\(x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)=-2\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

Ta lại có: 

\(x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)^3=1\)

\(\Leftrightarrow x+y+z=1\)

Làm nốt

biến đổi tương đương đi, nhân tung ngoặc ra

\(\left(a+b\right)\left(a^4+b^4\right)\ge\left(a^2+b^2\right)\left(a^3+b^3\right)\)

\(\Leftrightarrow a^5+ab^4+a^4b+b^5\ge a^5+a^2b^3+a^3b^2+b^5\)

\(\Leftrightarrow ab^4+a^4b-a^2b^3-a^3b^2\ge0\)

\(\Leftrightarrow ab\left(a^3+b^3-ab^2-a^2b\right)\ge0\)

\(\Leftrightarrow a^3+b^3-ab^2-a^2b\ge0\)(Do ab > 0)

\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)Luôn đúng do a,b dương

Dấu "='' khi a = b

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

rồi quy đồng biến            đổi ra : (x + y ) ( y + z ) ( z + x ) = 0

=> x = -y hoặc y = -z hoặc z = -x

kết hợp với 2x² + y = 1 rồi giải ra sau đó thay vào x + y + z = 3 rồi tìm ra x,y,z

như thế bạn làm nhá

\(A=x^4-14x^3+71x^2-154x+120\)

\(=x^3\left(x-2\right)-12x^2\left(x-2\right)+47x\left(x-2\right)-60\left(x-2\right)\)

\(=\left(x-2\right)\left(x^3-12x^2+47x-60\right)\)

\(=\left(x-2\right)\left[x^2\left(x-3\right)-9x\left(x-3\right)+20\left(x-3\right)\right]\)

\(=\left(x-2\right)\left(x-3\right)\left(x^2-9x+20\right)=\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)\)

b, Vì A là tích của 4 số nguyên liên tiếp nên A chia hết cho 24

=-2016,00099

\(\frac{x}{2017}+\frac{x-1}{2017}=\frac{x-2}{2019}-1\)

\(\Leftrightarrow\frac{2x-1}{2017}=\frac{x-2021}{2019}\)

\(\Leftrightarrow4038x-2019=2017x-4076357\)

\(\Leftrightarrow2021x=4074338\)

\(\Leftrightarrow x=\frac{4074338}{2021}\)(nếu sai thì ib vs mik nha)

\(\left(x+1\right)^4+\left(x+3\right)^4=16\)

\(\Leftrightarrow x+1+x+3=2\)

\(\Leftrightarrow x=-1\)

\(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}=\frac{1}{a-b+c}\)

\(\frac{bc-ac+ab}{abc}=\frac{1}{a-b+c}\)

\(\left(bc-ac+ab\right)\left(a-b+c\right)=abc\)

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

\(\left(abc+a^2b-b^2c-ab^2\right)-\left(a^2c+ac^2-bc^2-abc\right)=0\)

\(b.\left(ac+a^2-bc-ab\right)-c\left(a^2+ac-bc-ab\right)=0\)

\(\left(b-c\right)\left(a-b\right)\left(a+c\right)=0\)

vì a\(\ne\)b\(\ne\)c nên a + c = 0 suy ra a = -c 

a³ + c³ = a³ + ( -a )³ = 0

\(a,x^2-10x-39=0\)

\(\Leftrightarrow x^2-10x-39+64=64\)

\(\Leftrightarrow x^2-10x+25=64\)

\(\Leftrightarrow\left(x-5\right)^2=64\)

làm nốt

\(x^2-10x-39=0\Leftrightarrow x^2-13x+3x-39=0\Leftrightarrow x\left(x-13\right)+3\left(x-13\right)=0\)

\(\Leftrightarrow\left(x-13\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=13\\x=-3\end{cases}}\)