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1) \(\left(3x-2\right)^2=9x^2-12x+4\)
\(\left(\dfrac{1}{2}x^2+\dfrac{1}{3}\right)^2=\dfrac{1}{4}x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\)
\(\left(a+b\sqrt{3}\right)^2=a^2+2\sqrt{3}ab+3b^2\)
2) \(4a^2+4a+1=\left(2a+1\right)^2\)
\(9x^2-6x+1=\left(3x-1\right)^2\)
\(\dfrac{1}{4}x^2-\dfrac{1}{3}xy+\dfrac{1}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{1}{3}y\right)^2\)
a: \(=\dfrac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{6a^2+6a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-12a}{\left(a-1\right)\left(a^2+a+1\right)}\)
b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{5a^2-5a+5+10a+10-15}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
2. Đặt c + d = x
Ta có: \(a+b+c+d=0\Rightarrow a+b+x=0\Rightarrow a^3+b^3+c^3+d^3=3abx\)
\(\Rightarrow a^3+b^3+c^3+d^3+3cd\left(c+d\right)=3ab\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)=3\left(ab-cd\right)\left(c+d\right)\)
Câu 4:
\(a^{2016}+b^{2016}+c^{2016}=a^{1008}b^{1008}+b^{1008}c^{1008}+c^{1008}+a^{1008}\)
\(\Rightarrow2a^{2016}+2b^{2016}+2c^{2016}-2a^{1008}b^{1008}-2b^{1008}c^{1008}-2c^{1008}a^{1008}=0\)
\(\Rightarrow\left(a^{1008}-b^{1008}\right)^2+\left(b^{1008}-c^{1008}\right)^2+\left(c^{1008}-a^{1008}\right)^2=0\)
\(\Rightarrow a^{1008}=b^{1008},b^{1008}=c^{1008},c^{1008}=a^{1008}\)
\(\Rightarrow a=b,b=c,c=a\) (vì a,b,c > 0 nên \(a\ne-b,b\ne-c,c\ne-a\) )
\(\Rightarrow a-b=0,b-c=0,a-c=0\)
Thay vào A ta tính được A = 0
\(a,\left(2x+y+3\right)^2=4x^2+y^2+9+4xy+12x+6y\)
\(b,\left(x-2y+1\right)^2=x^2+4y^2+1-4xy+2x-4y\)
\(c,\left(x^2-2xy^2-3\right)^2=x^4+2x^2y^4+9-4x^3y^2-6x^2+12xy^2\)
Ý 3 bạn bỏ dòng áp dụng....ta có nhé
\(a^2+b^2+c^2+d^2\ge a\left(b+c+d\right)\)
\(\Leftrightarrow\left(\frac{a^2}{4}-2.\frac{a}{2}b+b^2\right)+\left(\frac{a^2}{4}-2.\frac{a}{2}c+c^2\right)+\)\(\left(\frac{a^2}{4}-2.\frac{a}{d}d+d^2\right)+\frac{a^2}{4}\ge0\forall a;b;c;d\)
\(\Leftrightarrow\left(\frac{a}{2}-b\right)+\left(\frac{a}{2}-c\right)+\)\(\left(\frac{a}{2}-d\right)^2+\frac{a^2}{4}\ge0\forall a;b;c;d\)( luôn đúng )
Dấu " = " xảy ra <=> a=b=c=d=0
6) Sai đề
Sửa thành:\(x^2-4x+5>0\)
\(\Leftrightarrow\left(x-2\right)^2+1>0\)
7) Áp dụng BĐT AM-GM ta có:
\(a+b\ge2.\sqrt{ab}\)
Dấu " = " xảy ra <=> a=b
\(\Leftrightarrow\frac{ab}{a+b}\le\frac{ab}{2.\sqrt{ab}}=\frac{\sqrt{ab}}{2}\)
Chứng minh tương tự ta có:
\(\frac{cb}{c+b}\le\frac{cb}{2.\sqrt{cb}}=\frac{\sqrt{cb}}{2}\)
\(\frac{ca}{c+a}\le\frac{ca}{2.\sqrt{ca}}=\frac{\sqrt{ca}}{2}\)
Dấu " = " xảy ra <=> a=b=c
Cộng vế với vế của các BĐT trên ta có:
\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\)
Áp dụng BĐT AM-GM ta có:
\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\le\frac{\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}}{2}=\frac{2\left(a+b+c\right)}{4}=\frac{a+b+c}{2}\)
Dấu " = " xảy ra <=> a=b=c
1)\(x^3+y^3\ge x^2y+xy^2\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)\ge xy\left(x+y\right)\)
\(\Leftrightarrow x^2-xy+y^2\ge xy\) ( vì x;y\(\ge0\))
\(\Leftrightarrow x^2-2xy+y^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng )
\(\Rightarrow x^3+y^3\ge x^2y+xy^2\)
Dấu " = " xảy ra <=> x=y
2) \(x^4+y^4\ge x^3y+xy^3\)
\(\Leftrightarrow x^4-x^3y+y^4-xy^3\ge0\)
\(\Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)( luôn đúng )
Dấu " = " xảy ra <=> x=y
3) Áp dụng BĐT AM-GM ta có:
\(\left(a-1\right)^2\ge0\forall a\Leftrightarrow a^2-2a+1\ge0\)\(\forall a\Leftrightarrow\frac{a^2}{2}+\frac{1}{2}\ge a\forall a\)
\(\left(b-1\right)^2\ge0\forall b\Leftrightarrow b^2-2b+1\ge0\)\(\forall b\Leftrightarrow\frac{b^2}{2}+\frac{1}{2}\ge b\forall b\)
\(\left(a-b\right)^2\ge0\forall a;b\Leftrightarrow a^2-2ab+b^2\ge0\)\(\forall a;b\Leftrightarrow\frac{a^2}{2}+\frac{b^2}{2}\ge ab\forall a;b\)
Cộng vế với vế của các bất đẳng thức trên ta được:
\(a^2+b^2+1\ge ab+a+b\)
Dấu " = " xảy ra <=> a=b=1
4) \(a^2+b^2+c^2+\frac{3}{4}\ge a+b+c\)
\(\Leftrightarrow\left[a^2-2.a.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]\)\(+\left[b^2-2.b.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]\)\(+\left[c^2-2.c.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]\ge0\forall a;b;c\)
\(\Leftrightarrow\left(a-\frac{1}{2}\right)^2\)\(+\left(b-\frac{1}{2}\right)^2\)\(+\left(c-\frac{1}{2}\right)^2\ge0\forall a;b;c\)( luôn đúng)
Dấu " = " xảy ra <=> a=b=c=1/2
a. 25 - \(x^2\) = (5-x) (5+x)
b) -196 + 4\(x^2\) = 196 - 4\(x^2\) = (14- 2x) (14+2x)
c)\(5^4-81x^4\) = \(\left[\left(5^2\right)^2\right]-\left[\left(81x^2\right)^2\right]\) = (\(\left(5^2-81x^2\right)\left(5^2+81x^2\right)\)
\(a,25-e=\left(5-\sqrt{e}\right)\left(5+\sqrt{e}\right)\)
\(b,-196+g=-\left(196-g\right)=-\left(14-\sqrt{g}\right)\left(14+\sqrt{g}\right)\)
\(c,2^6-47^2=\left(2^3\right)^2-47^2=\left(2^3-47\right)\left(2^3+47\right)\)
\(d,5^4-81x^4=\left(5^2\right)^2-\left(9x^2\right)^2=\left(5^2-9x^2\right)\left(5^2+9x^2\right)=\left(25-9x^2\right)\left(25+9x^2\right)\)
\(i,\dfrac{25}{16}-9y^2=\left(\dfrac{5}{4}-3y\right)\left(\dfrac{5}{4}+3y\right)\)
a) y 2 – 4 xy 2 + 4 x 2 y 2 . b) x 2 − 32 5 xy + 256 25 y 2 .
c) 1 9 a 2 b 4 − c 6 . d) a 4 − 8 9 a + 16 81 .