a, (4x - 3)2-(x + 1)3- 2x(x2-1)
b, 12 - 22 + 32 - 42 + 52 - 62+ ........... + 20172 -20122
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\(\frac{4}{9}x^4-16x^2=\left(\frac{2}{3}x^2\right)^2-\left(4x\right)^2=\left(\frac{2}{3}x^2+4x\right).\left(\frac{2}{3}x^2-4x\right)\)
b. Sử dụng các hằng đẳng thức
\(a^3+b^3+c^2-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=3\left(a^2+b^2+c^2-ab-bc-ca\right)\)
và \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
nên \(A=\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{1}{2}.\frac{\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Do (a - b) + (b - c) + (c - a) = 0 nên áp dụng hđt \(X^2+Y^2+Z^2=-2\left(XY+YZ+ZX\right)\)khi X + Y + Z = 0, ta có:
\(A=-2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right).\)
Bài 1 :
\(b,ax^2+3ax+9=a^2\)
\(\Leftrightarrow a^2x+3ax+9-a^2=0\)
\(\Leftrightarrow ax\left(a+3\right)+\left(a+3\right)\left(3-a\right)=0\)
\(\Leftrightarrow\left(a+3\right)\left(ax+3-a\right)=0\)
Vì \(a\ne3\Rightarrow\left(a+3\right)\ne0\Rightarrow ax+3-a=0\)
\(\Leftrightarrow ax=a-3\)
Vì \(a\ne0\Rightarrow x=\frac{a-3}{a}\)
a) Do \(1010\le n\le2016\)nên:
\(\sqrt{20203+21\times1010}\le a_n\le20203+21\times2016\)\(\Leftrightarrow204\le a_n\le250\)
b) Ta có:
\(a^2_n=20203+21n=\left(21\times962+1\right)+21n\)
\(\Leftrightarrow a^2_n-1=21\times\left(962+n\right)=3\times7\times\left(962+n\right)\)
\(\Rightarrow\left(a_n-1\right)\left(a_n+1\right)⋮7\Leftrightarrow\hept{\begin{cases}\left(a_n-1\right)⋮7\\\left(a_n+1\right)⋮7\end{cases}}\)
Hay \(a_n+1=7k\)hoặc \(a_n-1=7k\)\(\Rightarrow a_n=7k-1\)hoặc \(a_n=7k+1\left(k\in N\right)\)
\(\Rightarrow dpcm\)
{\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}\,}
{\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}\,}
{\displaystyle a^{2}-b^{2}=(a-b)(a+b)\,}
{\displaystyle (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\,}
{\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}\,}
{\displaystyle a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})=(a+b)^{3}-3a^{2}b-3ab^{2}=(a+b)^{3}-3ab(a+b)}
{\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})=(a-b)^{3}+3a^{2}b-3ab^{2}=(a-b)^{3}+3ab(a-b)}
\(B=2x^2+y^2+2xy+6x+2y+2015\)
\(=x^2+y^2+1+2xy+2y+2x+x^2+4x+4+2011\)
\(=\left(x^2+y^2+1+2xy+2y+2x\right)+\left(x^2+4x+4\right)+2011\)
\(=\left(x+y+1\right)^2+\left(x+2\right)^2+2011\)
Vì \(\left(x+y+1\right)^2+\left(x+2\right)^2\ge0\)nên \(\left(x+y+1\right)^2+\left(x+2\right)^2+2011\ge2011\)
Vậy \(MinB=2011\Leftrightarrow\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(x+2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y+1=0\\x+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=1\end{cases}}\)
Ta có:2x2-4x+10=2x2-4x+2+8
=2(x2-2x+1)+8=2(x-1)2+8.Vì \(\left(x-1\right)^2\ge0\Rightarrow2\left(x-1\right)^2\ge0\)
\(\Rightarrow2\left(x-1\right)^2+8\ge8\)\(\Rightarrow\)GTNN của A=8 đạt được khi \(\left(x-1\right)^2=0\Leftrightarrow x=1\)
Ta có : 2x2 - 4x + 10
= 2(x2 - 2x + 5)
= 2(x2 - 2x + 1 + 4)
= 2[(x - 1)2 + 4 ]
= 2(x - 1)2 + 4
Mà 2(x - 1)2 \(\ge0\forall x\)
Nên : 2(x - 1)2 + 4 \(\ge4\forall x\)
Vậy Amin = 4 , dấu "=" xảy ra khi và chỉ khi x = 1
Câu b giống tính tổng nhi?
12 - 22 + 32 - 42 + 52 - 62 +... + 20172 - 20122
= ( 12 - 22 ) + ( 32 -42 ) + ( 52 - 62 ) +.... + ( 20172 -20122 )
= -3 + -7 + -11 + ... + 20145
= ( 20145 + ( -3 ) . 5038 / 2
= 50737698
Cbht!!!!