bài 26
tìm giá trị lớp nhất của biểu thức
\(M=4-5x-4x^2\)
\(N=\frac{x^2-x+1}{x^2+x+1}\)
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\(f\left(x\right)=\left(x-a\right)\left(x-b\right)\left(x-c\right)\)
\(=\left(x^2-ax-bx+ab\right)\left(x-c\right)\)
\(=x^3-\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)
\(=x^3+\left(ab+bc+ca\right)x+abc\)
\(-f\left(-x\right)=-\left[\left(-x-a\right)\left(-x-b\right)\left(-x-c\right)\right]\)
\(=-\left[\left(x^2+ax+bx+ab\right)\left(-x-c\right)\right]\)
\(=-\left[-x^3-\left(a+b+c\right)x^2-\left(ab+bc+ca\right)x-abc\right]\)
\(=-\left[-x^3-\left(ab+bc+ca\right)x-abc\right]\left(a+b+c=0\right)\)
\(=x^3+\left(ab+bc+ca\right)x+abc\)
Sửa đề:
\((2x^2+x-2015)^2+4(x^2-5x-2016)^2=4(2x^2+x-2015)(x^2-5x-2016)\)
\(\Rightarrow\left(2x^2+x-2015\right)^2-2.\left(2x^2+x-2015\right).2.\left(x^2-5x-2016\right)+[2.\left(x^2-5x-2016\right)]^2=0\)
\(\Rightarrow[2x^2+x-2015-2.\left(x^2-5x-2016\right)]^2=0\)
\(\Rightarrow11x+2017=0\)
\(\Rightarrow x=\frac{-2017}{11}\)
1,\(A=2x^2-6x+7\)
\(=2\left(x^2-3x+\frac{9}{4}\right)+\frac{5}{2}\)
\(=2\left(x-\frac{3}{2}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)
Dấu "=" khi \(x=\frac{3}{2}\)
2,\(B=\frac{2x^2-6x+5}{x^2-2x+1}\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow Bx^2-2Bx+B=2x^2-6x+5\)
\(\Leftrightarrow x^2\left(B-2\right)+2x\left(3-B\right)+B-5=0\)(1)
*Với B = 2 thì \(\left(1\right)\Leftrightarrow x^2\left(2-2\right)+2x\left(3-2\right)+2-5=0\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\frac{3}{2}\left(TmĐKXĐ\right)\)
*Với \(B\ne2\)thì pt (1) là pt bậc 2 ẩn x tham số B
Pt (1) có nghiệm khi \(\Delta\ge0\)
\(\Leftrightarrow\left(3-B\right)^2-\left(B-2\right)\left(B-5\right)\ge0\)
\(\Leftrightarrow9-6B+B^2-B^2+7B-10\ge0\)
\(\Leftrightarrow B\ge1\)
Dấu "=" xảy ra khi \(\left(1\right)\Leftrightarrow-x^2+4x-4=0\)
\(\Leftrightarrow-\left(x-2\right)^2=0\)
\(\Leftrightarrow x=2\left(TmĐKXĐ\right)\)
Thấy 1 < 2 nên BMin = 1<=> x = 2
Vậy ....
A=(9x2-6x+1)+(7x2+7)-1=(3x2+1)2+7(x2+7)-1
Vì: (3x2+1)2\(\ge\)0 và 7(x2+7)\(\ge\)0
Nên:A\(\ge\) -1
B=\(\frac{A-2}{\left(x-1\right)^2}\)\(\ge\) -3
\(x^3-12x-16=0\Leftrightarrow x^2\left(x+2\right)-2x\left(x+2\right)-8\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x-8\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x\left(x-4\right)+2\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)^2\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=4\end{cases}}\)
\(2\left(ab+bc+ca\right)=\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow2\left(ab+bc+ca\right)=2^2-2\)
\(\Leftrightarrow2\left(ab+bc+ca\right)=2\Leftrightarrow ab+bc+ca=1\)
\(M=\left(a^2+ab+bc+ca\right)\left(b^2+ab+bc+ca\right)\left(c^2+ab+bc+ca\right)\)
\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(a+b\right)\right]\left[c\left(b+c\right)+a\left(b+c\right)\right]\)
\(=\left(a+b\right)^2\left(b+c\right)^2\left(a+c\right)^2=\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2\)
Ta có : theo điều kiện cho trước:
a + b + c =2
<=> \(\left(a+b+c\right)^2=4\)
<=> \(a^2+b^2+c^2+2ab+2ac+2bc=4\)
<=> \(2+2\left(ab+ac+bc\right)=4\)
<=> \(2\left(ab+ac+bc\right)=2\)
<=> \(ab+ac+bc=1\)
<=> \(\left(ab+ac+bc\right)^2=1\)
<=> \(a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+a^2bc+abc^2\right)=1\)
<=> \(a^2b^2+b^2c^2+a^2c^2=1-2\left(ab^2c+a^2bc+abc^2\right)\)
Theo đề bài ta có :
M = \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)
<=> \(\left(a^2b^2+a^2+b^2+1\right)\left(c^2+1\right)\)
<=> \(a^2b^2c^2+a^2b^2+a^2c^2+a^2+b^2c^2+b^2+c^2+1\)
<=> \(a^2b^2c^2+1-2ab^2c-2a^2bc-2abc^2+3\)
<=> \(a^2b^2c^2-2ab^2c-2a^2bc-2abc^2+4\)
<=> \(abc\left(abc-2b-2a-2c\right)+4\)
<=> \(abc\left\{abc-2\left(a+b+c\right)\right\}+4\)
<=> \(abc\left(abc-4\right)+4\)
<=> \(a^2b^2c^2-4abc+4\)
<=> \(\left(abc\right)^2-4abc+4\)
<=> \(\left(abc-2\right)^2\left(đpcm\right)\)
\(M=4-5x-4x^2\)
\(=-\left(4x^2+5x-4\right)\)
\(=-\left(4x^2+4x+1-5\right)\)
\(=-\left[\left(2x+1\right)^2-4\right]\)
\(=-\left(2x+1\right)^2+4\)
Vì \(-\left(2x+1\right)^2\le0\)với mọi x
\(\Rightarrow-\left(2x+1\right)^2+4\le4\)với mọi x
Dấu "=" xảy ra \(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy max M=4 khi \(x=-\frac{1}{2}\)