Tìm x
a, \(x+3+\frac{4-3a^2}{a^2-9}=\frac{5}{2a^2+6a}\left(a\ne0,a\ne3\right).\)
b, \(\frac{a^2-2ab+b^2}{a^4-b^4}.x=\frac{a^2-b^2}{a^2+b^2}\)
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1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
\(3a-b=3m\Rightarrow b=3a-3m\)
\(\Rightarrow2a+\left(3a-3m\right)=7m\Rightarrow5a=10m\Rightarrow a=2m\)
\(\Rightarrow b=3a-3m=6m-3m=3m\)
\(\Rightarrow P=\frac{\left(2m\right)^2-2.2m.3m}{\left(2m\right)^2+\left(3m\right)^2}=\frac{4m^2-12m^2}{4m^2+9m^2}=\frac{-8m^2}{13m^2}=\frac{-8}{13}\)
Đặt \(x^2=a\ge0;y^2=b\ge0\)
Ta có BĐT phụ:\(4ab\le\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\left(true\right)\)
Ta có:\(\frac{4ab}{\left(a+b\right)^2}+\frac{a}{b}+\frac{b}{a}\ge\frac{\left(a+b\right)^2}{\left(a+b\right)^2}+2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=3\) ( BĐT AM-GM )
Ta có đpcm
Câu 2:
\(\frac{a^2b}{2a^3+b^3}-\frac{1}{3}+1-\frac{a^2+2ab}{2a^2+b^2}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2}{2a^2+b^2}-\frac{\left(a-b\right)^2\left(2a+b\right)}{3\left(2a^3+b^3\right)}\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left[\frac{1}{2a^2+b^2}-\frac{\left(2a+b\right)}{3\left(2a^3+b^3\right)}\right]\ge0\)
\(\Leftrightarrow\frac{2\left(a-b\right)^4\left(a+b\right)}{3\left(2a^2+b^2\right)\left(2a^3+b^3\right)}\ge0\left(ok!\right)\)
Em tính/ quy đồng/ phân tích thành nhân tử sai chỗ nào thì chị tự check nhá:)
\(B=\frac{2}{x^2-y^2}\cdot\sqrt{\frac{9\left(x^2+2xy+y^2\right)}{4}}=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\sqrt{\frac{9\left(x+y\right)^2}{4}}\)
\(=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\frac{\sqrt{9\left(x+y\right)^2}}{\sqrt{4}}=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\frac{3\left(x+y\right)}{2}\)(vì x > -y <=> x + y > 0)
\(=\frac{3}{x-y}\)
\(C=\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}=\sqrt{\frac{2a}{3}\cdot\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\frac{a}{2}\)(vì a > = 0)
\(D=\frac{1}{a-b}\cdot\sqrt{a^4\left(a-b\right)^2}=\frac{1}{a-b}\cdot a^2\left(a-b\right)=a^2\)(a > b > 0)
câu cuối điều kiện là a>b
\(\frac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}=\frac{a^2\left|a-b\right|}{a-b}=\frac{a^2\left(a-b\right)}{a-b}=a^2\) (vì a>b)
Lời giải:
a) \(\frac{x^2-16}{4x-x^2}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
c)
\(\frac{(x+y)^2-z^2}{x+y+z}=\frac{(x+y-z)(x+y+z)}{x+y+z}=x+y-z\)
d)
Biểu thức không rút gọn được
e)
\(\frac{a^3+b^3+c^3}{a^2+b^2+c^2-ab-bc-ac}=\frac{(a+b)^3-3ab(a+b)+c^3}{a^2+b^2+c^2-ab-bc-ac}=\frac{(a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\frac{(a+b+c)(a^2+b^2+c^2-ac-bc+2ab)-3ab(a+b+c)+3abc}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc}{a^2+b^2+c^2-ab-bc-ac}=a+b+c+\frac{3abc}{a^2+b^2+c^2-ab-bc-ac}\)