ĐKXĐ : \(x\ge0\)

\(A=\frac{2}{3}.\frac{2+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)^2+\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2}{\left[1+\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2\right]\left[1+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)^2\right]}.\frac{2010}{x+1}\)

\(A=\frac{2}{3}.\frac{2+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}+\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2-2\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)}{\left[1+\frac{\left(2\sqrt{x}+1\right)^2}{3}\right]\left[1+\frac{\left(2\sqrt{x}-1\right)^2}{3}\right]}.\frac{2010}{x+1}\)

\(A=\frac{2}{3}.\frac{2+\left(\frac{4\sqrt{x}}{\sqrt{3}}\right)^2-\frac{2\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{3}}{\left(\frac{4x+4\sqrt{x}+4}{3}\right)\left(\frac{4x-4\sqrt{x}+4}{3}\right)}.\frac{2010}{x+1}\)

\(A=\frac{2}{3}.\frac{2+\frac{16x}{3}-\frac{2\left(4x-1\right)}{3}}{\frac{16\left(x+1+\sqrt{x}\right)\left(x+1-\sqrt{x}\right)}{9}}.\frac{2010}{x+1}\)

\(A=\frac{2}{3}.\frac{\frac{6+16x-8x+2}{3}}{\frac{16\left(x+1\right)^2-16x}{9}}.\frac{2010}{x+1}\)

\(A=\frac{x+1}{x^2+x+1}.\frac{2010}{x+1}=\frac{2010}{x^2+x+1}\le2010\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=0\)

... 

\(A\le\frac{4.2010}{3}\) ma ban quan

Ta có : \(x^2+x+1\ge1\)vì \(x\ge0\)

Nên \(M=\frac{2020}{x^2+x+1}\le\frac{2020}{1}=2020\)

Vậy Max của M là 2020 khi x = 0

a) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)\(5\)

=> \(\frac{2}{3}-\left(\frac{1}{3}x-\frac{1}{2}\right)-\left(x+\frac{1}{2}\right)=5\)

=>\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)

=>\(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5\)

=>\(\frac{2}{3}-\frac{4}{3}x=5\)

=>\(\frac{4}{3}x=\frac{2}{3}-5=-\frac{13}{3}\)

=>\(x=-\frac{13}{3}:\frac{4}{3}=-\frac{13}{4}\)

b)\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)

=>\(4x-x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)

=> \(3x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)

=>\(x=-\left(-\frac{9}{2}\right)+\frac{1}{2}=5\)

\(\frac{3-x+x}{3-x}=\frac{5x\left(x+2\right)+2\left(x+2\right)\left(3-x\right)}{\left(x+2\right)^2\left(3-x\right)}\)

\(\frac{3}{3-x}=\frac{\left(5x+2\left(3-x\right)\right)\left(x+2\right)}{\left(x+2\right)^2\left(3-x\right)}\)

\(\frac{3}{3-x}=\frac{5x+2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}\)

\(\frac{3}{3-x}=\frac{5x}{\left(x+2\right)\left(3-x\right)}+2\)

\(\frac{3}{3-x}-2=\frac{5x}{\left(x+2\right)\left(3-x\right)}\)

\(\frac{3-2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=\frac{5x}{\left(x+2\right)\left(3-x\right)}\)

\(3-2X\left(3-x\right)=5x\)

\(3-6+2x=5x\)

chị có thể tự giải tiếp ạ

e là hs lớp 7

cảm ơn e "dang long vu'' chị làm xong thấy cái j nó sai sai nhưng k biết sai chỗ nào nên muốn dò lại bài thôi cảm ơn e nha 

\(P=\left(\frac{9}{x^2-3x}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3-3x}\)

a,\(ĐKXĐ:x\ne0;x\ne3;x\ne1\)

\(P=\left(\frac{9}{x^2-3x}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3-3x}=\left(\frac{9}{x\left(x-3\right)}+\frac{x-2}{x}-\frac{x}{x-3}\right).\frac{x}{3\left(1-x\right)}\)

\(=\left(\frac{9+\left(x-2\right)\left(x-3\right)-x.x}{x\left(x-3\right)}\right).\frac{x}{3\left(1-x\right)}=\frac{9+x^2-5x+6-x^2}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}\)

\(=\frac{-5x+15}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}=\frac{-5\left(x-3\right)}{x\left(x-3\right)}.\frac{x}{3\left(1-x\right)}=-\frac{5}{3\left(1-x\right)}\)

b, \(x=\frac{1}{2}\)

\(\Rightarrow P=-\frac{5}{3\left(1-\frac{1}{2}\right)}=-\frac{5}{3.\frac{1}{2}}=-5:\frac{3}{2}=-\frac{10}{3}\)

c, Để \(P\in z\)thì \(3\left(1-x\right)\inƯ\left(5\right)=\left(-5;-1;1;5\right)\)

\(3\left(1-x\right)=-5\Rightarrow1-x=-\frac{5}{3}\Rightarrow x=\frac{8}{3}\)

\(3\left(1-x\right)=-1\Rightarrow1-x=-\frac{1}{3}\Rightarrow x=\frac{4}{3}\)

\(3\left(1-x\right)=1\Rightarrow1-x=\frac{1}{3}\Rightarrow x=\frac{2}{3}\)

\(3\left(1-x\right)=5\Rightarrow1-x=\frac{5}{3}\Rightarrow x=-\frac{2}{3}\)

mai mình đi học cô kiểm tra nên ai đó giúp mk vs

a) \(\frac{3x-6}{x+4}=\frac{2\left(x+5\right)+\left(x-3\right)}{x-2}\)

\(\frac{3\left(x-2\right)}{x+4}=\frac{2\left(x+5\right)+x-3}{x-2}\)

\(\frac{3\left(x-4\right)}{x+4}=\frac{3x+7}{x-2}\)

\(3\left(x-2\right)\left(x-2\right)=\left(3x+7\right)\left(x+4\right)\)

\(3\left(x-2\right)^2=\left(3x+7\right)\left(x+4\right)\)

\(3x^2-12x+12=3x^2+12x+7x+28\)

\(3x^2-12x+12=3x^2+19x+28\)

\(-12x+12=19x+28\)

\(12=19x+28+12x\)

\(19x+28+12x=12\) (chuyển vế)

\(31x+28=12\)

\(31x=12-28\)

\(31x=-16\)

\(x=-\frac{16}{31}\)

\(\Rightarrow x=-\frac{16}{31}\)

B1:

Ta có: a - b = ab => a = ab + b = b(a + 1)

Thay a = b(a + 1) vào a  - b  = a : b ta có: \(a-b=\frac{b\left(a+1\right)}{b}=a+1\)

=> a - b = a + 1 => a - a - b = 1 => -b = 1 => b = -1 

Lại có: ab = a - b

<=> a x (-1) = a - (-1) <=> -a = a + 1 <=> -a - a = 1 <=> -2a = 1 <=> a = -1/2

Vậy...

B2:

a, \(3y\left(y-\frac{2}{5}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3y=0\\y-\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0\\y=\frac{2}{5}\end{cases}}}\)

b, \(7\left(y-1\right)+2y\left(y-1\right)=0\)

\(\Rightarrow\left(y-1\right)\left(7+2y\right)=0\)

\(\Rightarrow\orbr{\begin{cases}y-1=0\\7+2y=0\end{cases}\Rightarrow}\orbr{\begin{cases}y=1\\2y=7\end{cases}\Rightarrow}\orbr{\begin{cases}y=1\\y=\frac{7}{2}\end{cases}}\)

B3: \(K=\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}\)

\(K=\left(-\frac{2}{3}+\frac{1}{6}\right)+\left(\frac{3}{4}-\frac{2}{5}\right)\)

\(K=\left(\frac{-4}{6}+\frac{1}{6}\right)+\left(\frac{15}{20}-\frac{8}{20}\right)\)

\(K=\frac{-1}{2}+\frac{7}{20}=\frac{-10}{20}+\frac{7}{20}=\frac{-3}{20}\)