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a: \(=-\left(x^2+10x-11\right)\)
\(=-\left(x^2+10x+25-36\right)\)
\(=-\left(x+5\right)^2+36< =36\)
Dấu '=' xảy ra khi x=-5
b: \(=-\left(x^2-6x+5\right)\)
\(=-\left(x^2-6x+9-4\right)\)
\(=-\left(x-3\right)^2+4< =4\)
Dấu '=' xảy ra khi x=3
c: \(=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)
Dấu '=' xảy ra khi x=1/2
d: \(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9< =9\)
Dấu '=' xảy ra khi x=-1
\(A=\left[\left(2x\right)^2+2.2x.y+y^2\right]+\left(16y^2-8y+1\right)\)
\(=\left(2x+y\right)^2+\left(4y-1\right)^2\ge0\)
Đẳng thức xảy ra khi \(x=-\frac{1}{8};y=\frac{1}{4}\)
\(B=\frac{2x^2-\left(x^2+2\right)}{x^2+2}=\frac{2x^2}{x^2+2}-2\ge-1\)
Đẳng thức xảy ra khi x =0
Tí làm tiếp
\(A=x^2+3x+7\)
\(=x^2+2.1,5x+2,25+4,75\)
\(=\left(x+1,5\right)^2+4,75\ge4,75\)
Vậy \(A_{min}=4,75\Leftrightarrow x=-1,5\)
\(B=2x^2-8x\)
\(=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4-4\right)\)
\(=2\left[\left(x-2\right)^2-4\right]\)
\(=2\left(x-2\right)^2-8\ge-8\)
Vậy \(B_{min}=-8\Leftrightarrow x=2\)
\(x^2-4x+1=x^2-2\cdot x\cdot2+4-4+1=\left(x-2\right)^2-4+1\)
\(=\left(x-2\right)^2-3\) \(\forall x\in Z\)
\(\Rightarrow A_{min}=-3khix=2\)
\(a,A=x^2-4x+1=x^2-2.2.x+2^2-3=\left(x-2\right)^2-3\ge-3\)
dấu = xảy ra khi x-2=0
=> x=2
Vậy MinA=-3 khi x=2
\(b,B=5-8x-x^2=-\left(x^2+8x+5\right)=-\left(x^2+2.4.x+4^2\right)+9=-\left(x+4\right)^2+9\le9\)
dấu = xảy ra khi x+4=0
=> x=-4
Vậy MaxB=9 khi x=-4
\(c,C=5x-x^2=-\left(x^2-5x\right)=-\left(x^2-\frac{2.x.5}{2}+\frac{25}{4}\right)+\frac{25}{4}=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
dấu = xảy ra khi \(x-\frac{5}{2}=0\)
=> x=\(\frac{5}{2}\)
Vậy Max C=\(\frac{25}{4}\)khi x=\(\frac{5}{2}\)
\(E=\frac{1}{x^2+5x+14}=\frac{1}{x^2+\frac{2.x.5}{2}+\frac{25}{4}+\frac{31}{4}}=\frac{1}{\left(x+\frac{5}{2}\right)^2+\frac{31}{4}}\)
\(\left(x+\frac{5}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}\)
dấu = xảy ra khi \(x+\frac{5}{2}=0\)
=> x\(=-\frac{5}{2}\)
vì tử thức >0,mẫu thức nhỏ nhất và lớn hơn 0 => E lớnnhất khi mẫu thức nhỏ nhất
Vậy \(MaxE=\frac{31}{4}\)khi x\(=-\frac{5}{2}\)
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
a) \(A=25x^2-10x+9\)
\(A=\left(5x\right)^2-2\cdot5x\cdot1+1^2+9\)
\(A=\left(5x-1\right)^2+9\ge9\)
Dấu "=" xảy ra \(\Leftrightarrow5x-1=0\Leftrightarrow x=\frac{1}{5}\)
a)\(2x^2-4x+7=2x^2-4x+2+5=2\left(x^2-2x+1\right)+5=2\left(x-1\right)^2+5\ge5\)
Dấu "=" xảy ra khi x=1
b)\(9x^2-6x+5=\left(3x\right)^2-2.3x.1+1+4=\left(3x-1\right)^2+4\ge5\)
Dấu "=" xảy ra khi x=1/3
c)\(3x^2-5x+2=3\left(x^2-\frac{5}{3}x+\frac{2}{3}\right)=3\left(x^2-2.\frac{5}{6}.x+\frac{25}{36}-\frac{1}{36}\right)\)
\(=3\left[\left(x-\frac{5}{6}\right)^2-\frac{1}{36}\right]=3\left(x-\frac{5}{6}\right)^2-\frac{1}{12}\ge-\frac{1}{12}\)
Dấu "=" xảy ra khi x=5/6
mấy câu sau tương tự
a) A= \(\frac{3x^2+5x-2}{3x^2-7x+2}=0\)
\(ĐK:3x^2-7x+2\ne0\)
\(\Leftrightarrow\orbr{\begin{cases}x\ne\frac{1}{3}\\x\ne2\end{cases}\left(^∗\right)}\)
=> 3x2 + 5x + 2 =0
<=> 3x2 + 3x + 2x +2 = 0
<=> 3x .( x + 1 ) + 2 .( x + 1 ) =0
<=> ( x + 1 )(3x + 2 ) =0
<=> \(\orbr{\begin{cases}x+1=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{-2}{3}\left(t/m\left(^∗\right)\right)\end{cases}}}\)
Vậy x = -2/3
b) \(B=\frac{2x^2+10x+12}{x^3-4x}=0\left(ĐK:x\ne0;x^2\ne4\Leftrightarrow x\ne0;x\ne\pm2\right)\)
<=> 2x2+ 10x + 12 = 0
<=> x2 + 5x+ 6 =0
<=> ( x + 2 ) ( x + 3 ) =0\(\Leftrightarrow\orbr{\begin{cases}x=-2\left(L\right)\\x=-3\left(t/m\right)\end{cases}}\)
Vậy x = -3
c)\(C=\frac{x^3+x^2-x-1}{x^3+2x-5}=0\) \(ĐK:x^3+2x-5\ne0\left(^∗\right)\)
<=> x3 + x2 -x -1 =0
<=> ( x - 1 )(x2 + 2x + 1 )
<=> ( x-1 ) (x+1)2 = 0
<=> \(\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(t/m\left(^∗\right)\right)\\x=-1\left(t/m\left(^∗\right)\right)\end{cases}}}\)
Vậy x = { 1 ; -1 }
a) A = \(\frac{3x^2+5x-2}{3x^2-7x+2}=0\) (ĐKXĐ: x khác 1/3, x khác 2)
<=> 3x^2 + 5x - 2 = 0
<=> (3x - 1)(x + 2) = 0
<=> 3x - 1 = 0 hoặc x + 2 = 0
<=> 3x = 1 hoặc x = -2
<=> x = 1/3 (ktm) hoặc x = -2 (tm)
=> x = -2
b) B = \(\frac{2x^2+10x+12}{x^3-4x}=0\) (ĐKXĐ: x khác 0, x khác +-2)
<=> \(\frac{2\left(x^2+5x+6\right)}{x\left(x^2-4\right)}=0\)
<=> \(\frac{2\left(x+2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=0\)
<=> \(\frac{2\left(x+3\right)}{x\left(x-2\right)}=0\)
<=> 2(x + 3) = 0
<=> x + 3 = 0
<=> x = -3
c) C = \(\frac{x^3+x^2-x-1}{x^3+2x-5}=0\) (ĐKXĐ: x khác x^3 + 2x - 5)
<=> \(\frac{x^2\left(x+1\right)-\left(x+1\right)}{x^3+2x-5}=0\)
<=> \(\frac{\left(x+1\right)\left(x^2-1\right)}{x^3+2x-5}=0\)
<=> \(\frac{\left(x+1\right)\left(x-1\right)\left(x+1\right)}{x^3+2x-5}=0\)
<=> (x + 1)(x - 1) = 0
<=> x + 1 = 0 hoặc x - 1 = 0
<=> x = -1 hoặc x = 1
a) \(A=\left(x^2-10x+25\right)\)\(-28\)
\(A=\left(x-5\right)^2-28\)\(>=\)-28
MinA = -28 <=> x-5=0 <=> x=5
b)\(B=-\left(x^2+2x+1\right)+6\)
\(B=-\left(x+1\right)^2+6\)\(< =\)6
MaxB = 6 <=> x+1=0 <=> x=-1
c)\(C=-5\left(x^2-\frac{6}{5}x+\frac{9}{25}\right)-\frac{26}{5}\)
\(C=-5\left(x-\frac{3}{5}\right)^2-\frac{26}{5}\)\(< =-\frac{26}{5}\)
MaxC = \(-\frac{26}{5}\)<=> \(x-\frac{3}{5}=0\)<=> x=\(\frac{3}{5}\)
d)\(D=-3\left(x^2+\frac{1}{3}x+\frac{1}{36}\right)+\frac{61}{12}\)
\(D=-3\left(x+\frac{1}{6}\right)^2+\frac{61}{12}\)\(< =\frac{61}{12}\)
MacD = \(\frac{61}{12}\)<=> \(x+\frac{1}{6}=0\)<=> \(x=\frac{-1}{6}\)
Đúng thì nhớ tích cho minh nha