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Bài 4:
Ta có:
\(a^2-2a+b^2+4b+4c^2-4c+6=0\)
\(\Leftrightarrow a^2-2a+1+b^2+4b+4+4c^2-4c+1\)
\(\Leftrightarrow\left(a^2-2b+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\)
Mà \(\hept{\begin{cases}\left(a-1\right)^2\ge0\\\left(b+2\right)^2\ge0\\\left(2c-1\right)^2\ge0\end{cases}}\)
\(\Rightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+2\right)^2=0\\\left(2c-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=-2\\c=\frac{1}{2}\end{cases}}}\)
Vậy \(\left(a,b,c\right)=\left(1;-2;\frac{1}{2}\right)\)
1. CMR với mọi số nguyên dương ta có:
A= x5/120 + x4/12 + 7x3/24 + 5x2/12 + x/5 luôn luôn dương
2. Cho a3 + 3ab2 =14 và b3 + 3a2b =13 . Tính: P= a2 _ b2
bui hai nam: s cóp lại y nguyên đề trong phần trả lời z bn =='
\(\left(10^{12}+5^{11}.2^9-5^{13}.2^8\right):4.5^5.10^6\)
\(=\left(5^{12}.2^{12}+5^{11}.2^9-5^{13}.2^8\right):2^2.5^5.2^6.5^6\)
\(=\left[5^{11}.2^8\left(5.2^4+2-5^2\right)\right]:2^8.5^{11}\)
\(=\frac{5^{11}.2^8\left(5.16+2-25\right)}{5^{11}.2^8}\)
\(=80+2-25\)
\(=57\)
1 ) \(x=7\Rightarrow x+1=8\)
\(\Rightarrow B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...-\left(x+1\right)x^2+\left(x+1\right)x-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}+....-x^3-x^2+x^2+x-5\)
\(=x-5=7-5=2\)
2 ) Gọi 3 số tự nhiên liên tiếp đó là a; a + 1; a + 2 (a thuộc N)
theo đề bài ta có : \(\left(a+1\right)\left(a+2\right)-a\left(a+1\right)=50\)
\(\Leftrightarrow a^2+3a+2-a^2-a=50\)
\(\Leftrightarrow2a+2=50\)
\(\Rightarrow a=24\)
Vậy 3 số TN liên tiếp cần tìm là 24;25;26
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
\(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)
\(=1\)
\(4x^2-4x-5=4x^2-4x+1-6=\left(2x-1\right)^2-6\ge-6\)
\(Min=-6\Leftrightarrow x=\dfrac{1}{2}\)
\(4x^2+12x+10=4\left(x^2+3x+\dfrac{9}{4}\right)+1=4\left(x+\dfrac{3}{2}\right)^2+1\ge1\)
\(Min=1\Leftrightarrow x=-\dfrac{3}{2}\)
\(4x^2-12x-5=4\left(x^2-3x+\dfrac{9}{4}\right)-14=4\left(x-\dfrac{3}{2}\right)^2-14\ge-14\)
\(Min=-14\Leftrightarrow x=\dfrac{3}{2}\)
\(9x^2+12x+8=\left(9x^2+12x+4\right)+4=\left(3x+2\right)^2+4\ge4\)
\(Min=4\Leftrightarrow x=-\dfrac{2}{3}\)