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\(a,\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow4x^2-4x+1-4x^2-4x-1=4x-12\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
\(b,\left(\frac{5x-7}{2}\right)=\left(\frac{16x+1}{7}\right)\)
\(\Leftrightarrow35x-49=32x+2\)
\(\Leftrightarrow3x=51\Leftrightarrow x=17\)
\(A=\dfrac{x^2-1}{x^2+2x+1}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)^2}=\dfrac{x-1}{x+1}\)
A = \(\dfrac{x+1-2}{x+1}=1+\dfrac{-2}{x+1}\)
Để x có giá trị nguyên thì x + 1 thuộc ước của -2 ( có khi sai nha)
* x + 1 = - 2
--> x = - 3
* x + 1 = 2
--> x = 1
* x + 1 = - 1
--> x = - 2
* x + 1 = 1
--> x = 0
Vậy ,,....
(2x-1)(x2-x+1)-2x3+3x2=2
=>(2x3-2x2+2x-x2+x-1)-2x3+3x2=2
=>2x3-2x2+2x-x2+x-1-2x3+3x2=2
=>2x3-3x2+3x-1-2x3+3x2=2
=>3x-1=2
=>3x=2+1=3
=>x=3:3=1
b)(x+1)(x2+2x+4)-x3-3x2+16=0
=>(x3+2x2+4x+x2+2x+4)-x3-3x2+16=0
=>x3+2x2+4x+x2+2x+4-x3-3x2+16=0
=>x3+3x2+6x+4-x3-3x2+16=0
=>6x+4+16=0
=>6x+20=0
=>6x=0-20=-20
=>x=\(\frac{-20}{6}=\frac{-10}{3}\)
ĐKXĐ: x≠-3; x≠3; x≠6
Ta có: \(\frac{x}{x+3}-\frac{x-2}{x-6}=\frac{x+2}{x^2-9}\)
\(\Leftrightarrow\frac{x\cdot\left(x-3\right)\cdot\left(x-6\right)}{\left(x+3\right)\left(x-3\right)\left(x-6\right)}-\frac{\left(x-2\right)\left(x+3\right)\left(x-3\right)}{\left(x-6\right)\left(x+3\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-6\right)}{\left(x-3\right)\left(x+3\right)\left(x-6\right)}=0\)
\(\Leftrightarrow x^3-6x^2-3x^2+18x-\left(x^3-2x^2-9x+18\right)-\left(x^2-4x-12\right)=0\)
\(\Leftrightarrow x^3-6x^2-3x^2+18x-x^3+2x^2+9x-18-x^2+4x+12=0\)
\(\Leftrightarrow-8x^2+31x-6=0\)
Δ=\(31^2-4\cdot\left(-8\right)\cdot\left(-6\right)=769\)
Vì Δ>0
nên \(\left\{{}\begin{matrix}x_1=\frac{-31-\sqrt{769}}{2\cdot\left(-8\right)}=\frac{31+\sqrt{769}}{16}\\x_2=\frac{-31+\sqrt{769}}{2\cdot\left(-8\right)}=\frac{31-\sqrt{769}}{16}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{\frac{31+\sqrt{769}}{16};\frac{31-\sqrt{769}}{16}\right\}\)