cho x+y=1. tính giá trị biểu thức
\(A=3\left(x^2+y^2\right)-2\left(x^3+y^3\right)\)
\(B=x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\)
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a) \(A=x^2-4x+9=x^2-4x+4+5=\left(x-2\right)^2+5\ge5\)
Vậy \(A_{min}=5\Leftrightarrow x=2\)
b) \(B=x^2-x+1=x^2-x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy \(B_{min}=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)
Giả sử trong hai số a, b không đồng thời chia hết cho 3
=> a+b không chia hết cho 3 => m+2n+n+2m=3(m+n) không chia hết cho 3 ( vô lí )
=> điều giả sử sai => đpcm
\(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có: \(a^2+b^2+c^2+3=2.\left(a+b+c\right)\)
\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}}\text{vì }\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)
\(\Leftrightarrow a=b=c=1\)
Vậy a=b=c=1
=(502-492)+...+(22-12)
=(50-49)(50+49)+(48-47)(48+47)+...+(2-1)(2+1)
=1.99+1.95+1.91+...+1.3
=99+95+91+...+3
=(99+3)+(95+7)+...+
\(50^2-49^2+48^2-47^2+...+2^2-1^2\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=99.1+95.1+...+3.1\)
\(=99+95+...+3\)
\(=3+...+95+99\)
Từ 3 đến 99 có: \(\left(99-3\right):4+1=25\left(\text{số hạng}\right)\)
Tổng là: \(\frac{\left(99+3\right)\times25}{2}=1275\)
a)\(\left(2+1\right)\left(2^2+1\right)....\left(2^{256}+1\right)-1\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)-1\)
\(=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{256}+1\right)-1\)
Tiếp tục như thế, ta được:
\(=\left(2^{256}-1\right)\left(2^{256}+1\right)-1=2^{512}-1-1=2^{512}-2\)
b) \(24\left(5^2+1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^4-1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)
Tiếp tục như thế, ta được:
\(=\left(5^{32}-1\right)\left(5^{32}+1\right)-5^{64}=5^{64}-1-5^{64}=-1\)
\(\left(2+1\right).\left(2^2+1\right)....\left(2^{256}+1\right)-1\)
\(\left(2-1\right).\left(2+1\right).\left(2^2+1\right).....\left(2^{256}+1\right)-1\)
\(=\left(2^2-1\right).\left(2^2+1\right)....\left(2^{256}+1\right)-1\)
\(=\left(2^{256}-1\right).\left(2^{256}+1\right)+1=2^{512}+1\)
Xét các tam giác vuông BKC và BHC có:
BC chung
^KBC=^HBC
=>\(\Delta\)BKC=\(\Delta\)BHC ( ch-gn )
=> BK=HC;KC=BH ( 1 )
Mà AB=AC=>AK=AH
Xét tam giác cân AKH có ^AKH=1800-^KAH-^KHA=\(\frac{180^0-\widehat{A}}{2}\)
Mà tam giác \(ABC\) cân tại A nên \(\widehat{B}=\frac{180^0-\widehat{A}}{2}\)
=> KH//BC ( 2 )
Từ ( 1 );( 2 ) suy ra đpcm
\(A=3\left(x^2+y^2\right)-2\left(x^3+y^3\right)\)
\(=3x^2+3y^2-2\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=3x^2+3y^2-2.1\left(x^2-xy+y^2\right)\)
\(=3x^2+3y^2-2x^2+2xy-2y^2\)
\(=x^2+2xy+y^2=\left(x+y\right)^2=1^2=1\)
\(B=x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2.1\)
\(=x^3+y^3+3xy\left(x+y\right)^2-6x^2y^2+6x^2y^2\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\)
\(=x^2-xy+y^2+3xy\)
\(=x^2+2xy+y^2=\left(x+y\right)^2=1^2=1\)