Tìm x biết: a) \(\dfrac{a^2-3a}{a+4}x=\dfrac{a^2-9}{a^2+4a}\) với a khác 0; a khác 3; a khác -4 b) \(\dfrac{a^2-2ab}{a^2b}x=\dfrac{a^2b-4b}{3ab}\) với a khác 0; b khác 0 ; a khác 2b c) \(a^2x+5ax_{ }+25=a^2\) với a khác 0; a khác 5 giúp mk nhé
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\(a, x^3+5x^2-9x-45=0\\ \Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\left(x\ne-5\right)\\ \text{Với }x=3\Leftrightarrow A=\dfrac{9-9}{3\left(3+5\right)}=0\\ \text{Với }x=-3\Leftrightarrow A=\dfrac{9-9}{3\left(-3+5\right)}=0\\ \text{Vậy }A=0\\ b,B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:
\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).
Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).
2.
\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)
Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)
\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )
\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)
\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)
Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)
3. Chia 2 vế giả thiết cho \(x^2y^2\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)
\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
a: \(A=\left(\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{8\sqrt{x}}{4-x}\right):\frac{\sqrt{x}+2}{1-2\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-2\right)^2+8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{1-2\sqrt{x}}{\sqrt{x}+2}\)
\(=\frac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)^2}\cdot\frac{1-2\sqrt{x}}{\sqrt{x}-2}=\frac{-2\sqrt{x}+1}{\sqrt{x}-2}\)
b: \(A=-\frac13\)
=>\(\frac{-2\sqrt{x}+1}{\sqrt{x}-2}=\frac{-1}{3}\)
=>\(3\left(-2\sqrt{x}+1\right)=-\left(\sqrt{x}-2\right)\)
=>\(-6\sqrt{x}+3=-\sqrt{x}+2\)
=>\(-5\sqrt{x}=-1\)
=>\(\sqrt{x}=\frac15\)
=>x=1/25(nhận)
\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)
\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)
\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)
\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)
\(A=\sqrt{\left(a-3\right)^2}-3a\)
=3-a-3a
=3-4a
a: ĐKXĐ: a∉{-1/3;-3}
\(\frac{3a-1}{3a+1}+\frac{a-3}{a+3}=2\)
=>\(\frac{\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)}{\left(3a+1\right)\left(a+3\right)}=2\)
=>\(2\left(3a+1\right)\left(a+3\right)=\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)\)
=>\(2\left(3a^2+9a+a+3\right)=3a^2+9a-a-3+3a^2-9a+a-3\)
=>\(6a^2+20a+6=6a^2-6\)
=>20a=-12
=>a=-3/5(nhận)
b: ĐKXĐ: a∉{5/2;2/3}
\(\frac{2a-9}{2a-5}+\frac{3a}{3a-2}=2\)
=>\(\frac{2a-5-4}{2a-5}+\frac{3a-2+2}{3a-2}=2\)
=>\(1-\frac{4}{2a-5}+1+\frac{2}{3a-2}=2\)
=>\(\frac{2}{3a-2}=\frac{4}{2a-5}\)
=>\(\frac{4}{6a-4}=\frac{4}{2a-5}\)
=>6a-4=2a-5
=>4a=-1
=>a=-1/4(nhận)
c: ĐKXĐ: a<>-3
\(\frac{10}{3}-\frac{3a-1}{4a+12}-\frac{7a+2}{6a+18}=2\)
=>\(\frac{3a-1}{4a+12}+\frac{7a+2}{6a+18}=\frac{10}{3}-2=\frac43\)
=>\(\frac{3\left(3a-1\right)}{12\left(a+3\right)}+\frac{2\left(7a+2\right)}{12\left(a+3\right)}=\frac43\)
=>\(\frac{9a-3+14a+4}{12\left(a+3\right)}=\frac{4\cdot4\cdot\left(a+3\right)}{12\left(a+3\right)}\)
=>23a+1=16(a+3)=16a+48
=>7a=47
=>a=47/7(nhận)
Ta có:
\(VT=\left[\dfrac{16a-a^2-\left(3+2a\right)\left(a+2\right)-\left(2-3a\right)\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}\right]:\dfrac{a-1}{a^3+4a^2+4a}\)
\(=\dfrac{16a-a^2-3a-6-2a^2-4a-2a+4+3a^2-6a}{\left(a-2\right)\left(a+2\right)}.\dfrac{a\left(a+2\right)^2}{a-1}\)
\(=\dfrac{a-2}{\left(a-2\right)\left(a+2\right)}.\dfrac{a\left(a+2\right)^2}{a-1}=\dfrac{a\left(a+2\right)}{a-1}\left(a\ne\pm2;a\ne1\right)\)
\(=a-\dfrac{a\left(a+2\right)}{a-1}=\dfrac{a^2-a-a^2-2a}{-1}=\dfrac{-3a}{a-1}=\dfrac{3a}{1-a}=VP\left(đpcm\right)\)