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Ta có :
\(A=\left(x-1\right)^4+\left(x-3\right)^4+6\left(x-1\right)^2\left(x-3\right)^2\)
\(A=\left(x-1\right)^4+2\left(x-1\right)^2\left(x-3\right)^2+\left(x-3\right)^4+4\left(x-1\right)^2\left(x-3\right)^2\)
\(A=\left[\left(x-1\right)^2+\left(x-3\right)^2\right]^2+4\left(x-1\right)^2\left(x-3\right)^2\)
\(A=\left[2x^2-8x+10\right]^2+4\left(x^2-4x+3\right)^2\)
\(A=\left[2\left(x-2\right)^2+2\right]+4\left[\left(x-2\right)^2-1\right]^2\)
\(A=4\left(x-2\right)^4+8\left(x-2\right)^2+4+4\left(x-2\right)^4-8\left(x-2\right)^2+4\)
\(A=8\left(x-2\right)^4+8\ge8\)
Vậy GTNN của biểu thức A là 8 \(\Leftrightarrow x=2\)
Đặt x-2=y
=> \(A=\left(y+1\right)^4+\left(y-1\right)^4+6\left(y+1\right)^2\left(y-1\right)^2\)
Khai triển A ta được
\(A=2y^4+12y^2+2+6\left(y^4-2y^2+1\right)\)
\(=8y^4+8=8\left(y^4+1\right)\ge8\)
Dấu "=" xảy ra khi y=0 lúc đó x=0+2=2
Vậy Amin=8 khi x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x\left(x+1\right)\left(x^2+x-4\right)\)
\(=\left(x^2+x\right)\left(x^2+x-4\right)\)
Đặt \(x^2+x=k\)
Lúc đó \(A=k\left(k-4\right)\)
\(=k^2-4k+4-4=\left(k-2\right)^2-4\ge-4\)
(Dấu "=" xảy ra khi \(k=2\Leftrightarrow x^2+x=2\)
\(\Leftrightarrow x^2+x-2=0\)
Ta có: \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+3}{2}=1\\x=\frac{-1-3}{2}=-2\end{cases}}\))
![](https://rs.olm.vn/images/avt/0.png?1311)
Với \(x\ne0\), đặt \(\left|x\right|=a>0\)
\(A=\frac{\left(a^2+18a+32\right)\left(a^2+9a+8\right)}{a^2}=\frac{\left(a+2\right)\left(a+16\right)\left(a+1\right)\left(a+8\right)}{a^2}\)
\(A=\frac{\left(a+2\right)\left(a+8\right)\left(a+1\right)\left(a+16\right)}{a^2}=\frac{\left(a^2+10a+16\right)\left(a^2+17a+16\right)}{a^2}\)
\(A=\frac{\left(a^2+16+10a\right)}{a}.\frac{\left(a^2+16+17a\right)}{a}=\left(a+\frac{16}{a}+10\right)\left(a+\frac{16}{a}+17\right)\)
\(\Rightarrow A\ge\left(2\sqrt{a.\frac{16}{a}}+10\right)\left(2\sqrt{a.\frac{16}{a}}+17\right)=\left(8+10\right)\left(8+17\right)=450\)
\(\Rightarrow A_{min}=450\) khi \(a^2=16\Rightarrow a=4\Rightarrow x=\pm4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow A=\left(x^2+x\right)\left(x^2+x-4\right)\) Đặt x^2+x-a có
\(A=a\left(a-4\right)=a^2-4a\ge-4\)
Min A=-4 với a-2=0\(\Rightarrow x^2+x=2\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\frac{9}{4}\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{3}{2}\\x+\frac{1}{2}=-\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
theo nghiệm Fx=Gx mũ 2
suy ra x mũ 2 +1 mũ x 2
suy ra chịch chịch chịch
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=\left(x^2+3x\right)\left(x^2+3x+2\right)=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1-1=\left(x^2+3x+1\right)^2-1\ge-1\)
Vậy MIN bt là -1 với \(x^2+3x+1=0\Rightarrow\left[{}\begin{matrix}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{matrix}\right.\)
Ta có x(x+1)(x+2)(x+3)=\(\left(x^2+3x\right)\left(x^2+3x+2\right)\)
=\(\left(x^2+3x+1-1\right)\left(x^2+3x+1+1\right)\)
Đặt \(x^2+3x+1=a\)<=>(a-1)(a+1)=\(a^2-1\)≥-1
=\(\left(x^2+3x+1\right)^2-1\)≥-1
Vậy Min biểu thức là -1,với \(x^2+3x+1=0\)
<=>x=\(=\frac{-3+\sqrt{5}}{2}\),\(\frac{-3-\sqrt{5}}{2}\)
\(L=9\left|x-4\right|+\left|x-1\right|+x\)
\(L=\left|x-4\right|+\left|x-1\right|+\left|x-4\right|+x+7\left|x-4\right|\)
\(L=\left|4-x\right|+\left|x-1\right|+\left|4-x\right|+x+7\left|x-4\right|\)
Áp dụng liên tiếp 2 bất đẳng thức: \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\) và \(\left|a\right|\ge a\) ta có:
\(L\ge\left|4-x+x-1\right|+4-x+x+7\left|x-4\right|\)
\(L\ge3+4+7\left|x-4\right|=7+\left|x-4\right|\ge7\)
Dấu "=" xảy ra khi tất cả các bđt đều xảy ra dấu "=",nghĩa là:
\(\left\{{}\begin{matrix}1\le x\le4\\x\le4\\x=4\end{matrix}\right.\Leftrightarrow x=4\).Vậy \(min_M=7\) khi \(x=4\)
7+7|x-4|