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a)
(x-2).(x+2)-(x+2)^2=4
<=>(x^2-2^2)-(x^2+4x+4)=4
<=> x^2-4-x^2-4x-4=4
<=> -4x=12
<=> x=-3
a) ( x - 2 )( x + 2 ) - ( x + 2 )2 = 4
<=> x2 - 4 - ( x2 + 4x + 4 ) = 4
<=> x2 - 4 - x2 - 4x - 4 = 4
<=> -4x - 8 = 4
<=> -4x = 12
<=> x = -3
b) 4( x + 1 )2 + ( 2x - 1 )2 - 8( x - 1 )( x + 1 ) = 11
<=> 4( x2 + 2x + 1 ) + 4x2 - 4x + 1 - 8( x2 - 1 )
<=> 4x2 + 8x + 4 + 4x2 - 4x + 1 - 8x2 + 8 = 11
<=> 4x + 13 = 11
<=> 4x = -2
<=> x = -2/4 = -1/2

\(x^2-3x+1\)
\(=x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{5}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\)
\(=\left(x-\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{5}}{2}\right)\)

\(10,\\ a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\\ \Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2\ge4ab+4ac+4ad+4ae\\ \Leftrightarrow\left(a^2-4ab+4b^2\right)+\left(a^2-4ac+4c^2\right)+\left(a^2-4ad+4d^2\right)+\left(a^2-4ae+4e^2\right)\ge0\\ \Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+\left(a-2e\right)^2\ge0\left(luôn.đúng\right)\)
Dấu \("="\Leftrightarrow\dfrac{a}{2}=b=c=d=e\)
\(4,\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-\dfrac{1}{4}\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)\ge3ab+3bc+3ca\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-\dfrac{1}{2}a^2-\dfrac{1}{2}b^2-\dfrac{1}{2}c^2-ab-bc-ac\ge0\\ \Leftrightarrow\dfrac{1}{2}a^2+\dfrac{1}{2}b^2+\dfrac{1}{2}c^2+ab+ac+bc\ge0\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge0\\ \Leftrightarrow\left(a+b+c\right)^2\ge0\left(luôn.đúng\right)\)
Dấu \("="\Leftrightarrow a+b+c=0\)



a) \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\\\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{15}=\dfrac{z}{21}\end{matrix}\right.\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y+z}{10+15+21}=\dfrac{92}{46}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{10}=2\Rightarrow x=2.10=20\\\dfrac{y}{15}=2\Rightarrow y=2.15=30\\\dfrac{z}{21}=2\Rightarrow z=2.21=42\end{matrix}\right.\)

Ta có : D=(x^2 - 2x + 1) + (4x^2 + 4x +1)
= x^2 - 2x + 1 + 4x^2 +4x + 1
= 5x^2 + 2x + 2
=5(x^2 + 2/5x + 2/5)
=5(x^2 + 2/5x + 1/25 + 9/25)
=5(x^2 +2/5x +1/25) + 9/5 >= 9/5
Vậy MinD=9/5 khi x=-1/5
Mình bổ sung 1 chút ở chỗ 5(x^2 + 2/5x + 1/25) + 9/5 = 5(x+1/5)^2 + 9/5 >= 9/5
1+1=2 nha em
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