b\(\left(a^2b-3ab^2\right):\left(\frac{1}{2}ab\right)+\left(6b^3-5ab^2\right):b^2\)

\(=ab\left(a-3b\right).\frac{2}{ab}+b^2\left(6b-5a\right).\frac{1}{b^2}\)

\(=\left(a-3b\right).2+\left(6b-5a\right)\)

\(=2a-6b+6b-5a\)

\(=-3a\)

\(y\ge xy+1\ge2\sqrt{xy}\Rightarrow\sqrt{\dfrac{y}{x}}\ge2\Rightarrow\dfrac{y}{x}\ge4\)

\(Q=\dfrac{1-\dfrac{2y}{x}+2\left(\dfrac{y}{x}\right)^2}{\dfrac{y}{x}+\left(\dfrac{y}{x}\right)^2}\)

Đặt \(\dfrac{y}{x}=a\ge4\)

\(Q=\dfrac{2a^2-2a+1}{a^2+a}=\dfrac{2a^2-2a+1}{a^2+a}-\dfrac{5}{4}+\dfrac{5}{4}=\dfrac{\left(a-4\right)\left(3a-1\right)}{4\left(a^2+1\right)}+\dfrac{5}{4}\ge\dfrac{5}{4}\)

\(Q_{min}=\dfrac{5}{4}\) khi \(a=4\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

55rt5re3e4rt6huy6rrer55t6yy7

e, \(\sqrt{6-2\sqrt{5}}+\sqrt{8+2\sqrt{15}}-\sqrt{3}\)

\(=\sqrt{5}-1+\sqrt{5}+\sqrt{3}-\sqrt{3}=2\sqrt{5}-1\)

7d, Cho x = 0 => \(y=3-3k\)

=> \(A\left(0;-3k+3\right)\)thuộc d1 => d1 cắt trục Oy tại OA = \(\left|3-3k\right|\)

Cho y = 0 => \(x=\frac{3k-3}{k-3}\)

=> \(B\left(\frac{3k-3}{k-3};0\right)\)thuộc d1 => d1 cắt trục Ox tại OB = \(\left|\frac{3k-3}{k-3}\right|\)

Ta có : \(S_{OAB}=\frac{1}{2}\left|\frac{3k-3}{k-3}.\left(3-3k\right)\right|=1\)

\(\Leftrightarrow\left|\frac{\left(3k-3\right)\left(3-3k\right)}{k-3}\right|=1\)

\(\Leftrightarrow\left|\frac{-3k^2+9k}{k-3}\right|=1\Leftrightarrow\left|\frac{-3k\left(k-3\right)}{k-3}\right|=1\Leftrightarrow\left|-3k\right|=1\)

đk : \(-3k\ge0\Leftrightarrow k\le0\)

TH1 : \(-3k=1\Leftrightarrow k=-\frac{1}{3}\)(ktm)

TH2 : \(-3k=-1\Leftrightarrow k=\frac{1}{3}\)(tm) 

sửa dòng 5 từ dưới lên nhé 

\(\Leftrightarrow\left|\frac{\left(3k-3\right)\left(3-3k\right)}{k-3}\right|=2\Leftrightarrow\left|\frac{-\left(3k-3\right)^2}{k-3}\right|=2\)

\(\Leftrightarrow\frac{9\left(k-1\right)^2}{\left|k-3\right|}=2\Leftrightarrow\left|k-3\right|=\frac{9}{2}\left(k-1\right)^2\Leftrightarrow\left(k-3\right)^2=\frac{81}{4}\left(k-1\right)^4\)

\(\Leftrightarrow\frac{81}{4}\left(k-1\right)^4-\left(k-3\right)^2=0\Leftrightarrow k=1,56;k=0,21\) 

=81

nhe ban

Căn bậc hai cuẩ 9 là 3

500000000 +30000000+6000000+100000+30000+4000+600+50+4

536134654 = 500000000 + 30000000+ 6000000+100000+30000+ 4000+ 600 + 50 + 4

HT

\(x^2-3x+1=0\)\(\left(a=1;b=-3;c=1\right)\)

Ta thấy \(\Delta=b^2-4ac=\left(-3\right)^2-4.1.1=5>0\)nên phương trình đã cho có 2 nghiệm phân biệt:

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-\left(-3\right)+\sqrt{5}}{2.1}=\frac{3+\sqrt{5}}{2}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-\left(-3\right)-\sqrt{5}}{2.1}=\frac{3-\sqrt{5}}{2}\end{cases}}\)

Vậy tập nghiệm của phương trình đã cho là \(S=\left\{\frac{3+\sqrt{5}}{2};\frac{3-\sqrt{5}}{2}\right\}\)