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Nhiều quá làm 1 bài tiêu biểu thôi nhé:
a/ \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)}\)
\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)\left(c+a\right)\left(b+c\right)\left(a+b\right)\left(c+a\right)\left(b+c\right)}=1\)
1) Thay xyz = 1 , ta có :
\(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}=\frac{z}{z+xz+xyz}+\frac{xz}{xz+xyz+xyz^2}+\frac{1}{1+z+xz}\)
\(=\frac{z}{z+xz+1}+\frac{xz}{xz+1+z}+\frac{1}{z+xz+1}=\frac{z+xz+1}{z+xz+1}=1\)
2) Phân tích A thành nhân tử được \(A=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
Vì a + b + c = 0 nên A = 0
3) Phân tích A thành \(\frac{\left(b-a\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
1
a) Ta có \(\frac{b^2-c^2}{\left(a+b\right).\left(a+c\right)}=\frac{\left(b+c\right)\left(b-c\right)}{\left(a+b\right).\left(a+c\right)}=\frac{\left(b+c\right)\left(a+b-a-c\right)}{\left(a+b\right).\left(a+c\right)}\)
\(=\frac{\left(b+c\right)\left(a+b\right)-\left(b+c\right).\left(a+c\right)}{\left(a+b\right).\left(a+c\right)}=\frac{b+c}{a+c}-\frac{b+c}{a+b}\)
Tương tự \(\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}=\frac{c+a}{b+a}-\frac{c+a}{b+c}\)
\(\frac{a^2-b^2}{\left(c+a\right).\left(c+b\right)}=\frac{a+b}{c+b}-\frac{a+b}{c+a}\)
Do đó \(\frac{b^2-c^2}{\left(a+b\right)\left(a+c\right)}+\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}+\frac{a^2-b^2}{\left(c+a\right).\left(c+b\right)}\)
\(=\frac{b+c}{a+c}-\frac{b+c}{a+b}+\frac{c+a}{b+a}-\frac{c+a}{b+c}+\frac{a+b}{c+b}-\frac{a+b}{c+a}\)
\(=\frac{b+c-a-b}{a+c}+\frac{a+b-c-a}{b+c}+\frac{c+a-b-c}{a+b}\)
\(=\frac{c-a}{a+c}+\frac{b-c}{b+c}+\frac{a-b}{a+b}\)
bđt cần c/m <=>
\(\frac{1}{\left(a+c-b-c\right)^2}+\frac{\left(b+c\right)^2}{\left(a+c\right)^2\left(b+c\right)^2}+\frac{\left(a+c\right)^2}{\left(b+c\right)^2\left(a+c\right)^2}\ge4\\ \)
\(\frac{1}{\left(a+c\right)^2+\left(b+c\right)^2-2}+\left(b+c\right)^2+\left(a+c\right)^2\ge4\\ \)
\(\frac{1}{\left(a+c\right)^2+\left(b+c\right)^2-2}+\left(b+c\right)^2+\left(a+c\right)^2-2\ge2\)(đúng , theo cô-si)
ok
\(2x^2+2y^2=5xy\Leftrightarrow2x^2+2y^2-5xy=0\)
\(\Leftrightarrow\left(2x-y\right)\left(x-2y\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{y}{2}\\x=2y\end{cases}}\)
Mặt khác : x > y > 0 \(\Rightarrow x=2y\)
Ta có : \(E=\frac{x+y}{x-y}=\frac{2y+y}{2y-y}=\frac{3y}{y}=3\)
a) Dễ tự làm đi
b) Xét 1 + a2 = ab + bc + ca + a2
= b(c + a) + a(c + a)
= (c + a)(b + a)
Cmtt ta có : 1 + b2 = (c + b)(a + b)
1 + c2 = (b+c)( a + c)
Do đó : A = \(\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)\left(c+b\right)\left(b+a\right)\left(c+a\right)\left(a+c\right)\left(b+c\right)}\)= 1
Xét a2 + 2bc - 1 = a2 + 2bc - ab - bc - ca
= a2 - ab + bc - ca
= a(a-b) - c(a-b)
= (a-b)(a-c)
Cmtt ta cũng có : b2 + 2ac - 1 = (b-c)(b-a)
c2 + 2ab - 1 = (c-a)(c-b)
Do đó : \(B=\frac{\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ba-1\right)}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)
\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(b-a\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)
= -1
Ta có: \(\frac{1}{x\left(a-b\right)\left(a-c\right)}+\frac{1}{y\left(b-a\right)\left(b-c\right)}+\frac{1}{z\left(c-a\right)\left(c-b\right)}\)
\(=\frac{1}{x\left(a-b\right)\left(a-c\right)}-\frac{1}{y\left(a-b\right)\left(b-c\right)}+\frac{1}{z\left(a-c\right)\left(b-c\right)}\)
\(=\frac{yz\left(b-c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{xz\left(a-c\right)}{yxz\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{xy\left(a-b\right)}{zxy\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)
\(=\frac{yz\left(b-c\right)-xz\left(a-c\right)+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)\(=\frac{yz\left(b-c\right)-xz\left[\left(b-c\right)+\left(a-b\right)\right]+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{yz\left(b-c\right)-xz\left(b-c\right)-xz\left(a-b\right)+xy\left(a-b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)z\left(y-x\right)-\left(a-b\right)x\left(z-y\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)z\left(c+a-b-b-c+a\right)-\left(a-b\right)x\left(a+b-c-c-a+b\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)z\left(2a-2b\right)-\left(a-b\right)x\left(2b-2c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)2z\left(a-b\right)-\left(a-b\right)2x\left(b-c\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(2z-2x\right)}{xyz\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{2\left(z-x\right)}{xyz\left(a-c\right)}=\frac{2\left(a+b-c-b-c+a\right)}{xyz\left(a-c\right)}\)
\(=\frac{2\left(2a-2c\right)}{xyz\left(a-c\right)}=\frac{2.2\left(a-c\right)}{xyz\left(a-c\right)}=\frac{4}{xyz}\Rightarrowđpcm\)