chứng minh rằng:
a) x2+y2≥(\(\left(\dfrac{x+y}{2}\right)\)2
Cho x2 + y2 = 1 và bx2 = ay2
Chứng minh rằng : \(\dfrac{x^{2000}}{a^{1000}}+\dfrac{y^{2000}}{b^{1000}}=\dfrac{2}{\left(a+b\right)^{1000}}\)
\(bx^2=ay^2\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)
\(\Rightarrow\left(\dfrac{x^2}{a}\right)^{1000}=\left(\dfrac{y^2}{b}\right)^{1000}=\left(\dfrac{1}{a+b}\right)^{1000}\)
\(\Rightarrow\dfrac{x^{2000}}{a^{1000}}=\dfrac{y^{2000}}{b^{1000}}=\dfrac{1}{\left(a+b\right)^{1000}}\)
\(\Rightarrow\dfrac{x^{2000}}{a^{1000}}+\dfrac{y^{2000}}{b^{1000}}=\dfrac{1}{\left(a+b\right)^{1000}}+\dfrac{1}{\left(a+b\right)^{1000}}=\dfrac{2}{\left(a+b\right)^{1000}}\)
Chứng minh giá trị của biểu thức không phụ thuộc vào biến
a) 2(x3 + y3) - 3(x2 + y2) với x + y = 1
b) \(\dfrac{\left(x+5\right)^2+\left(x-5\right)^2}{x^2+25}\)
Bạn xem lại đề bài b nhé.
a) \(2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)
\(=2\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]-3\left[\left(x+y\right)^2-2xy\right]\)
\(=2\left(1-3xy\right)-3\left(1-2xy\right)\)
\(=2-6xy-3+6xy=-1\)
\(\Rightarrow\) Giá trị của biểu thức không phụ thuộc vào biến \(x,y\)
b) \(\dfrac{\left(x+5\right)^2+\left(x-5\right)^2}{x^2+25}\)
\(=\dfrac{x^2+10x+25+x^2-10x+25}{x^2+25}\)
\(=\dfrac{2x^2+50}{x^2+25}=\dfrac{2\left(x^2+25\right)}{x^2+25}=2\)
\(\Rightarrow\) Giá trị của biểu thức không phụ thuộc vào biến \(x\)
a) Ta có: \(2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)
\(=2\left(x+y\right)^3-6xy\left(x+y\right)-3\left(x+y\right)^2+6xy\)
\(=\left(x+y\right)^2\left[2\left(x+y\right)-3\right]-6xy\left(x+y-1\right)\)
\(=2\cdot1-3-6xy\left(1-1\right)\)
=-1
b) Ta có: \(\dfrac{\left(x+5\right)^2+\left(x-5\right)^2}{x^2+25}\)
\(=\dfrac{x^2+10x+25+x^2-10x+25}{x^2+25}\)
=2
Chứng minh rằng:
A=\(\dfrac{x^2+y^2}{\left(x-y\right)^2}-\dfrac{2}{xy}:\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^2=1\)
cho các số dương x,y,z chứng minh rằng:
\(\dfrac{x^2}{\left(x+y\right)\left(x+z\right)}\)+\(\dfrac{y^2}{\left(y+z\right)\left(y+x\right)}\)+\(\dfrac{z^2}{\left(z+x\right)\left(z+y\right)}\)≥\(\dfrac{3}{4}\)
Chứng minh rằng giá trị của các biểu thức sau ko phụ thuộc vào biến:
a) y.(x2-y2).(x2+y2)-y.(x4-y4)
b) (\(\dfrac{1}{3}\)+2x).(4x2-\(\dfrac{2}{3}\)x+\(\dfrac{1}{9}\))-(8x3-\(\dfrac{1}{27}\))
c) (x-1)3-(x-1).(x2+x+1)-3.(1-x).x
a: Ta có: \(y\left(x^2-y^2\right)\cdot\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)
\(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)\)
=0
b: Ta có: \(\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\left(8x^3-\dfrac{1}{27}\right)\)
\(=8x^3+\dfrac{1}{27}-8x^3+\dfrac{1}{27}\)
\(=\dfrac{2}{27}\)
c: Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0
\(A=\dfrac{x^2+y^2}{\left(x-y\right)^2}-\dfrac{2}{xy}:\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^2=1\)Chứng minh rằng:
Cho \(a+b+c=a^2+b^2+c^2=1\) và \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) \(\left(a\ne0,b\ne0,c\ne0\right)\)
Chứng minh rằng: \(\left(x+y+z\right)^2=x^2+y^2+z^2\)
Lời giải:
Đặt $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=t$
$\Rightarrow x=at; y=bt; z=ct$. Ta có:
$(x+y+z)^2=(at+bt+ct)^2=t^2(a+b+c)^2=t^2(*)$
Mặt khác:
$x^2+y^2+z^2=(at)^2+(bt)^2+(ct)^2=t^2(a^2+b^2+c^2)=t^2(**)$
Từ $(*); (**)\Rightarrow (x+y+z)^2=x^2+y^2+z^2$ (đpcm)
Giả thiết x, y, z > 0 và xy + y2 + zx = a. Chứng minh rằng :
\(x\sqrt{\dfrac{\left(a+y^2\right)\left(a+z^2\right)}{a+x^2}}+y\sqrt{\dfrac{\left(a+z^2\right)\left(a+x^2\right)}{a+y^2}}+z\sqrt{\dfrac{\left(a+x^2\right)\left(a+y^2\right)}{a+z^2}}=2a\)
Ta có :
\(\left\{{}\begin{matrix}a+y^2=xy+yz+zx+y^2=\left(x+y\right)\left(y+z\right)\\a+z^2=xy+yz+zx+z^2=\left(x+z\right)\left(y+z\right)\\a+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\end{matrix}\right.\)
Do đó :
\(VT=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\left(x+z\right)\right)}}+y\sqrt{\dfrac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(x+z\right)\left(x+y\right)\left(y+z\right)}{\left(x+z\right)\left(y+z\right)}}\)
\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(=2\left(xy+yz+zx\right)\)
\(=2a\) ( đpcm )
cho 3 số x,y,z đôi 1 khác nhau và chứng minh rằng :
\(\dfrac{y-z}{\left(x-y\right)\cdot\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\cdot\left(y-x\right)}+\dfrac{y-x}{\left(z-x\right)\cdot\left(z-y\right)}=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)
Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)
Tương tự:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)
Chứng minh rằng nếu x+y=1 thì x2 + y2 \(\ge\) \(\dfrac{1}{2}\)
Mong mn giúp đỡ
\(x+y=1\)
Áp dụng BĐT AM-GM, ta có:
\(\dfrac{x^2}{1}+\dfrac{y^2}{1}\ge\dfrac{\left(x+y\right)^2}{2}=\dfrac{1^2}{2}=\dfrac{1}{2}\)
--> \(x^2+y^2\ge\dfrac{1}{2}\)