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\(\left[\left(x+1\right).\left(x+4\right)\right].\left[\left(x+2\right).\left(x+3\right)\right]-24\)
\(=\left(x^2+5x+4\right).\left(x^2+5x+6\right)-24\)
Đặt m=x2+5x+4, ta có:
\(m.\left(m+2\right)-24=m^2+2m-24=m^2+6m-4m-24\)
\(=m.\left(m+6\right)-4.\left(m+6\right)=\left(m-4\right).\left(m+6\right)\)
Tự làm tiếp :v
\(1.a\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
\(=\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)-24\)
\(=\left(x^2+5x+5\right)^2-1-24\)
\(=\left(x^2+5x+5\right)^2-25\)
\(=\left(x^2+5x+5+5\right)\left(x^2+5x+5-5\right)\)
\(=\left(x^2+5x+10\right)\left(x^2+5x\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
\(b.x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)
\(2.a\) Đặt \(a=\frac{x+3}{x-2},b=\frac{x-3}{x+2}\)
Thay vào PT ta được:\(a^2+6b^2=7ab\)
\(\Leftrightarrow a^2-7ab+6b^2=0\)
\(\Leftrightarrow a^2-ab-6ab+6b^2=0\)
\(\Leftrightarrow a\left(a-b\right)-6b\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-6b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a-b=0\\a-6b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=b\\a=6b\end{cases}\Leftrightarrow}\orbr{\begin{cases}\frac{x+3}{x-2}=\frac{x-3}{x+2}\\\frac{x+3}{x-2}=6.\frac{x-3}{x+2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x+3\right)\left(x+2\right)=\left(x-3\right)\left(x-2\right)\\\left(x+3\right)\left(x+2\right)=\left(6x-18\right)\left(x-2\right)\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1hayx=6\end{cases}}\) (bước kia dài bạn tự làm nhé)
\(f\left(-1\right)=-4\Rightarrow-1+a-b+c=-4\)
\(\Rightarrow a-b+c=-3\)
\(f\left(2\right)=5\Rightarrow8+4a+2b+c=5\Rightarrow4a+2b+c=-3\)
\(\Rightarrow3a+3b=0\Rightarrow a=-b\)
\(\Rightarrow a^{2019}=-b^{2019}\Rightarrow a^{2019}+b^{2019}=0\)
\(\Rightarrow A=0\)
Ta thấy
\(f\left(x\right):g\left(x\right)\)
\(\Rightarrow\left(x^{100}+x^{99}+x^{98}+x^5+2020\right):\left(x^2-1\right)\)
\(=\left(x^{98}+x^{97}+2x^{96}+2x^{95}+...2x^4+3x^3+2x^2+3x+2\right)\) có số dư là \(R\left(x\right)=3x+2022\)
\(\Rightarrow R\left(2021\right)=3.2021+2022=8085\)
Èo,phân tích ra tưởng cái hệ 3 ẩn r định bỏ cuộc và cái kết:(
Ta có:
\(f\left(x\right)=\left(x-2\right)\cdot Q\left(x\right)+5\)
\(f\left(x\right)=\left(x+1\right)\cdot K\left(x\right)-4\)
Theo định lý Huy ĐZ ta có:
\(f\left(2\right)=5\Rightarrow8+4a+2b+c=5\left(1\right)\)
\(\Rightarrow f\left(-1\right)=-4\Rightarrow-1+a-b+c=-4\left(2\right)\)
Lấy \(\left(1\right)-\left(2\right)\) ta được:
\(9+3a+3b=9\Leftrightarrow a+b=0\)
Khi đó:
\(\left(a^3+b^3\right)\left(b^5+c^5\right)\left(c^7+d^7\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)\left(b^5+c^5\right)\left(c^7+a^7\right)\)
\(=0\) ( theo Huy ĐZ thì \(a+b=0\) )
Ap dung dinh ly Bozout ta co
\(f\left(2\right)=2^3+a.2^2+b.2+c=5\)
<=> \(4a+2b+c=-3\) (1)
tuong tu \(f\left(-1\right)=\left(-1\right)^3+a-b+c=-4\)
<=> \(a-b+c=-3\) (2)
tu (1) va (2) => \(4a+2b=a-b=-3\)
=> a=b+-3
=> \(4\left(b-3\right)+2b=-3\Rightarrow b=\frac{3}{2}\)
=> \(a=-\frac{3}{2}\)
=> \(\left(a^3+b^3\right)=\left(a+b\right)\left(a^2-ab+b^2\right)=\left(\frac{3}{2}-\frac{3}{2}\right)\left(a^2-ab+b^2\right)=0\)
=> gia tri bieu thuc =0
Vì \(f\left(x\right)⋮x-2;f\left(x\right):x^2-1\) dư 1\(\Rightarrow\left\{{}\begin{matrix}f\left(x\right)=g\left(x\right)\cdot\left(x-2\right)\\f\left(x\right)=q\left(x\right)\left(x^2-1\right)+x=q\left(x\right)\left(x-1\right)\left(x+1\right)+x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=0\\f\left(1\right)=1\\f\left(-1\right)=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}32+4a+2b+c=0\\2+a+b+c=1\\2+a-b+c=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}4a+2b+c=-32\left(1\right)\\a+b+c=-1\left(2\right)\\a-b+c=-3\left(3\right)\end{matrix}\right.\)
Trừ từng vế của (2) cho (3) ta được:
\(\Rightarrow2b=2\Rightarrow b=1\)
Thay b=1 vào lần lượt (1) ,(2),(3) ta được:
\(\Rightarrow\left\{{}\begin{matrix}4a+2+c=-32\\a+1+c=-1\\a-1+c=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+c=-34\\a+c=-2\\a+c=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4a+c=-34\left(4\right)\\a+c=-2\left(5\right)\end{matrix}\right.\)
Trừ từng vế của (4) cho (5) ta được:
\(\Rightarrow3a=-32\Rightarrow a=-\dfrac{32}{3}\Rightarrow c=-2+\dfrac{32}{3}=\dfrac{26}{3}\) Vậy...
Theo đề bài ta có :
\(F\left(x\right)=\left(x-1\right)\cdot Q\left(x\right)-4\) (1)
\(F\left(x\right)=\left(x+2\right)\cdot R\left(x\right)+5\) (2)
Thay \(x=1\) vào (1) ta có :
\(F\left(1\right)=-4\)
\(\Leftrightarrow1+a+b+c=-4\)
\(\Leftrightarrow a+b+c=-5\)
Thay \(x=-2\) vào (2) ta có :
\(F\left(-2\right)=5\)
\(\Leftrightarrow-8+4a-2b+c=5\)
\(\Leftrightarrow4a-2b+c=13\)
Do đó ta có : \(\hept{\begin{cases}a+b+c=-4\\4a-2b+c=13\end{cases}}\)
....
Nguyễn Lê Phước Thịnh White Hold HangBich2001 Phạm Vũ Trí Dũng Nguyễn Huyền Trâm