+ Viết chương trình

+ Chạy thử chương trình

1) \(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(P=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

\(P=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(P=\left(x^2+5x\right)^2-6^2\)

\(P=\left(x^2+5x\right)^2-36\)

Vì \(\left(x^2+5x\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)

\(\Rightarrow Pmin=-36\Leftrightarrow x^2+5x=0\)

\(\Rightarrow x\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

a) \(\dfrac{2}{3x+9}-\dfrac{x-3}{3x^2+9x}\)

\(=\dfrac{2}{3\left(x+3\right)}-\dfrac{x-3}{3x\left(x+3\right)}\)

\(=\dfrac{2x}{3x\left(x+3\right)}-\dfrac{x-3}{3x\left(x+3\right)}\)

\(=\dfrac{2x-x+3}{3x\left(x+3\right)}\)

\(=\dfrac{x+3}{3x\left(x+3\right)}\)

\(=\dfrac{1}{3x}\)

b) \(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)

\(=\dfrac{x\left(x+1\right)}{5\left(x^2-2x+1\right)}:\dfrac{3\left(x+1\right)}{5\left(x-1\right)}\)

\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}:\dfrac{3\left(x+1\right)}{5\left(x-1\right)}\)

\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}.\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\)

\(=\dfrac{x}{\left(x-1\right).3}\)

\(=\dfrac{x}{3x-3}\)

c) \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+99}-\dfrac{1}{x+100}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+100}\)

\(=\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)

\(=\dfrac{x+100-x}{x\left(x+100\right)}\)

\(=\dfrac{100}{x\left(x+100\right)}\)

a)\(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)

\(\Rightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2=x^2+6x+64\)

\(\Rightarrow\left(2x+3-2x+5\right)^2=x^2+6x+64\)

\(\Rightarrow8^2=x^2+6x+64\)

\(\Rightarrow64=x^2+6x+64\)

\(\Rightarrow x^2+6x=0\)

\(\Rightarrow x\left(x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

b) \(\left(x^4+2x^3+10x-25\right):\left(x^2+5\right)=3\)

\(\Rightarrow\left(x^4+5x^2-5x^2-25+2x^3+10x\right):\left(x^2+5\right)=3\)

\(\Rightarrow\left[x^2\left(x^2+5\right)-5\left(x^2+5\right)+2x\left(x^2+5\right)\right]:\left(x^2+5\right)=3\)

\(\Rightarrow\left(x^2+5\right)\left(x^2-5+2x\right):\left(x^2+5\right)=3\)

\(\Rightarrow x^2+2x-5=3\)

\(\Rightarrow x^2+2x-5-3=0\)

\(\Rightarrow x^2+2x-8=0\)

\(\Rightarrow x^2+4x-2x-8=0\)

\(\Rightarrow x\left(x+4\right)-2\left(x+4\right)=0\)

\(\Rightarrow\left(x+4\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

Ta có:

\(c+d=4\)

\(\Rightarrow\left(c+d\right)^2=4^2\)

\(\Rightarrow c^2+2cd+d^2=16\)

\(\Rightarrow4a^2+b^2+c^2+2cd+d^2=2+16=18\left(1\right)\)

Áp dụng bất đẳng thức Cauchy, ta lại có:

\(4a^2+c^2\ge2.2a.c=4ac\)

\(b^2+d^2\ge2bd\)

\(\Rightarrow4a^2+b^2+c^2+d^2\ge4ac+2bd\)

\(\Rightarrow4a^2+b^2+c^2+2cd+d^2\ge4ac+2bd+2cd\)

\(\Rightarrow18\ge4ac+2bd+2cd\) ( Theo (1) )

\(\Rightarrow18\ge2\left(2ac+bd+cd\right)\)

\(\Rightarrow9\ge2ac+bd+cd\)

\(\Rightarrow2ac+bd+cd\le9\)

\(\Rightarrow Amax=9\Leftrightarrow2a=c;b=d\)

Đề tìm Max mới đúng

\(\left(x-1\right)\left(x+2\right)-x-2=0\)

\(\Rightarrow\left(x-1\right)\left(x+2\right)-\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x-1-1\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)

a) \(\left(27x^2+a\right):\left(3x+2\right)\) được thương là 9x -16 và dư a + 12

Để \(\left(27x^2+a\right)⋮\left(3x+2\right)\) thì số dư phải bằng 0

=> a + 12 = 0

=> a = -12

Bài b và c tham khảo cách làm tương tự ở đây

Giải toán trên mạng - Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath

\(4\left(x-3y\right).\left(x+3y\right)-\left(2x-6y\right)^2\)

\(=4\left(x^2-9y^2\right)-2^2\left(x-3y\right)^2\)

\(=4x^2-36y^2-4\left(x^2-6xy+9y^2\right)\)

\(=4x^2-36y^2-4x^2+24xy-36y^2\)

\(=24xy\)