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ax^3+ bx^2- 11x+30 x^2-3x-10 ax+(b-3a) ax^3+ 3ax^2-10ax x^2(b-3a)-x(11-10a)+30 x^2(b-3a)-3x(b-3a)-10(b-3a) x(3b-a-11)+10(b-3a-3)
Phần còn lại dành cho bạn ;) Đến đây nắm vững lý thuyết làm oke
thay 11=x+1 ta có:
f(x)= \(x^{10}\)-11\(x^9\)+11\(x^8\)-11\(x^7\)+....+11\(x^2\)-11x+100
=\(x^{10}\)-(x+1)\(x^9\)+(x+1)\(x^8\)-(x+1)\(x^7\)+...+(x+1)\(x^2\)-(x+1)x+100
=\(x^{10}\)-\(x^{10}\)-\(x^9\)+\(x^9\)+\(x^8\)-\(x^8\)-\(x^7\)+......+\(x^3\)+\(x^2\)-\(x^2\)-x+100
=-x+100
=> f(10)=-10+100=90
Thay 11 = x + 1 ta có:
f(x) = \(x^{10}-11x^9+11x^8-11x^7+...+11x^2-11x+100\)
\(=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...+\left(x+1\right)x^2-\left(x+1\right)x^2-\left(x+1\right)x+100\)
= \(x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^3+x^2-x^2-x+100\)
= -x+100
=>f(10)= - 10 + 100 = 90
Đặt \(f\left(x\right)=ax^3+bx^2-11x+30\)
Ta có : \(x^2-3x-10=\left(x+2\right)\left(x-5\right)\)
+) \(f\left(x\right)⋮x+2\)
\(\Leftrightarrow f\left(-2\right)=0\)
\(\Leftrightarrow-8a+4b-11.\left(-2\right)+30=0\)
\(\Leftrightarrow-8a+4b+22+30=0\)
\(\Leftrightarrow-8a+4b+52=0\)
\(\Leftrightarrow-2a+b+13=0\)( * )
+) \(f\left(x\right)⋮x-5\)
\(\Leftrightarrow f\left(5\right)=0\)
\(\Leftrightarrow125a+25b-11.5+30=0\)
\(\Leftrightarrow125a+25b-25=0\)
\(\Leftrightarrow5a+b-1=0\)
\(\Leftrightarrow-2a+7a+b+13-14=0\)
\(\Leftrightarrow\left(-2a+b+13\right)+\left(7a-14\right)=0\)( ** )
Từ ( * ) ; ( ** )
\(\Rightarrow7a-14=0\)
\(\Rightarrow7a=14\)
\(\Rightarrow a=2\)
\(\Rightarrow b=-9\)
Vậy với \(a=2;b=-9\) thì \(ax^3+bx^2-11x+30⋮x^2-3x-10\)
a)P(x)=5x^3+3x^2-2x-5
Q(x)=5x^3+2x^2-2x+4
b)P(x)+Q(x)=10x^3+5x^2-4x-1
P(x)-Q(x)=x^2-9
c)x=3
_CÓ AI GIỎI TOÁN THÌ GIÚP MK NHA RẤT CẢM ƠN ______________
Ta có : \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}\right|\ge0\forall x\)
=> 11x \(\ge\)0
=> x \(\ge\)0
Khi đó \(\orbr{\begin{cases}x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=11x\left(10\text{ số hạng x }\right)\\x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=-11x\left(10\text{ số hạng x}\right)\end{cases}}\)
=> \(\orbr{\begin{cases}10x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=11x\\10x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=-11x\end{cases}}\)
=> \(\orbr{\begin{cases}10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\\10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=-11x\end{cases}}\)
=> \(\orbr{\begin{cases}10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\\10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=-11x\end{cases}}\)
=> \(\orbr{\begin{cases}10x+\left(1-\frac{1}{11}\right)=11x\\10x+\left(1-\frac{1}{11}\right)=-11x\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{10}{11}\\21x=-\frac{10}{11}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{10}{11}\left(\text{tm}\right)\\x=-\frac{10}{231}\left(\text{loại}\right)\end{cases}}}\)
Vậy \(x=\frac{10}{11}\)
o ti k hieu