Nên đk không ta -.-

Làm ơn ra khỏi web này, pls :)

Ta có:

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

\(\Rightarrow xyz>0\)

Áp dụng BĐT Cauchy cho 3 số dương ta có:

\(3=\dfrac{xyz}{z^2}+\dfrac{xyz}{y^2}+\dfrac{xyz}{z^2}\ge3\sqrt[3]{xyz}\)

\(\Rightarrow1\ge\sqrt[3]{xyz}\)

\(\Rightarrow1\ge xyz\)

\(\Rightarrow0< xyz\le1\)

Vì x, y, z nguyên nên \(xyz=1\)

Mà \(xyz=1\) \(\Rightarrow\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\ge3\sqrt[3]{xyz}=3\)

Dấu "=" xảy ra khi \(x^2+y^2+z^2\)

\(\Leftrightarrow\left|x\right|=\left|y\right|=\left|z\right|\)

Từ đó tìm ra 4 nghiệm là \(\left(x,y,z\right)=\left(1,1,1\right);\left(1,-1,-1\right);\left(-1,1,-1\right);\left(-1,-1,1\right).\)

Ta có: \(a+b+c=0\)

\(\Rightarrow\)\((a+b+c)^2=0\)

\(\Rightarrow\)\(a^2+b^2+c^2+2ab+2ac+2bc=0\)

\(\Rightarrow\)\(1+2(ab+bc+ac)=0\) ( Vì \(a^2+b^2+c^2=1\) )

\(\Rightarrow\)\(ab+bc+cd=\)\(-\dfrac{1}{2}\)

\(\Rightarrow\)\((ab+bc+cd)^2=\)\(\dfrac{1}{4}\)

\(\Rightarrow\)\(a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2\)\(=\)\(\dfrac{1}{4}\)

\(\Rightarrow\)\(a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c)\)\(=\dfrac{1}{4}\)

\(\Rightarrow\)\(a^2b^2 +a^2c^2+b^2c^2\)\(=\dfrac{1}{4}\) ( Vì \(a+b+c=0 \)) \((1)\)

Mặt khác: \(a^2+b^2+c^2=1\)

\(\Rightarrow\)\((a^2+b^2+c^2)^2=1\)

\(\Rightarrow\)\(a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1\)

\(\Rightarrow\)\(a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)=1\)

\(\Rightarrow\)\(a^4+b^4+c^4+2.\)\(\dfrac{1}{4}=1\) (Theo \(1\))

\(\Rightarrow\)\(a^4+b^4+c^4 \)\(=1-\dfrac{1}{2}=\dfrac{1}{2}\)

\(\Rightarrow\) Đpcm.

\(x^2\) chứ không phải \(x^3\) nha bạn.

\(x^2+12x+36\)

\(=x^2+2.x.6+6^2\)

\(=\left(x+6\right)^2\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{x+y+1+x+z+2+x+y-3}{x+y+z}=\dfrac{\left(x+y+z\right)+\left(x+y+z\right)+\left(1+2-3\right)}{x+y+z}=\dfrac{2.\left(x+y+z\right)}{x+y+z}=2\)

Lại có:

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

\(\Rightarrow2=\dfrac{1}{x+y+z}\)

\(\Rightarrow2.\left(x+y+z\right)=1\)

\(\Rightarrow x+y+z=\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{y+z+1}{x}=2\\\dfrac{x+z+2}{y}=2\\\dfrac{x+y-3}{z}=2\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+y+z+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}+1=3x\\\dfrac{1}{2}+2=3y\\\dfrac{1}{2}-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\dfrac{1}{2}}{3}\\y=\dfrac{\dfrac{1}{2}+2}{3}\\z=\dfrac{\dfrac{1}{2}-3}{3}\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=-\dfrac{5}{6}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=-\dfrac{5}{6}\end{matrix}\right.\) .

=200. Tich cho mik trước nhé !

họ 2;3;4;5;6;7;8;9;10;11;1213;14;15;1611

+x² -5 =-4 => x =1 => x =1 hoặc x =-1

+x² -5 = 4 => x² =9 => x =3 hoặc x =-3

\(49.\left(y-4\right)^2-9y^2-36y-36\)

\(=7^2\left(y-4\right)^2-\left(9y^2+36y+36\right)\)

\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)

\(=\left(7y-28+3y+6\right).\left(7y-28-3y-6\right)\)

\(=\left(10y-22\right).\left(4y-34\right)\)

\(=4.\left(5y-11\right).\left(2y-17\right)\)