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a) Ta có:
\(VT=\left(a-b\right)^2\)
\(=a^2-2\cdot a\cdot b+b^2\)
\(=a^2-2ab+b^2\)
\(=a^2-4ab+2ab+b^2\)
\(=\left(a^2+2ab+b^2\right)-4ab\)
\(=\left(a+b\right)^2-4ab=VP\)
⇒ Đpcm
b) Ta có:
\(VT=\left(x+y\right)^2+\left(x-y\right)^2\)
\(=x^2+2\cdot x\cdot y+y^2+x^2-2\cdot x\cdot y+y^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2\)
\(=\left(x^2+x^2\right)+\left(2xy-2xy\right)+\left(y^2+y^2\right)\)
\(=2x^2+0+2y^2\)
\(=2x^2+2y^2\)
\(=2\left(x^2+y^2\right)=VP\)
⇒ Đpcm
a: (a-b)^2
=a^2-2ab+b^2
=a^2+2ab+b^2-4ab
=(a+b)^2-4ab
b: (x+y)^2+(x-y)^2
=x^2+2xy+y^2+x^2-2xy+y^2
=2x^2+2y^2
=2(x^2+y^2)
\(a,VT=\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(VP=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(\Rightarrow VT=a^2c^2+b^2c^2+a^2d^2+b^2d^2=VP\left(đpcm\right)\)
b, Tham khảo:Chứng minh hằng đẳng thức:(a+b+c)3= a3 + b3 + c3 + 3(a+b)(b+c)(c+a) - Hoc24
(a-b)^2=(a-b)(a-b)=a^2-ab-ab+b^2=a^2-2ba+b^2
(a-b)(a+b)=a^2+ab-ab-b^2=a^2-b^2
(a+3)^3=(a+b)^2*(a+b)
=(a^2+2ab+b^2)(a+b)
=a^3+a^2b+2a^2b+2ab^2+b^2a+b^3
=a^3+3a^2b+3ab^2+b^3
a) Ta có:
\(VT=\left(a-b\right)^2\)
\(=a^2-2ab+b^2\)
\(=a^2+2ab+b^2-4ab\)
\(=\left(a+b\right)^2-4ab=VP\left(dpcm\right)\)
b) Ta có:
\(VT=\left(x+y\right)^2+\left(x-y\right)^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2\)
\(=\left(x^2+y^2\right)+\left(x^2+y^2\right)\)
\(=2\left(x^2+y^2\right)=VP\left(dpcm\right)\)
a) \(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)
\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(b+c\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2+bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ac\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
a)Có \(a^2+1\ge2a\) với mọi a; \(b^2+1\ge2b\) với mọi b
Cộng vế với vế \(\Rightarrow a^2+b^2+2\ge2\left(a+b\right)\)
Dấu = xảy ra <=> a=b=1
b) Áp dụng BĐT bunhiacopxki có:
\(\left(x+y\right)^2\le\left(1+1\right)\left(x^2+y^2\right)\Leftrightarrow\left(x+y\right)^2\le2\)
\(\Leftrightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)
\(\Rightarrow\left(x+y\right)_{max}=\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=\dfrac{\sqrt{2}}{2}\)
\(\left(x+y\right)_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=-\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=-\dfrac{\sqrt{2}}{2}\)
c) \(S=\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\)
Với x,y>0, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) (1)
Thật vậy (1) \(\Leftrightarrow\dfrac{y+x}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)\(\Leftrightarrow\left(x-y\right)^2\ge0\) (lđ)
Áp dụng (1) vào S ta được:
\(S\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\)
Lại có: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\Leftrightarrow2ab\le\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow2ab\le\dfrac{1}{2}\)\(\Rightarrow\dfrac{1}{2ab}\ge2\)
\(\Rightarrow S\ge\dfrac{4}{\left(a+b\right)^2}+2=6\)
\(\Rightarrow S_{min}=6\Leftrightarrow a=b=\dfrac{1}{2}\)
a) Áp dụng BĐT Cosi với ab>0, ta có:
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}=2\)(đpcm)
b) Ta có: \(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
a) (a2 +2ab +b2)-(a2-2ab+b2)= a^2+2ab+b^2-a^2+2ab-b^2=2ab+2ab=4ab
b) \(a^2+2ab+b^2+a^2-2ab+b^2=a^2+b^2+a^2+b^2=2a^2+2b^2=2\left(a^2+b^2\right)\)
c)\(a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2\)