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Bài 1:
a) Ta có: \(A=-x^2-4x-2\)
\(=-\left(x^2+4x+2\right)\)
\(=-\left(x^2+4x+4-2\right)\)
\(=-\left(x+2\right)^2+2\le2\forall x\)
Dấu '=' xảy ra khi x=-2
b) Ta có: \(B=-2x^2-3x+5\)
\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)
\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)
\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)
c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)
\(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9\le9\forall x\)
Dấu '=' xảy ra khi x=-1
Bài 2:
a) Ta có: \(=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)
b) Ta có: \(B=9x^2-6xy+2y^2+1\)
\(=9x^2-6xy+y^2+y^2+1\)
\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)
c) Ta có: \(E=x^2-2x+y^2-4y+6\)
\(=x^2-2x+1+y^2-4y+4+1\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)
\(A=-x^2+3x-5\)\(=-\dfrac{11}{4}-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)=-\dfrac{11}{4}-\left(x-\dfrac{3}{2}\right)^2\le-\dfrac{11}{4}\) với mọi x
\(\Rightarrow A_{max}=-\dfrac{11}{4}\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
\(B=5x-4x^2-3=-\dfrac{23}{16}-\left(4x^2-2.\dfrac{5}{4}.2x+\dfrac{25}{16}\right)\)\(=-\dfrac{23}{16}-\left(2x-\dfrac{5}{4}\right)^2\)\(\le-\dfrac{23}{16}\forall x\)
\(\Rightarrow B_{max}=-\dfrac{23}{16}\Leftrightarrow2x-\dfrac{5}{4}=0\Leftrightarrow x=\dfrac{5}{8}\)
\(C=5-4x-25x^2=\dfrac{129}{25}-\left(25x^2+2.5x.\dfrac{2}{5}+\dfrac{4}{25}\right)\)\(=\dfrac{129}{25}-\left(5x+\dfrac{2}{5}\right)^2\le\dfrac{129}{25}\forall x\)
\(\Rightarrow C_{max}=\dfrac{129}{25}\Leftrightarrow5x+\dfrac{2}{5}=0\Leftrightarrow x=-\dfrac{2}{25}\)
\(D=3x-2x^2=-2\left(x^2-\dfrac{3}{2}x\right)=-2\left(x^2-2.\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{9}{8}\)\(=\dfrac{9}{8}-2\left(x-\dfrac{3}{4}\right)^2\le\dfrac{9}{8}\) với mọi x
\(\Rightarrow D_{max}=\dfrac{9}{8}\Leftrightarrow x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{4}\)
\(E=2+6x-\dfrac{1}{4}x^2=-\dfrac{1}{4}\left(x^2-24x\right)+2=-\dfrac{1}{4}\left(x^2-2.12x+144\right)+38\)\(=38-\dfrac{1}{4}\left(x-12\right)^2\le38\forall x\)
\(\Rightarrow E_{max}=38\Leftrightarrow x-12=0\Leftrightarrow x=12\)
\(F=-5x^2+4x=-5\left(x^2-\dfrac{4}{5}x\right)=-5\left(x^2-2.\dfrac{2}{5}x+\dfrac{4}{25}\right)+\dfrac{4}{5}\)\(=\dfrac{4}{5}-5\left(x-\dfrac{2}{5}\right)^2\le\dfrac{4}{5}\forall x\)
\(\Rightarrow F_{max}=\dfrac{4}{5}\Leftrightarrow x-\dfrac{2}{5}=0\Leftrightarrow x=\dfrac{2}{5}\)
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3