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a: Ta có: \(A=x^2-7x+11\)
\(=x^2-2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{5}{4}\)
\(=\left(x-\dfrac{7}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{7}{2}\)
b: ta có: \(A=9x^2+6x+11\)
\(=9x^2+6x+1+10\)
\(=\left(3x+1\right)^2+10\ge10\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =4x^2-4x+1+9
=(2x-1)^2+9>=9
Dấu = xảy ra khi x=1/2
b: =2(x^2+3x)
=2(x^2+3x+9/4-9/4)
=2(x+3/2)^2-9/2>=-9/2
Dấu = xảy ra khi x=-3/2
c: =x^2-x+1/4-1/4
=(x-1/2)^2-1/4>=-1/4
Dấu = xảy ra khi x=1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
C = -( 9x2 -2x +1) -17
= -(3x-1)2-17
ta có -(3x-1)2 bé hơn hoặc bằng 0 với mọi x
nên -(3x-1)2 -17 bé hơn hoặc bằng -17 với mọi x
vậy.............
\(C=-9x^2+2x-17\)
\(=-9\left(x^2-2.\dfrac{1}{9}x+\dfrac{1}{81}\right)-\dfrac{152}{9}\)
\(=-9\left(x-\dfrac{1}{9}\right)^2-\dfrac{152}{9}\)
Vì \(-9\left(x-\dfrac{1}{9}\right)^2\le0\)
Nên \(-9\left(x-\dfrac{1}{9}\right)^2-\dfrac{152}{2}\le0\)
Vậy C luôn âm với mọi giá trị của biến
\(D=-5x^2-6x-11\)
\(=-5\left(x^2+2.\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{46}{5}\)
\(=-5\left(x+\dfrac{3}{5}\right)^2-\dfrac{46}{5}\)
Vì \(-5\left(x+\dfrac{3}{5}\right)^2\le0\)
Nên \(-5\left(x+\dfrac{3}{5}\right)^2-\dfrac{46}{5}\le0\)
vậy D luôn âm với mọi giá trị của biến
\(E=\dfrac{-1}{4}x^2+3x-15\)
\(=-\dfrac{1}{4}\left(x^2-12x+36\right)-6\)
\(=-\dfrac{1}{4}\left(x-6\right)^2-6\le0\)
Vậy E luôn âm với mọi giá trị
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhân chéo, chuyển vế đưa về dạng pt bậc 2, xét đenta cho nó >=0 rồi giải
\(A=\frac{4x-11}{x^2+3}=\frac{4x^2+4x+1-4x^2-12}{x^2+3}=\frac{\left(2x+1\right)^2-4\left(x^2+3\right)}{x^2+3}=\frac{\left(2x+1\right)^2}{x^2+3}-4\)
Phân số \(\frac{\left(2x+1\right)^2}{x^2+3}\ge0\forall x\Rightarrow A=\frac{\left(2x+1\right)^2}{x^2+3}-4\ge-4\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2x+1\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy \(A_{min}=-4\Leftrightarrow x=-\frac{1}{2}\)
Chúc bạn học tốt.
![](https://rs.olm.vn/images/avt/0.png?1311)
A = (x^4-2x^2+1)+(3x^2-6x+3)+5
= (x^2-1)^2+3.(x-1)^2+5 >= 5
Dấu "=" xảy ra <=> x^2-1=0 và x-1=0 <=> x=1
Vậy Min A = 5 <=> x=1
k mk nha
A=\(x^4+x^2-6x+9\)
\(=\left(x^4-2x^2+1\right)\left(3x^2-6x+3\right)+5\)
\(=\left[\left(x^2\right)^2-2x^2.1+1^2\right]+3.\left(x^2-2x+1\right)+5\)
\(=\left(x^2-1\right)^2+3.\left(x-1\right)^2+5\ge5\)
Min A=5 khi \(\hept{\begin{cases}x^2-1=0\\x-1=0\end{cases}}\)=> x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\frac{5}{2}\)
\(\Leftrightarrow3x^3-x^2+x-3x^2+x-1+4x^2-3x^3=\frac{5}{2}\)
\(\Leftrightarrow2x-1=\frac{5}{2}\Leftrightarrow2x=1+\frac{5}{2}=\frac{7}{2}\Leftrightarrow x=\frac{7}{4}\)
b)
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
\(\Leftrightarrow4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(\Leftrightarrow8x+4-4x+1+8=11\Leftrightarrow4x+13=11\Leftrightarrow4x=-2\Leftrightarrow x=-\frac{1}{2}\)
c)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow-4x+1+6x+9+245=0\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{255}{2}\).
a ) ( 3x2 - x + 1 ) ( x + 1 ) + x2 ( 4 - 3x ) = 5/2
=> 3x3 + 3x2 - x2 - x + x + 1 + 4x2 - 3x3 = 5/2
=> 6x2 + 1 = 5/2
=> 6x2 = 1,5
=> x2 = 0,25
=> x = 0,5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+1\right)2-4\left(x+2\right)2=9\)
\(4x+2-8x-16=9\)
\(4x-8x=9+16-2\)
\(-4x=23\)
\(x=-\frac{23}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: Ta có: \(4x^2-36=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)
\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)
\(\Leftrightarrow2x=10\)
hay x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+2\right).\left(3x-2\right)-\left(3x-1\right).\left(x-5\right)=11\)
\(\Rightarrow3x^2-2x+6x-4-\left(3x^2-15x-x+5\right)=11\)
\(\Rightarrow3x^2-2x+6x-4-3x^2+15x+x-5=11\)
\(\Rightarrow20x-9=11\)
\(\Rightarrow20x=20\Rightarrow x=1\)
(x + 2)(3x - 2) - (3x - 1)(x - 5) = 11
=> 3x2 - 2x + 6x - 4 - 3x2 + 15x + x - 5 = 11
=> 20x - 9 = 11
=> 20x = 11 + 9
=> 20x = 20
=> x = 20 : 20
=> x = 1
\(A=x^2-10x+25+2x^2-4x+2+11=3x^2-14x+38=3\left(x^2-\frac{14}{3}x+\frac{38}{3}\right)\)
\(=3\left(x^2-2\cdot\frac{7}{3}x+\frac{49}{9}-\frac{49}{9}+\frac{38}{3}\right)=3\left(x-\frac{7}{3}\right)^2+\frac{65}{3}\ge\frac{65}{3}\)
Vậy \(Min_A=\frac{65}{3}\) khi x=7/3
(kiểm tra lại nhé, hôm nay tớ làm bài dễ bị sai lắm)