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\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
\(S=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{2a^2+2b^2+2c^2-2ab-2bc-2ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{2a^2+2b^2+2c^2-2ab-2bc-2ac}\)
\(=\dfrac{3\cdot\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)\cdot\dfrac{1}{2}}{2a^2+2b^2+2c^2-2ab-2bc-2ac}=\dfrac{3}{2}\)
Queen Material Giải:
\(25\left(x+y\right)^2-16\left(x-y\right)^2\)
\(=25\left(x^2+2xy+y^2\right)-16\left(x^2-2xy+y^2\right)\)
\(=25x^2+50xy+25y^2-16x^2+32xy-16y^2\)
\(=9x^2+82xy+9y^2\)
\(=x\left(9x+y\right)+9y\left(9x+y\right)\)
\(=\left(x+9y\right)\left(9x+y\right)\).
\(\left\{{}\begin{matrix}25=5^2\\16=4^2\\25\left(x+y\right)^2=\left[5\left(x+y\right)\right]^2\\16\left(x-y\right)^2=\left[4\left(x-y\right)\right]^2\end{matrix}\right.\)
\(A=\left[5\left(x+y\right)-4\left(x-y\right)\right]\left[5\left(x+y\right)+4\left(x-y\right)\right]\)
\(A=\left(x+9y\right)\left(9x+y\right)\)
\(x^5+y^5=\left(x+y\right)^3\)
\(pt\Leftrightarrow\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=\left(x+y\right)^3\)
\(\Leftrightarrow\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4-\left(x+y\right)^2\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4-x^2-2xy-y^2\right)=0\)
*)Xét \(x+y=0\Rightarrow x=-y\)
\(\Rightarrow\left(-y\right)^5+y^5=\left(-y+y\right)^3\Rightarrow0=0\) (luôn đúng)
*)Xét \(x^4-x^3y+x^2y^2-xy^3+y^4-x^2-2xy-y^2=0\)
+)Xét \(x=0 \Rightarrow y=±1; y=0 \Rightarrow x=±1\)
Và \(x=y=0\)
+)Xét \(x;y\ne 0\) ta có:\(x^4-x^3y+x^2y^2-xy^3+y^4-x^2-2xy-y^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^4+y^4-x^3y-xy^3=0\\x^2y^2-x^2-2xy-y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2\left(x^2+xy+y^2\right)=0\\\left(xy+x+y\right)\left(xy-x-y\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\\\left(x^2+2x\right)\left(x^2-2x\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\\x^2\left(x-2\right)\left(x+2\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\\x=2;x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=y=-2\\x=y=2\end{matrix}\right.\)
Trình bày chắc sai chỗ nào đó nhưng kq thì đúng 100%-WF đã chứng minh :V
\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Rightarrow2x^2+3\left(x^2-1\right)=5x^2+5x\)
\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Rightarrow5x^2-3=5x^2+5x\)
\(\Rightarrow-3=5x\)
\(\Rightarrow5x=-3\)
\(\Rightarrow x=-\dfrac{3}{5}\)
Vậy ....
P/s : Làm bừa !