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\(A=\dfrac{6x^2+21x+22}{x^2+4x+4}\)
\(=\dfrac{6\left(x^2+4x+4\right)-3x-2}{x^2+4x+4}\)
\(=6+\dfrac{-3x-2}{\left(x+2\right)^2}\)
\(=6+\dfrac{-3\left(x+2\right)+4}{\left(x+2\right)^2}\)
\(=6-\dfrac{3}{x+2}+\dfrac{4}{\left(x+2\right)^2}\)
-Đặt \(a=\dfrac{1}{x+2}\) thì:
\(A=6-3a+4a^2=\left(2a\right)^2-2.2a.\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{87}{16}=\left(2a-\dfrac{3}{4}\right)^2+\dfrac{87}{16}\ge\dfrac{87}{16}\)
\(A_{min}=\dfrac{87}{16}\)\(\Leftrightarrow\left(2a-\dfrac{3}{4}\right)^2=0\Leftrightarrow2a-\dfrac{3}{4}=0\Leftrightarrow2a=\dfrac{3}{4}\)
\(\Leftrightarrow2.\dfrac{1}{x+2}=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{x+2}=\dfrac{3}{8}\Leftrightarrow x+2=\dfrac{8}{3}\Leftrightarrow x=\dfrac{2}{3}\)
\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)
\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)
Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)
Dấu \("="\Leftrightarrow x=-5\)
\(A=-x^2-4x-2\)
\(\Leftrightarrow-A=x^2+4x+2\)
\(\Leftrightarrow-A=x^2+4x+4-2\)
\(\Leftrightarrow-A=\left(x+2\right)^2-2\)
Vì \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2-2\ge-2\)hay \(-A\ge-2\)
\(\Rightarrow A\le2\)
Vậy GTLN của A là 2\(\Leftrightarrow x=-2\)
1a) 8xy(8-12x+6x*x-x*x*x)
chú thích x*x là x bình phương
x*x*x là x lập phương
2. a) 3x (x-5)- (x-1)(2+3x)=30
3x*x-15x-2x-3x*x+2+3x=30
14x=28
x=2
b) (x+2)(x-3)-(x-2)(x+5)=0
x*x-3x+2x-6-x*x-5x+2x+10=0
2x=-4
x=-2
còn mấy bài còn lại mình không biết
a) (x-1)*(x+2)-(x-3)*(-x+4)=19
\(\Leftrightarrow x^2+2x-x-2-\left(-x^2+4x+3-12\right)=19\)
\(\Leftrightarrow x^2+2x-x-2+x^2-4x-3+12=19\)
\(\Leftrightarrow2x^2-3x+7-19=0\)
\(\Leftrightarrow2x^2-3x-12=0\)
Đề sai??
b) (2x -1)*(3x+5)-(6x-1)*(6x+1)=(-17)
\(\Leftrightarrow6x^2+10x-3x-5-\left(36x^2+6x-6x-1\right)=-17\)
\(\Leftrightarrow6x^2+10x-3x-5-36x^2-6x+6x+1=-17\)
\(\Leftrightarrow-30x^2+7x-4+17=0\)
\(\Leftrightarrow-30x^2+7x+13=0\)
???
2/
a, \(A=2x^2+6x-5=2\left(x^2+3x-\frac{5}{2}\right)=2\left(x^2+2x\cdot\frac{3}{2}+\frac{9}{4}-\frac{19}{4}\right)=2\left[\left(x+\frac{3}{2}\right)^2-\frac{19}{4}\right]=2\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\)
Vì \(\left(x+\frac{3}{2}\right)^2\ge0\Rightarrow A=\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\ge-\frac{19}{2}\)
Dấu "=" xảy ra khi x=-3/2
Vậy Amin=-19/2 khi x=-3/2
b,bài này phải tìm min
\(B=\left(2x-x\right)\left(x+4\right)=x\left(x+4\right)=x^2+4x=x^2+4x+4-4=\left(x+2\right)^2-4\)
Vì \(\left(x-2\right)^2\ge0\Rightarrow B=\left(x-2\right)^2+4\ge4\)
Dấu "=" xảy ra khi x = 2
Vậy Bmin=4 khi x=2
\(f\left(x\right)⋮g\left(x\right)\)
\(\Leftrightarrow x^4-3x^3+4x^2-x^2+3x-4+\left(a-3\right)x+\left(b+4\right)⋮x^2-3x+4\)
\(\Leftrightarrow\left(a,b\right)=\left(3;-4\right)\)