1) \(x^3+y^3+z^3-3xyz=\left(x^3+3x^2y+3xy^2+y^3\right)+z^3-3xyz-3x^2y-3xy^2=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

2) Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)

\(\Leftrightarrow ab+bc+ac=0\)

\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)

\(\Leftrightarrow\dfrac{a^3b^3+b^3c^3+a^3c^3}{a^3b^3c^3}=\dfrac{3}{abc}\)

\(\Leftrightarrow\dfrac{a^3b^3+b^3c^3+a^3c^3}{a^2b^2c^2}=3\)

\(\Leftrightarrow a^3b^3+b^3c^3+a^3c^3=3a^2b^3c^2\)

\(\Leftrightarrow\left(ab+bc\right)^3-3ab^2c\left(ab+bc\right)+a^3b^3-3a^2b^2c^2=0\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left[\left(ab+bc\right)^2-\left(ab+bc\right)ac+a^2c^2\right]-3ab^2c\left(ab+bc+ac\right)=0\)

\(\Leftrightarrow0+0=0\left(đúng\right)\)

Đăng 5 -6 câu từng lần ha bạn!

\(1,7x-8=4x+7\)

\(\Leftrightarrow7x-8-4x=7\)

\(\Leftrightarrow7x-4x=7+8\)

\(\Leftrightarrow3x=15\)

\(\Rightarrow x=5\)

\(2,3-2x=3\left(x+1\right)-x-2\)

\(\Leftrightarrow3-2x=2x+1\)

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

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

\(\Leftrightarrow-4x=-2\)

\(\Rightarrow x=\dfrac{1}{2}\)

\(3,5\left(3x+2\right)=4x+1\)

\(\Leftrightarrow5.3x+5.2=4x+1\)

\(\Leftrightarrow15x+10=4x+1\)

\(\Leftrightarrow15x-4x=1-10\)

\(\Leftrightarrow11x=-9\)

\(\Rightarrow x=\dfrac{-9}{11}\)

Câu 6,7,9,10 cùng dạng là PT tích, anh làm 1 câu em làm các câu còn lại nhá

\(C.6:\\ \left(3x+2\right).\left(4x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=0\\4x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=-2\\4x=5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\\ \Rightarrow S=\left\{-\dfrac{2}{3};\dfrac{5}{4}\right\}\)

Câu 4,5,8 nhân ra chuyển về nên anh nghĩ em làm được

Câu 11,12,13,14 nó là mẫu số không có ẩn. Nên anh làm mẫu 2 câu trong số này nhé. Các câu khác em cứ dùng cách tương tự

\(C11\\ \dfrac{2x-1}{3}-x+2=\dfrac{3x+5}{4}\\ \Leftrightarrow\dfrac{4.\left(2x-1\right)}{12}-\dfrac{12x}{12}+\dfrac{24}{12}=\dfrac{3.\left(3x+5\right)}{12}\\\Leftrightarrow 8x-4-12x+24=9x+15\\ \Leftrightarrow8x-12x-9x=15-24+4\\ \Leftrightarrow-13x=-5\\ \Leftrightarrow x=\dfrac{-5}{-13}=\dfrac{5}{13}\\ \Rightarrow S=\left\{\dfrac{5}{13}\right\}\\ C14.\\ \dfrac{x-3}{6}+\dfrac{x+2}{4}=\dfrac{x-5}{3}+1\\ \Leftrightarrow\dfrac{2.\left(x-3\right)}{12}+\dfrac{3.\left(x+2\right)}{12}=\dfrac{4.\left(x-5\right)}{12}+\dfrac{12}{12}\\ \Leftrightarrow2x-6+3x+6=4x-20+12\\ \Leftrightarrow2x+3x-4x=-20+12+6-6\\ \Leftrightarrow x=-8\\ \Rightarrow S=\left\{8\right\}\)

Câu 15 nó có GTTĐ, anh sẽ làm như này em nha!

\(6-\left|2x-1\right|=3\\ \Leftrightarrow\left|2x-1\right|=6-3=3\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=3+1=4\\2x=-3+1=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{2}=2\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\\ \Rightarrow S=\left\{-1;2\right\}\)