-x+14=x^2-3x-4x+12+7x-21\)

\(\Leftrightarrow x=-5\) (nhận)

Vậy ....

b)  \(\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)=1680\)

\(\Leftrightarrow\left(x-4\right)\left(x-7\right)\left(x-5\right)\left(x-6\right)=1680\)

\(\Leftrightarrow\left(x^2-11x+28\right)\left(x^2-11x+28+2\right)-1680=0\)

\(\Leftrightarrow\left(x^2-11x+28\right)^2+2\left(x^2-11x+28\right)+1-1681=0\)

\(\Leftrightarrow\left(x^2-11x+28+1\right)^2-41^2=0\)

\(\Leftrightarrow\left(x^2-11x+29-41\right)\left(x^2-11x+29+41\right)=0\)

\(\Leftrightarrow\left(x^2-11x-12\right)\left(x^2-11x+70\right)=0\)

    Th1:  \(x^2-11x-12=0\Leftrightarrow x^2+x-12x-12=0\Leftrightarrow\left(x-12\right)\left(x+1\right)=0\)

                \(\Leftrightarrow x-12=0\Leftrightarrow x=12\)   hoặc    \(x+1=0\Leftrightarrow x=-1\)

   Th2:\(x^2-11x+70=0\Leftrightarrow x^2-2.x.\frac{11}{2}+\left(\frac{11}{2}\right)^2+\frac{159}{4}=0\Leftrightarrow\left(x-\frac{11}{2}\right)^2+\frac{159}{4}=0\)

           Vì\(\left(x-\frac{11}{2}\right)^2\ge0\Rightarrow\left(x+\frac{11}{2}\right)^2+\frac{159}{4}\ge\frac{159}{4}\)

         Mà ta có   \(\left(x+\frac{11}{2}\right)^2+\frac{159}{4}=0\)    Nên k có giá trị của x

Vậy tập nghiệm của phương trình là   \(S=\left\{12;-1\right\}\)

a) x=-3,

x=2;

x = -(căn bậc hai(3)*căn bậc hai(5)*i+1)/2;

x = (căn bậc hai(3)*căn bậc hai(5)*i-1)/2;

a) x=-3 hoặc x=2

b) x=-1 hoặc x=12

đặt ẩn phụ đi là nhah nhất

Áp dụng BĐT:\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

Ta có: \(\left|\sqrt{x-1}+2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}+2+3-\sqrt{x-1}\right|=5\)

Dấu \(=\)xảy ra khi \(AB\ge0\)

dat \(\sqrt{x-1}\) = t

ta có: \(\sqrt{x+3+4t}\)+ \(\sqrt{x+8-6t}\)= 5

     x + 3 + 4t + x + 8 - 6t = 25

   2x - 2t = 14 ( chia cả 2 vế cho 2)

   x - t = 7

   t = x - 7

  thay t = \(\sqrt{x}-1\)vào ta được:

 x - 7 = \(\sqrt{x-1}\)

( x - 7 )² = x - 1

x² -14x + 49 = x - 1

x² - 15x + 50 = 0

​k biết đúng hay k

OoO Ledegill2 OoO. Ban co the giai thich ro hon giup minh duoc khong. hi