Cho x,y>0.Chứng tỏ \(\frac{x}{y}+\frac{y}{x}\)lớn hơn hoặc bằng 1/2
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100 + 120 + 451 + 115 x 0
= 100 + 120 + 451 + 0
= 220 + 451
= 671
a)
\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)
\(=16\)
\(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=\left(a-b+c+b-c\right)\left(a-b+c-b+c\right)+2ab-2ac\)
\(=a\left(a-2b+2c\right)+2ab-2ac\)
\(=a^2-2ab+2ac+2ab-2ac\)
\(=a^2\)
\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)
\(=\left(3x+1-3x-5\right)^2\)
\(=\left(-4\right)^2=16\)
Sửa đề: Cho \(a^2+b^2+c^2=m\)
Tính: \(A=\left(2a+2b-c\right)^2+\left(2b+2c-a\right)^2+\left(2c+2a-b\right)^2\)
Giải:
Ta có: \(\left(x+y-z\right)^2=\left(x+y\right)^2-2\left(x+y\right).z+z^2=x^2+y^2+z^2+2xy-2xz-2yz\)
Ứng dụng vào bài trên:
\(A=\left[\left(2a\right)^2+\left(2b\right)^2+c^2+2\left(2a\right)\left(2b\right)-2\left(2a\right)c-2\left(2b\right)c\right]\)
\(+\left[\left(2b\right)^2+\left(2c\right)^2+a^2+2\left(2b\right)\left(2c\right)-2\left(2b\right)a-2\left(2c\right)a\right]\)
\(+\left[\left(2c\right)^2+\left(2a\right)^2+b^2+2\left(2c\right)\left(2a\right)-2\left(2c\right)b-2\left(2a\right)b\right]\)
\(=4a^2+4b^2+c^2+8ab-4ac-4bc\)
\(+4b^2+4c^2+a^2+8bc-4ba-4ca\)
\(+4c^2+4a^2+b^2+8ca-4cb-4ab\)
\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)
\(=9m\).
\(A=x^2+2xy+y^2-4x-4y+q\)
\(=\left(x+y\right)^2-4\left(x+y\right)+q\)
\(=3^2-4.3+q\)
\(=q-3\)
\(4x-4y+x^2-y^2\)
\(=4\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y+4\right)\)
Ta có:
\(\left(\frac{x}{y}+\frac{y}{x}\right)^2=\frac{x^2}{y^2}+2.\frac{x}{y}.\frac{y}{x}+\frac{y^2}{x^2}=\left(\frac{x}{y}-\frac{y}{x}\right)^2+4.\frac{x}{y}.\frac{y}{x}\)
\(=\left(\frac{x}{y}-\frac{y}{x}\right)^2+4\ge4\) với mọi x y >0
Vì x, y >0 => \(\frac{x}{y}+\frac{y}{x}>0\) mà \(\left(\frac{x}{y}+\frac{y}{x}\right)^2\ge4\)
=> \(\frac{x}{y}+\frac{y}{x}\ge2>\frac{1}{2}\)với mọi x, y >0
"=" xảy ra <=> x =y
Em kiểm tra lại đề bài nha.