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\(\dfrac{4a-4b}{5a+5b}.x=\dfrac{a^2-b^2}{a^2+2ab+b^2}\)
\(\Leftrightarrow\dfrac{4\left(a-b\right)}{5\left(a+b\right)}.x=\dfrac{\left(a-b\right)\left(a+b\right)}{\left(a+b\right)^2}\)
\(\Leftrightarrow x=\dfrac{a-b}{a+b}.\dfrac{5\left(a+b\right)}{4\left(a-b\right)}\)
\(\Leftrightarrow x=\dfrac{5}{4}\)
Vậy \(x=\dfrac{5}{4}\)
Ta có: \(\dfrac{4\left(a-b\right)}{5\left(a+b\right)}x=\dfrac{\left(a-b\right)\left(a+b\right)}{\left(a+b\right)^2}\)
\(\Rightarrow\)\(\dfrac{4\left(a-b\right)}{5\left(a+b\right)}x=\dfrac{a-b}{a+b}\)\(\Rightarrow\dfrac{4}{5}x=\dfrac{a-b}{a+b}.\dfrac{a+b}{a-b}=1\)
\(\Rightarrow\)x = \(1:\dfrac{4}{5}=\dfrac{5}{4}\)
Chúc các bạn học tốt
Ta có :
\(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\)
\(\Leftrightarrow\) \(\dfrac{1}{1+a^2}-\dfrac{1}{1+ab}+\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\ge0\)
\(\Leftrightarrow\) \(\dfrac{1+ab-1-a^2}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{1+ab-1-b^2}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)
\(\Leftrightarrow\) \(\dfrac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)
\(\Leftrightarrow a\left(b-a\right)\left(1+b^2\right)+b\left(a-b\right)\left(1+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left[-a\left(1+b^2\right)+b\left(1+a^2\right)\right]\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(-a-ab^2+b+a^2b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left[ab\left(a-b\right)-\left(a-b\right)\right]\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a-b\right)\left(ab-1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(ab-1\right)\ge0\) (*)
Vì \(a.b=1\Rightarrow ab-1=0,\left(a-b\right)^2\ge0\)
Do đó (*) đúng . Vậy \(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\left(đpcm\right)\)
a) \(\dfrac{20+10a+5a^2}{a^3-8}=\dfrac{5\left(a^2+2a+4\right)}{\left(a-2\right)\left(a^2+2a+4\right)}=\dfrac{5}{a-2}\)
b) \(\dfrac{x\left(2-x\right)}{x^2-5x+6}=\dfrac{-x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{-x}{x-3}=\dfrac{x}{3-x}\)
c) \(\dfrac{y^2-x^2}{x^2-3xy+2y^2}=\dfrac{\left(y-x\right)\left(y+x\right)}{\left(x-y\right)\left(x-2y\right)}=\dfrac{x+y}{2y-x}\)
Theo AM-GM ta có:
\(\left\{{}\begin{matrix}b^2+1\ge2\sqrt{b^2}=2b\\a^2+b^2\ge2\sqrt{a^2b^2}=2ab\end{matrix}\right.\)
\(\Rightarrow a^2+2b^2+1\ge2ab+2b\Rightarrow a^2+2b^2+3\ge2ab+2b+2\)
\(=2\left(ab+b+1\right)\Rightarrow\dfrac{1}{a^2+2b^2+3}\le\dfrac{1}{2\left(ab+b+1\right)}\)
Tương tự cho 2 BĐT còn lại ta có:
\(\dfrac{1}{b^2+2c^2+3}\le\dfrac{1}{2\left(bc+c+1\right)};\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2\left(ca+a+1\right)}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{ab}{ab+b+1}+\dfrac{b}{ab+b+1}+\dfrac{1}{ab+b+1}\right)\left(abc=1\right)\)
\(=\dfrac{1}{2}\left(\dfrac{ab+b+1}{ab+b+1}\right)=\dfrac{1}{2}=VP\)
1a)\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
b)\(\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{9}\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
2a)\(a^2+\dfrac{b^2}{4}\ge ab\)
\(\Leftrightarrow a^2-ab+\dfrac{b^2}{4}\ge0\)
\(\Leftrightarrow a^2-2\cdot\dfrac{1}{2}b\cdot a+\left(\dfrac{1}{2}b\right)^2\ge0\)
\(\Leftrightarrow\left(a-\dfrac{1}{2}b\right)^2\ge0\)(luôn đúng)
b)Đã cm
c)\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)
Dấu bằng xảy ra khi a=b=1
\(\left\{{}\begin{matrix}a.b\ne0\left(!\right)\\9a^2-b\ne0\left(!!\right)\\10a^2-3b^2-5ab=0\left(1\right)\\A=\dfrac{2a-b}{3a-b}+\dfrac{5b-a}{3a+b}-3\left(2\right)\end{matrix}\right.\)
Từ (!) \(\Rightarrow\left(1\right)\Leftrightarrow10-3\left(\dfrac{b}{a}\right)^2-5\left(\dfrac{b}{a}\right)=0\)(3)
Đặt b/a =x
\(\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}3x^2+5x-10=0\\\left[{}\begin{matrix}x_1=\dfrac{-5-\sqrt{5.29}}{6}\\x_2=\dfrac{-5+\sqrt{5.29}}{6}\end{matrix}\right.\end{matrix}\right.\)(4)
Từ (!) \(\Rightarrow\left(2\right)\Leftrightarrow A=\dfrac{2-x}{3-x}+\dfrac{5x-1}{3+x}-3=\left(1-\dfrac{1}{3-x}\right)+\left(5-\dfrac{16}{x+3}\right)-3=B+3\)
\(B=\dfrac{1}{x-3}-\dfrac{16}{x+3}=\dfrac{x+3-16x+48}{x^2-9}=\dfrac{-15x+51}{x^2-9}=\dfrac{3\left(17-5x\right)}{x^2-9}\)
Từ (4)\(\Rightarrow\left\{{}\begin{matrix}17-5x=3x^2+7\\B=\dfrac{3\left(3x^2+7\right)}{x^2-9}\end{matrix}\right.\) \(B=9+\dfrac{81+27}{x^2-9}\)
\(A=12+\dfrac{108}{x^2-9}\)
Bạn tự thay vào :\(\begin{matrix}A\left(x_1\right)=\\A\left(x_2\right)=\end{matrix}\) chú ý bp => x^2 --> mới thay vào
Mình nghi đề của bạn nhầm dấu: biểu thức (1)
\(10a^2-3b^2-5ab=0\Rightarrow10\left(a-\dfrac{b}{4}\right)^2-\dfrac{29b^2}{8}=0\)
\(\Rightarrow a=b=0\)
tự làm tiếp nhé, phần khó nhất mk đã giúp bn r`h thay vào thôi
\(\dfrac{ab+a^2}{b^2-5b+5a-a^2}\cdot\dfrac{a^2-10a+25-b^2}{a^2-b^2}\)
\(=\dfrac{a\left(a+b\right)}{\left(b^2-a^2\right)-\left(5b-5a\right)}\cdot\dfrac{\left(a-5\right)^2-b^2}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a\left(a+b\right)}{\left(b-a\right)\left(b+a\right)-5\left(b-a\right)}\cdot\dfrac{\left(a-5-b\right)\left(a-5+b\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a}{a-b}\cdot\dfrac{\left(a-b-5\right)\left(a+b-5\right)}{\left(b-a\right)\left(b+a-5\right)}\)
\(=\dfrac{a}{a-b}\cdot\dfrac{a-b-5}{b-a}=\dfrac{-a\left(a-b-5\right)}{\left(a-b\right)^2}\)