Đặt B=\(\frac{x-1}{x^2-5x+7}\)=>\(\frac{1}{B}\)=\(\frac{x^2-5x+7}{x-1}\)=\(x-4\)\(+\frac{3}{x-1}\)

Để B nguyên thì x-1 thuộc ư(3)={-3;-1;1;3}

Vậy để B thuộc Z thì x={-1;0;2;4)

\(2x^4-21x^3+74x^2-105x+50=0\)

\(< =>2x^4-10x^3-11x^3+55x^2+19x^2-95x^2-10x+50=0\)

\(< =>2x^3\left(x-5\right)-11x^2\left(x-5\right)+19x\left(x-5\right)-10\left(x-5\right)=0\)

\(< =>\left(x-5\right).\left(2x^3-11x^2+19x-10\right)=0\)

\(< =>\left(x-5\right).\left(2x^3-2x^2-9x^2+9x+10x-10\right)=0\)

\(< =>\left(x-5\right).\left(x-1\right).\left(2x^2-9x+10\right)=0\)

\(2x^2-9x+10\ge0\)

\(< =>x=5\)hoặc \(x=1\)

Vậy S = 1 hoặc 5

a)\(\sqrt{x+1}\left(x+4\right)=\left(x+18\right)\sqrt{6+x}-3x-40\)

\(pt\Leftrightarrow\sqrt{x+1}\left(x+4\right)-14=\left(x+18\right)\sqrt{6+x}-63-3x-9\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+4\right)^2-196}{\sqrt{x+1}\left(x+4\right)+14}=\frac{\left(x+18\right)^2\left(x+6\right)-3969}{\left(x+18\right)\sqrt{6+x}+63}-3\left(x-3\right)\)

\(\Leftrightarrow\frac{x^3+9x^2+24x-180}{\sqrt{x+1}\left(x+4\right)+14}-\frac{x^3+42x^2+540x-2025}{\left(x+18\right)\sqrt{6+x}+63}+3\left(x-3\right)=0\)

\(\Leftrightarrow\frac{\left(x-3\right)\left(x^2+12x+60\right)}{\sqrt{x+1}\left(x+4\right)+14}-\frac{\left(x-3\right)\left(x^2+45x+675\right)}{\left(x+18\right)\sqrt{6+x}+63}+3\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{x^2+12x+60}{\sqrt{x+1}\left(x+4\right)+14}-\frac{x^2+45x+675}{\left(x+18\right)\sqrt{6+x}+63}+3\right)=0\)

Pt trong ngoặc to to kia vô nghiệm

Suy ra x=3

b)\(3\left(\sqrt{x+9}-\sqrt{x+1}\right)=4-4x\)

\(pt\Leftrightarrow\sqrt{x+9}-\sqrt{x+1}=\frac{4-4x}{3}\)

\(\Leftrightarrow2x+10-2\sqrt{\left(x+1\right)\left(x+9\right)}=\frac{16x^2-32x+16}{9}\)

\(\Leftrightarrow-2\sqrt{\left(x+1\right)\left(x+9\right)}=\frac{16x^2-32x+16}{9}-\left(2x+10\right)\)

\(\Leftrightarrow4\left(x+1\right)\left(x+9\right)=\frac{256x^4-1600x^3+132x^2+7400x+5476}{81}\)

\(\Leftrightarrow\frac{-64\left(x^2-5x-5\right)\left(4x^2-5x-8\right)}{81}=0\)

mỗi lần bình phương tự rút ra điều kiện mà khử nghiệm nhé :v

hi, ^.^, thanks bn nhìu nha >3

a: \(A=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)

\(=\sqrt{2}-1-\sqrt{2}-1\)

=-2

b: \(B=2\left(\sqrt{5}-2\right)-\left(2\sqrt{5}-1\right)\)

\(=2\sqrt{5}-4-2\sqrt{5}+1=-3\)