Bài 1:
1: \(P=\left(\dfrac{4}{x-1}-\dfrac{7x+5}{x^3-1}\right):\left(1-\dfrac{x-4}{x^2+x+1}\right)\)
\(=\left(\dfrac{4}{x-1}-\dfrac{7x+5}{\left(x-1\right)\left(x^2+x+1\right)}\right):\dfrac{x^2+x+1-x+4}{x^2+x+1}\)
\(=\dfrac{4\left(x^2+x+1\right)-7x-5}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x^2+5}\)
\(=\dfrac{4x^2+4x+4-7x-5}{\left(x-1\right)\left(x^2+5\right)}\)
\(=\dfrac{4x^2-3x-1}{\left(x-1\right)\left(x^2+5\right)}\)
\(=\dfrac{4x^2-4x+x-1}{\left(x-1\right)\left(x^2+5\right)}=\dfrac{\left(x-1\right)\left(4x+1\right)}{\left(x-1\right)\left(x^2+5\right)}\)
\(=\dfrac{4x+1}{x^2+5}\)
2: \(x^3-x^2-4=0\)
=>\(x^3-2x^2+x^2-4=0\)
=>\(x^2\left(x-2\right)+\left(x-2\right)\left(x+2\right)=0\)
=>\(\left(x-2\right)\left(x^2+x+2\right)=0\)
mà \(x^2+x+2=x^2+x+\dfrac{1}{4}+\dfrac{7}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}>\dfrac{7}{4}>0\forall x\)
nên x-2=0
=>x=2
Thay x=2 vào P, ta được:
\(P=\dfrac{4\cdot2+1}{2^2+5}=\dfrac{9}{9}=1\)
3: \(P=\dfrac{4x+1}{x^2+5}\)
=>\(P-1=\dfrac{4x+1-x^2-5}{x^2+5}=\dfrac{-x^2+4x-4}{x^2+5}\)
\(=-\dfrac{\left(x-2\right)^2}{x^2+5}< =0\forall x\) thỏa mãn ĐKXĐ
=>\(P< =1\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x-2=0
=>x=2
Bài 2:
1: \(2x^2+3y^2+4x=19\)
=>\(\left(2x^2+4x+2\right)+3y^2-21=0\)
=>\(2\left(x+1\right)^2+3y^2=21\)
mà x,y nguyên
nên \(\left[2\left(x+1\right)^2;3y^2\right]\in\left(18;3\right)\)
=>\(\left[\left(x+1\right)^2;y^2\right]\in\left(9;1\right)\)
=>\(\left(x+1;y\right)\in\left\{\left(3;1\right);\left(-3;-1\right);\left(3;-1\right);\left(-3;1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(2;1\right);\left(-4;-1\right);\left(2;-1\right);\left(-4;1\right)\right\}\)
