ok de ot

\(A=\frac{1}{3589}.7\frac{1}{297}-3\frac{3588}{3589}.\frac{2}{297}-\frac{7}{3589}-\frac{3}{3589.297}\)

\(A=\frac{1}{3589}.\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{3589}\right).2.\frac{1}{297}-7.\frac{1}{3589}-3.\frac{1}{3689}.\frac{1}{297}\)

\(A=7.\frac{1}{3689}+\frac{1}{3589}.\frac{1}{297}-8.\frac{1}{297}+2.\frac{1}{3589}.\frac{1}{297}-7.\frac{1}{3589}\)

\(A=-8.\frac{1}{297}\)

\(A=\frac{-8}{297}\)

a, \(A=\frac{x-1}{x+1}=\frac{x+1-1-1}{x+1}=\frac{x+1-2}{x+1}=1-\frac{2}{x+1}\)

Để  \(A\in z\) thì \(x+1\inƯ\left(2\right)=\left(-2;-1:1;2\right)\)

\(x+1=-2\Rightarrow x=-3\)

\(x+1=-1\Rightarrow x=-2\)

\(x+1=1\Rightarrow x=0\)

\(x+1=2\Rightarrow x=1\)

Vậy \(x=\left(-3;-2;0;1\right)\)thì \(A\in z\)

b, \(A=\frac{x+1}{x-2}=1+\frac{3}{x-2}\)

Để \(A\in z\)thì \(x-2\inƯ\left(3\right)=\left(-3;-1;1;3\right)\)

\(x-2=-3\Rightarrow x=-1\)

\(x-2=-1\Rightarrow x=1\)

\(x-2=1\Rightarrow x=3\)

\(x-2=3\Rightarrow x=5\)

Vậy \(x=\left(-1;1;3;5\right)\)thì \(A\in z\)

c, \(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)\(ĐK:\)\(x\ge0;x\ne9\)

\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để \(A\in z\)thì \(\sqrt{x}-3\inƯ\left(4\right)=\left(-4;-2;-1;1;2;4\right)\)

\(\sqrt{x}-3=-4\Rightarrow\sqrt{x}=-1VN\)

\(\sqrt{x}-3=-2\Rightarrow\sqrt{x}=1\Rightarrow x=1\) 

\(\sqrt{x}-3=-1\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(\sqrt{x}-3=1\Rightarrow\sqrt{x}=4\Rightarrow x=16\)

\(\sqrt{x}-3=2\Rightarrow\sqrt{x}=5\Rightarrow x=25\)

\(\sqrt{x}-3=4\Rightarrow\sqrt{x}=7\Rightarrow x=49\)

Vậy \(x=\left(1;4;16;25;49\right)\)thì \(A\in z\)

d, \(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}\) \(ĐK:\)\(x\ge0;x\ne1\)

\(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)

Để \(A\in z\) thì \(\sqrt{x}-1\inƯ\left(2\right)=\left(-2;-1;1;2\right)\)

\(\sqrt{x}-1=-2\Rightarrow\sqrt{x}=-1VN\)

\(\sqrt{x}-1=-1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)

\(\sqrt{x}-1=1\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(\sqrt{x}-1=2\Rightarrow\sqrt{x}=3\Rightarrow x=9\)

Vậy \(x=\left(0,4,9\right)\)thì \(A\in z\)

\(a,A=\frac{x-1}{x+1}\)

Để \(A\in Z\)

\(\Rightarrow\frac{x-1}{x+1}\in Z\)

\(\Rightarrow\frac{x+1-2}{x+1}\in Z\)

\(\Rightarrow1-\frac{2}{x+1}\in Z\)

\(\Rightarrow\frac{2}{x+1}\in Z\)

\(\Rightarrow x+1\in U_{\left(2\right)}\)

\(\Rightarrow x+1=\left\{-2,-1,1,2\right\}\)

\(\Rightarrow x=\left\{-3,-2,0,1\right\}\)

Bạn ghi lại đề đi bạn

Nếu x = y = z = t thì:
x⁴ + y⁴ + z⁴ + t⁴ = 4xyzt
=> x⁴ + x⁴ + x⁴ + x⁴ = 4xxxx
=> 4x⁴ = 4x⁴    ( đpcm )

Ta có:\(x^4+y^4+z^4+t^4=4xyzt\)

\(\Rightarrow\left(x^4-2x^2y^2+y^4\right)+\left(z^4-2z^2t^2+t^4\right)+2\left(x^2y^2-2xyzt+z^2t^2\right)=0\)

\(\Leftrightarrow\left(x^2-y^2\right)+\left(z^2-t^2\right)+2\left(x^2y^2-z^2t^2\right)=0\)

Mà x,y,z,t là các số nguyên dương nên x=y=z=t(đpcm)

đi

\(\dfrac{3x^2}{2}+y^2+z^2+yz=1\)

\(\Leftrightarrow\dfrac{3}{2}x^2+\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}=1\)

Áp dụng BĐT Bunhiacopxki:

\(\left(\dfrac{2}{3}+1+\dfrac{1}{3}\right)\left(\dfrac{3}{2}x^2+\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}\right)\ge\left(\sqrt{\dfrac{2}{3}.\dfrac{3}{2}x^2}+\sqrt{1.\left(y+\dfrac{z}{2}\right)^2}+\sqrt{\dfrac{1}{3}.\dfrac{3z^2}{4}}\right)^2\)

\(\Leftrightarrow2.1\ge\left(x+y+\dfrac{z}{2}+\dfrac{z}{2}\right)^2=\left(x+y+z\right)^2\)

\(\Rightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)

\(\frac{3x^2}{2}+y^2+z^2+yz=1\)

\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2\)

\(\Rightarrow\left(x+y+z\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)

Olm chào em, Olm rất vui khi em học tập tiến bộ và tiếp thu được nhiều kiến thức bổ ích trên Olm. Cảm ơn em đã đồng hành cùng Ol, chúc em học tập hiệu quả và vui vẻ cùng Olm, em nhé.

de ma :))))))))))))))))))))))))))))))

bài này đụng đến tam giác đồng dạng. Qua tết hãy nói nhé!

Cảm ơn em :)