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\(M=x^6-20x^5-20x^4-20x^3-20x^2-20x+3\)
\(M=x^6-\left(x-1\right)x^5-\left(x-1\right)x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x+3\)
\(M=x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x+3\)
\(M=x+3\) (1)
Thay \(x=21\)vào (1) ta được:
\(M=21+3\)
\(M=24\)
Còn câu N bạn tham khảo tại link này nha:
Câu hỏi của Hoang Linh - Toán lớp 8 | Học trực tuyến
Chúc bạn học thật tốt!
1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x=t\)
\(\Rightarrow BT=\left(t+10\right)\left(t+12\right)-24\)
\(=t^2+22x+96=\left(t+11\right)^2-25\ge-25\)
Vậy GTNN của bt là - 25\(\Leftrightarrow x^2+7x+11=0\)
\(\Delta=7^2-4.11=5\)
\(\orbr{\begin{cases}x_1=\frac{-22+\sqrt{5}}{2}\\x_2=\frac{-22-\sqrt{5}}{2}\end{cases}}\)
2) \(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)
\(=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-20\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)
Đặt \(x^2-8x=t\)
\(\RightarrowĐT=\left(t+7\right)\left(t+15\right)-20\)
\(=t^2+22t+85=\left(t+11\right)^2-36\ge-36\)
Vậy GTNN của bt là - 36\(\Leftrightarrow x^2-8x+11=0\)
\(\Delta=\left(-8\right)^2-4.11=20\)
\(\orbr{\begin{cases}x_1=\frac{-22-\sqrt{20}}{2}\\x_2=\frac{-22+\sqrt{20}}{2}\end{cases}}\)
1/(x+2)(x+3)(x+4)(x+5)-24
=(x+2)(x+5)(x+3)(x+4)
=(x+2)(x-2+7)(x+3)(x-3+7)
=[(x+2)(x-2)+7x+14][(x+3)(x-3)+7x+21]
=(x2-4+7x+14)(x2-9+7x+21)
=(x2+10+7x)(x2+12+7x)
2/(x2+x)2+4(x2+x)-12
=(x2+x)2+4(x2+x)+22-16
=(x2+x+2)2-42
=(x2+x+2+4)(x2+x+2-4)
=(x2+x+6)(x2+x-2)
3/(x2+x+1)(x2+x+2)-12
=(x2+x+1)(x2+x+-1+3)-12
=(x2+x+1)(x2+x+-1)+3(x2+x+1)-12
=(x2+x)-1+3(x2+x)+3-12
=(x2+x)(x2+x+3)-10
làm đến đây thì mk bí, bạn giúp suy nghĩ nốt nha
4/nó là nhân tử sẵn rồi mà
\(3/\)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+1\right)-12\)
\(=\left(x^2+x+1\right)^2+x^2+x+1-12\)
\(=\left(x^2+x+1\right)^2+4\left(x^2+x+1\right)-3\left(x^2+x+1\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+4\right)-3\left(x^2+x+1+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+15\right)-24\)
\(=a\left(a+5\right)-24;a=x^2+7x+10\)
\(=a^2+5a-24=\left(a+8\right)\left(a-3\right)\)
\(=\left(x^2+7x+18\right)\left(x^2+7x+7\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)=24\)
\(\Leftrightarrow\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24=0\)
\(\Leftrightarrow\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24=0\)
\(\Leftrightarrow\left(x^2+5x+2x+10\right)\left(x^2+4x+3x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left[\left(x^2+7x+10\right)+2\right]-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)^2-4\left(x^2+7x+10\right)+6\left(x^2+7x+10\right)-24=0\)
\(\Leftrightarrow\left[\left(x^2+7x+10\right)^2-4\left(x^2+7x+10\right)\right]+\left[6\left(x^2+7x+10\right)-24\right]=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left[\left(x^2+7x+10\right)-4\right]+6\left[\left(x^2+7x+10\right)-4\right]=0\)
\(\Leftrightarrow\left[\left(x^2+7x+10\right)-4\right]\left[\left(x^2+7x+10\right)+6\right]=0\)
\(\Leftrightarrow\left(x^2+7x+6\right)\left(x^2+7x+16\right)=0\)
\(\Leftrightarrow\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)=0\)
\(\Leftrightarrow\left[\left(x^2+x\right)+\left(6x+6\right)\right]\left(x^2+7x+16\right)=0\)
\(\Leftrightarrow\left[x\left(x+1\right)+6\left(x+1\right)\right]\left(x^2+7x+16\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+6=0\\x^2+7x+16=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\\\left(x+\dfrac{7}{2}\right)^2+\dfrac{15}{4}>0\left(loai\right)\end{matrix}\right.\)
Vậy \(x=-1\) hoặc \(x=-6\)