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Bài làm:
a) Ta có: \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1\)
\(=-\left(2x+1\right)^2-1\le-1< 0\left(\forall x\right)\)
=> đpcm
b) \(x^2+4y^2+z^2-2x-6z+8y+15\)
\(=\left(x^2-2x+1\right)+\left(4y^2-8y+4\right)+\left(z^2-6z+9\right)+1\)
\(=\left(x-1\right)^2+4\left(y-1\right)^2+\left(z-3\right)^2+1\ge1>0\left(\forall x\right)\)
=> đpcm
a) Ta có: \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1\)
\(=-\left(2x+1\right)^2-1\)
Vì \(-\left(2x+1\right)^2\le0\forall x\)\(\Rightarrow\)\(-\left(2x+1\right)^2-1\le-1\forall x\)
\(\Rightarrow\)\(-\left(2x+1\right)^2-1< 0\forall x\)
\(\Rightarrow\)\(-4x^2-4x-2< 0\forall x\)( ĐPCM )
b) Ta có: \(x^2+4y^2+z^2-2x-6z+8y+15\)
\(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)
\(=\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\)
Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(2y+2\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{cases}}\)\(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\forall x,y,z\)
\(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\ge1\forall x,y,z\)
\(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1>0\forall x,y,z\)( ĐPCM )
đkxđ: \(a\ne\pm3\)
\(P=\left(\frac{a}{a+3}+\frac{3-a}{a+3}+\frac{a^2+3a+9}{a^2-9}\right)\div\frac{3}{a+3}\)
\(P=\left[\frac{3}{a+3}+\frac{a^2+3a+9}{\left(a-3\right)\left(a+3\right)}\right].\frac{a+3}{3}\)
\(P=\frac{3\left(a-3\right)+a^2+3a+9}{\left(a-3\right)\left(a+3\right)}.\frac{a+3}{3}\)
\(P=\frac{a^2+6a}{3\left(a-3\right)}\)
Đề nghị xem lại đề
Bài làm:
Ta có: \(x^2-22x+127=\left(x^2+22x+121\right)+6=\left(x+11\right)^2\ge6\left(\forall x\right)\)
Áp dụng bất đẳng thức Bunhiacopxki ta có:
\(\left(\sqrt{x-2}+\sqrt{20-x}\right)^2\le\left(1^2+1^2\right)\left[\left(\sqrt{x-2}\right)^2+\left(\sqrt{20-x}\right)^2\right]\)
\(=2\left(x-2+20-x\right)=2.18=36\)
\(\Rightarrow\sqrt{x-2}+\sqrt{20-x}\le\sqrt{36}=6\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-11\right)^2\\x-2=20-x\end{cases}}\Rightarrow x=11\)
a) x = [((n + 1)(n + 4)].[(n + 2)(n + 3)] + 1
= (n2 + 5n + 4)(n2 + 5n + 6) + 1
= (n2 + 5n + 5 - 1)(n2 + 5n + 5 + 1) + 1
= (n2 + 5n + 5)2 - 12 + 1 = (n2 + 5n + 5)2 (đpcm)
b) y = [n(n + 9)].[(n + 3)(n + 6)] + 81
= (n2 + 9n).(n2 + 9n + 18) + 81
= (n2 + 9n + 9 - 9)(n2 + 9n + 9 + 9) + 81
= (n2 + 9n + 9)2 - 92 + 81 = (n2 + 9n + 9)2 (đpcm)
a) \(x=\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)+1\)
\(=\left(n+1\right)\left(n+4\right)\left(n+2\right)\left(n+3\right)+1\)
\(=\left(n^2+5n+4\right)\left(n^2+5n+6\right)+1\) ( 1 )
Đặt \(t=n^2+5n\)
\(\left(1\right)\Leftrightarrow=\left(t+4\right)\left(t+6\right)+1\)
\(=t^2+10+24+1\)
\(=t^2+10t+25\)
\(=\left(t+5\right)^2\)
Vậy x là số chính phương
b) \(y=n\left(n+3\right)\left(n+6\right)\left(n+9\right)+81\)
\(=n\left(n+9\right)\left(n+3\right)\left(n+6\right)+81\)
\(=\left(n^2+9n\right)\left(n^2+9n+18\right)+81\) ( 1 )
Đặt \(a=n^2+9n\)
\(\Leftrightarrow\left(1\right)=a\left(a+18\right)+81\)
\(=a^2+18a+81\)
\(=\left(a+9\right)^2\)
Vậy y là số chính phương
A B C D
Ta có: Vì AB // CD
=> \(\widehat{D}=180^0-\widehat{A}=180^0-30^0=150^0\)
Vì \(\widehat{B}+\widehat{C}=180^0\Leftrightarrow2\widehat{C}+\widehat{C}=180^0\Leftrightarrow3\widehat{C}=180^0\)
\(\Rightarrow\widehat{C}=60^0\Rightarrow\widehat{B}=120^0\)
C1 : Cardano (mk chưa học )
C2 : Mode set up -> 5 -> ax^3 + bx^2 + cx + d = 0
PT <=> \(x_1=-1,209...;x_2=2,104....\)
a) \(5\left(x-2\right)>3\left(x-4\right)\)
\(\Leftrightarrow5x-10>3x-12\)
\(\Leftrightarrow2x>-2\)
\(\Rightarrow x>-1\)
b) \(7\left(x+3\right)< 9\left(x-1\right)\)
\(\Leftrightarrow7x+21< 9x-9\)
\(\Leftrightarrow2x>30\)
\(\Rightarrow x>15\)
c) Vì \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\left(\forall x\right)\)
=> \(2x-5>0\Rightarrow2x>5\Rightarrow x>\frac{5}{2}\)
d) \(x^2-2x+5=\left(x-1\right)^2+4>0\left(\forall x\right)\)
\(\Rightarrow3x-8< 0\Rightarrow3x< 8\Rightarrow x< \frac{8}{3}\)
cd
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