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U
1
DC
15 tháng 10 2015
= x13 -(7+1)x12 + (7+1)x11 -(7+1)x10 .... -(7+1)x12 +(7+1)x +8
= x13 -(x+1)x12 + (x+1)x11 -(x+1)x10 .... - (x+1)x2 +(x+1)x +8 ( Vì x=7)
=x13 - x13 - x12 + x12 + x11 - x11 - x11 - ..... -x3 - x2 +x2 +x+8
=x+8=7+8=15
4 tháng 2 2022
a: \(=\dfrac{2x^3-6x^2+10x-2x^2+6x-10}{x^2-3x+5}=2x-2\)
b: \(=\dfrac{x^3-2x^2+7x-2x^2+4x-14}{x^2-2x+7}=x-2\)
c: \(=2x^4+4x^3-4x^3-8x^2+6x^2+12x+44x+88\)
\(=2x^4-2x^2+56x+88\)
LN
2 tháng 4 2020
a. x3+5x2+3x-9
= x3-x2+6x2-6x+9x-9
= x2(x-1)+6x(x-1)+9(x-1)
= (x2+6x+9)(x-1)
= (x+3)2(x-1)
b. x3+9x2+11x-21
= x3-x2+10x2-10x+21x-21
= x2(x-1)+10x(x-1)+21(x-1)
= (x2+10x+21)(x-1)
= (x+7)(x+3)(x-1)
c. x3-7x+6
= x3-x2+x2-x-6x+6
= x2(x-1)+x(x-1)-6(x-1)
= (x2+x-6)(x-1)
= (x+3)(x-2)(x-1)
d. x3-5x2+8x-4
= x3-x2-4x2+4x+4x-4
= x2(x-1)-4x(x-1)+4(x-1)
= (x2-4x+4)(x-1)
= (x-2)2(x-1)
e. x3-3x+2
= x3+2x2-2x2-4x+x+2
= x2(x+2)-2x(x+2)+(x+2)
= (x2-2x+1)(x+2)
= (x-1)2(x+2)
f. x3+8x2+17x+10
= x3+5x2+3x2+15x+2x+10
= x2(x+5)+3x(x+5)+2(x+5)
= (x2+3x+2)(x+5)
= (x+1)(x+2)(x+5)
g. x3+3x2+6x+4
= x3+x2+2x2+2x+4x+4
= x2(x+1)+2x(x+1)+4(x+1)
= (x2+2x+4)(x+1)
h. x3-2x-4
= x3-2x2+2x2-4x+2x-4
= x2(x-2)+2x(x-2)+2(x-2)
= (x2+2x+2)(x-2)
k. x3+x2+4
= x3+2x2-x2-2x+2x+4
= x2(x+2)-x(x+2)+2(x+2)
= (x2-x+2)(x+2)
l. x3-12x+7x-2
= x3+2x2-2x2-4x-x-2
= x2(x+2)-2x(x+2)-(x+2)
= (x2-2x-1)(x+2)
KT
1
3 tháng 8 2020
1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)
\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left[x\left(x-8\right)+\left(x-8\right)\right]=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)
Vậy: S={0;-7;8;-1}
2) Ta có: \(x^3-8x^2+17x-10=0\)
\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)
\(\Leftrightarrow x^2\left(x-2\right)-6x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-6x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-x-5x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=5\end{matrix}\right.\)
Vậy: S={2;1;5}
3) Ta có: \(2x^3-5x^2-x+6=0\)
\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-3\right)+\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=3\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{2;\frac{3}{2};-1\right\}\)
4) Ta có: \(4x^4-4x^2-3=0\)
\(\Leftrightarrow4x^4-6x^2+2x^2-3=0\)
\(\Leftrightarrow2x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\)
\(\Leftrightarrow\left(2x^2-3\right)\left(2x^2+1\right)=0\)
mà \(2x^2+1>0\forall x\in R\)
nên \(2x^2-3=0\)
\(\Leftrightarrow2x^2=3\)
\(\Leftrightarrow x^2=\frac{3}{2}\)
hay \(x=\pm\sqrt{\frac{3}{2}}\)
Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)
\(\left(x-2\right)^3+\left(x+2\right)^3-x^3-8x^2+10\)
\(=x^3-6x^2+12x-8+x^3+6x^2+12x+8-x^3-8x^2+10\)
\(=x^3-2x^2+24x+10\)