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\(a=x^2-2x+15=x^2-2x+1+14=\left(x-1\right)^2+14>0\)
\(b=4a^2-4a+23=4a^2-4a+4+19=\left(2a-2\right)^2+19>0\)
\(c=2x^2-7x+37=2\left(x^2-\dfrac{7}{2}x+\dfrac{37}{2}\right)\)
\(c=2\left(x^2-\dfrac{7}{2}x+\dfrac{49}{16}+\dfrac{247}{16}\right)\)
\(c=2\left(x^2-\dfrac{7}{2}x+\dfrac{49}{16}\right)+\dfrac{247}{8}\)
\(c=2\left(x-\dfrac{7}{4}\right)^2+\dfrac{247}{8}>0\)
\(d=4x-9x^2-17=-9x^2+4x-17\)
\(d=-9x^2+4x-\dfrac{4}{9}-\dfrac{149}{9}\)
\(d=-\left(9x^2-4x+\dfrac{4}{9}\right)-\dfrac{149}{9}\)
\(d=-\left(3x-\dfrac{2}{3}\right)^2-\dfrac{149}{9}< 0\)
Mẫu câu đầu
\(4x^2+4x-5=4x^2+4x+1-6\)
\(=4\left(x^2+x+\frac{1}{4}\right)-9\)
\(=4\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)-6\)
\(=4\left(x+\frac{1}{2}\right)^2-6\ge-6\)
Vậy Min A=-6 dấu bằng xảy ra khi và chỉ khi \(x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)
a: \(=4a^2+4a+1-6=\left(2a+1\right)^2-6>=-6\)
Dấu = xảy ra khi a=-1/2
b: \(=-\left(y^2-4y-3\right)\)
\(=-\left(y^2-4y+4-7\right)\)
\(=-\left(y-2\right)^2+7< =7\)
Dấu = xảy ra khi y=2
c: \(=-25x^2+3x\)
\(=-25\left(x^2-\dfrac{3}{25}x\right)\)
\(=-25\left(x^2-2\cdot x\cdot\dfrac{3}{50}+\dfrac{9}{2500}-\dfrac{9}{2500}\right)\)
\(=-25\left(x-\dfrac{3}{50}\right)^2+\dfrac{9}{100}< =\dfrac{9}{100}\)
Dấu = xảy ra khi x=3/50
e: \(=3\left(x^2+\dfrac{7}{3}x+\dfrac{1}{3}\right)\)
\(=3\left(x^2+2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}\right)\)
\(=3\left(x+\dfrac{7}{6}\right)^2-\dfrac{37}{12}>=-\dfrac{37}{12}\)
Dấu = xảy ra khi x=-7/6
a, \(2x^2+2x+5x+5=2x\left(x+1\right)+5\left(x+1\right)=\left(2x+5\right)\left(x+1\right)\)
b,\(2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\)
c,\(x^3-3x^2+1-3x=\left(x^3+1\right)-3x\left(x+1\right)=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
d,\(x^2-4x-5=x^2+x-5x-5=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
e,\(\left(a^2+1\right)^2-4a^2=\left(a^2+1\right)^2-\left(2a\right)^2=\left(a^2-2a+1\right)\left(a^2+2a+1\right)=\left(a-1\right)^2\left(a+1\right)^2\)
\(\left(7x-4\right)\left(2x+3\right)-13x\)
\(=14x^2+21x-8x-12-13x\)
\(=14x^2-12\)
\(a^3-\left(a^2-3a\right)\left(a+3\right)\)
\(=a^3-\left(a^3+3a^2-3a^2-9a\right)\)
\(=a^3-a^3-3a^2+3a^2+9a\)
\(=9a\)
\(\left(2a-b\right)\left(b+4a\right)+2a\left(b-3a\right)\)
\(=2ab+8a^2-b^2-4ab+2ab-6a^2\)
\(=\)\(2a^2-b^2\)
\(5b\left(2x-b\right)+\left(x-6a\right)\left(5a+2x\right)\)
\(=10bx-5b^2+5ax+2x^2-30a^2-12ax\)
\(=2x^2-30a^2-5b^2+10bx-7ax\)
a) Ta có: \(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
b) Ta có: \(16x-5x^2-3\)
\(=-5x^2+16x-3\)
\(=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(-5x+1\right)\)
c) Ta có: \(2x^2+7x+5\)
\(=2x^2+2x+5x+5\)
\(=2x\left(x+1\right)+5\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+5\right)\)
d) Ta có: \(2x^2+3x-5\)
\(=2x^2+5x-2x-5\)
\(=x\left(2x+5\right)-\left(2x+5\right)\)
\(=\left(2x+5\right)\left(x-1\right)\)
e) Ta có: \(x^3-3x^2+1-3x\)
\(=\left(x+1\right)\cdot\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
f) Ta có: \(x^2-4x-5\)
\(=x^2-4x+4-9\)
\(=\left(x-2\right)^2-3^2\)
\(=\left(x-2-3\right)\left(x-2+3\right)\)
\(=\left(x-5\right)\left(x+1\right)\)
g) Ta có: \(\left(a^2+1\right)^2-4a^2\)
\(=\left(a^2+1\right)^2-\left(2a\right)^2\)
\(=\left(a^2+1-2a\right)\left(a^2+1+2a\right)\)
\(=\left(a-1\right)^2\cdot\left(a+1\right)^2\)
h) Ta có: \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) Ta có: \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
k) Ta có: \(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
l) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(=3x\left(x+2\right)\)
m) Ta có: \(x^4+4x^2-5\)
\(=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
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