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\(\frac{4x^2-6x+5}{2x-1}=2x-2+\frac{3}{2x-1}\)
Để biểu thức có giá trị nguyên thì \(\left(2x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
Với 2x - 1 = 1 => 2x = 2 => x = 1
2x - 1 = -1 => 2x = 0 => x = 0
2x - 1 = 3 => 2x = 4 => x = 2
2x - 1 = -3 => 2x = -2 => x = -1
Vậy x = {1;0;2;-1}
\(\Delta=\left(3+\sqrt{11}\right)^2-4.2.\left(-1\right)=20+6\sqrt{11}+8=28+6\sqrt{11}\)
=> phương trình có 2 nghiệm \(x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{3+\sqrt{11}\pm\sqrt{\Delta}}{4}\)
\(\Delta=\left(\sqrt{11}-3\right)^2-4.2.\left(-1\right)=20-6\sqrt{11}+8=28-6\sqrt{11}\)
\(\Rightarrow\Delta>0\)
\(\Rightarrow\)Phương trình có 2 nghiệm \(x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{\sqrt{11}-3\pm\sqrt{\Delta}}{4}\)
bạn đăng tách ra nhé
a, \(\left(2x+1\right)\left(x-4\right)=\left(2x+1\right)^2\)
\(\Leftrightarrow2x^2-7x-4=4x^2+4x+1\Leftrightarrow2x^2+11x+5=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)=0\Leftrightarrow x=-5;x=-\frac{1}{2}\)
b, sửa đề : \(\left(x-4\right)\left(x^2+4x+16\right)-\left(x^2-6\right)=2\)
\(\Leftrightarrow x^3-64-x^2+6=2\Leftrightarrow x^3-x^2-60=0\Leftrightarrow x=4,27...\)
c, \(\left(2x-1\right)^2-\left(3x+4\right)^2=0\Leftrightarrow\left(2x-1+3x+4\right)\left(2x-1-3x-4\right)=0\)
\(\Leftrightarrow\left(5x+3\right)\left(-x-5\right)=0\Leftrightarrow x=-\frac{3}{5};x=-5\)
d, \(\left(9x+2\right)\left(x-1\right)-\left(3x-1\right)^2=0\)
\(\Leftrightarrow9x^2-7x-2-9x^2+6x-1=0\Leftrightarrow-x-3=0\Leftrightarrow x=-3\)
e, \(\left(2x+3\right)^2-4\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow4x^2+12x+9-4\left(x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow4x^2+12x+9-4\left(x^3-x-x^2+1\right)=0\)
\(\Leftrightarrow4x^2+12x+9-4x^3+4x+4x^2-4=0\)
\(\Leftrightarrow-4x^3+8x^2+16x+5=0\Leftrightarrow x=-0,9...;x=-0,41...;x=3,31...\)
f, \(15x\left(x+4-6x-24\right)=0\Leftrightarrow15\left(-5x-20\right)=0\)
\(\Leftrightarrow-75x-300=0\Leftrightarrow x=-4\)
g, \(\left(4x-10\right)\left(2-3x\right)-30^2=0\)
\(\Leftrightarrow8x-12x^2-20+30x-900=0\Leftrightarrow-12x^2+38x-920=0\)
vô nghiệm
\(C=\left(\frac{x}{x+1}+\frac{1}{x-1}-\frac{2x}{1-x^2}\right):\left(\frac{x+1}{6}\right)\) (ĐK : x khác 1 và x khác -1)
\(=\frac{x\left(x-1\right)+x+1+2x}{\left(x-1\right)\left(x+1\right)}.\frac{6}{x+1}=\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{6}{x+1}=\frac{6.\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)}=\frac{6}{x-1}\)
Để C = x tức là \(\frac{6}{x-1}=x\Leftrightarrow x^2-x-6=0\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
(x+1)(x-2)(2x-1)=0
*)x+1=o>x=-1
*)x-2=0>x=2
*)2x-1=0>x=1/2