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

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

\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)

\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)

\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)

Bất đẳng thức à

ủa nhưng mà thỏa mãn cái gì mới c.m mấy cái kia chứ

a, Vì x² ≥ 0 , 2y² ≥ 0 với mọi x,y

=>x²+2y²+ 1 ≥ 1

=>Phân thức trên luôn có nghĩa

\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)

\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)

kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)

b. \(\)-\(3x-4\)

\(x^5+y^5=\left(x+y\right)^3\)

\(pt\Leftrightarrow\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=\left(x+y\right)^3\)

\(\Leftrightarrow\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4-\left(x+y\right)^2\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4-x^2-2xy-y^2\right)=0\)

*)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:\(x^4-x^3y+x^2y^2-xy^3+y^4-x^2-2xy-y^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^4+y^4-x^3y-xy^3=0\\x^2y^2-x^2-2xy-y^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2\left(x^2+xy+y^2\right)=0\\\left(xy+x+y\right)\left(xy-x-y\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=y\\\left(x^2+2x\right)\left(x^2-2x\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=y\\x^2\left(x-2\right)\left(x+2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=y\\x=2;x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=y=-2\\x=y=2\end{matrix}\right.\)

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c)(x²+x)²-2(x²+x)-15

đặt x²+x=a ta có

a²-2a-15

=a²+3a-5a-15

=(a²+3a)-(5a+15)

=a(a+3)-5(a+3)

=(a+3)(a-5)

thay a=x²+x

(x²+x+3)(x²+x-5)

x¹¹+x⁴+1

= x¹¹+x¹⁰+x⁹-x¹⁰-x⁹-x⁸+x⁸+x⁷+x⁶-x⁷-x⁶-x⁵+x⁵+x⁴+x³-x³-x²-x+x²+x+1

= x⁹(x²+x+1)-x⁸(x²+x+1)+x⁶(x²+x+1)-x⁵(x²+x+1)+x³(x²+x+1)-x(x²+x+1)+(x²+x+1)

= (x²+x+1)(x⁹-x⁸+x⁶-x⁵+x³-x+1)

x¹¹+x⁷+1

= x¹¹+x¹⁰+x⁹-x¹⁰-x⁹-x⁸+x⁸+x⁷+x⁶-x⁶-x⁵-x⁴+x⁵+x⁴+x³-x³-x²-x+x²+x+1

= x⁹(x²+x+1)-x⁸(x²+x+1)+x⁶(x²+x+1)-x⁴(x²+x+1)+x³(x²+x+1)-x(x²+x+1)+(x²+x+1)

= (x²+x+1)(x⁹-x⁸+x⁶-x⁴+x³-x+1)

\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)

\(\Rightarrow2x^2+3\left(x^2-1\right)=5x^2+5x\)

\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)

\(\Rightarrow5x^2-3=5x^2+5x\)

\(\Rightarrow-3=5x\)

\(\Rightarrow5x=-3\)

\(\Rightarrow x=-\dfrac{3}{5}\)

Vậy ....

P/s : Làm bừa !