Phần tô đậm là phần nào thế bạn 

Lời giải:

Áp dụng BĐT AM-GM:

$\frac{a^4}{b^2(c+a)}+\frac{b(c+a)}{4}+\frac{b}{2}+\frac{1}{2}\geq 4\sqrt[4]{\frac{a^4}{b^2(c+a)}.\frac{b(c+a)}{4}.\frac{b}{2}.\frac{1}{2}}=2a$

Làm tương tự với các phân thức còn lại và cộng theo vế, thu gọn thì được:

$A+\frac{ab+bc+ac}{2}+\frac{a+b+c}{2}+\frac{3}{2}\geq 2(a+b+c)$

$\Leftrightarrow A\geq \frac{3}{2}(a+b+c)-\frac{3}{2}-\frac{ab+bc+ac}{2}=\frac{9}{2}-\frac{3}{2}-\frac{ab+bc+ac}{2}=3-\frac{ab+bc+ac}{2}$
Theo hệ quả quen thuộc của BĐT AM-GM:

$ab+bc+ac\leq \frac{(a+b+c)^2}{3}=\frac{3^2}{3}=3$

$\Rightarrow A\geq 3-\frac{ab+bc+ac}{2}\geq 3-\frac{3}{2}=\frac{3}{2}$
Vậy ta có đpcm

Dấu "=" xảy ra khi $a=b=c=1$

\(2\sqrt{9-\sqrt{4}}\\ =2\sqrt{9-2}\\ =2\sqrt{7}\)

Đc bạn nhé ! 

1 If his younger brother had more free time, he would learn Chinese and Korean
2 Zhang Yiming was born on April 1st, 1983 in Longyan, Fujian, China
3 Our teacher has had this laptop for nearly seven years
4 Sarah has a good sense of humor. We are interested in her stories
5 Thomas asked Jane what time she would come to the party the next day
6 We wish people in the world resolved conflicts and lived in peace
7 In my family, my older sister studies English better than I do
8 Mark went camping with his friends last month, didn't he
9 Helen works as a(n) designer for a big fashion company in the city

1: Xét tứ giác SBOC có \(\hat{SBO}+\hat{SCO}=90^0+90^0=180^0\)

nên SBOC là tứ giác nội tiếp

2: Sửa đề: SO cắt BC tại E

Xét (O) có

SB,SC là các tiếp tuyến

Do đó: SB=SC

=>S nằm trên đường trung trực của BC(1)

Ta có: OB=OC

=>O nằm trên đường trung trực của BC(2)

Từ (1),(2) suy ra SO là đường trung trực của BC

=>SO⊥BC tại E và E là trung điểm của BC

Xét ΔSBO vuông tại B có BE là đường cao

nên \(BE^2=EO\cdot ES\)

a) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{5}\\\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{5}\\\dfrac{2}{x}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+\dfrac{1}{y}=\dfrac{4}{5}\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{4}{5}-\dfrac{1}{2}\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=1:\dfrac{3}{10}\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{10}{3}\\x=2\end{matrix}\right.\)

vậy: ....

b_ \(\left\{{}\begin{matrix}\dfrac{15}{x}-\dfrac{7}{y}=9\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{60}{x}-\dfrac{28}{y}=36\\\dfrac{60}{x}+\dfrac{135}{y}=525\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{163}{y}=489\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\\dfrac{4}{x}+9:\dfrac{1}{3}=35\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\\dfrac{4}{x}=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)

vậy: ... 

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\)

Hệ phương trình sẽ trở thành:

\(\left\{{}\begin{matrix}a+b=\dfrac{4}{5}\\a-b=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b+a-b=\dfrac{4}{5}+\dfrac{1}{5}\\a-b=\dfrac{1}{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2a=1\\b=a-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{2}\\\dfrac{1}{y}=\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{10}{3}\end{matrix}\right.\left(nhận\right)\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\)

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}15a-7b=9\\4a+9b=35\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}60a-28b=36\\60a+135b=525\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}163b=489\\4a+9b=35\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b=3\\4a=35-9b=35-9\cdot3=35-27=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=2\\\dfrac{1}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)(nhận)

c: ĐKXĐ: \(x\ne\pm y\)

\(\left\{{}\begin{matrix}\dfrac{1}{x+y}+\dfrac{1}{x-y}=\dfrac{5}{8}\\\dfrac{1}{x+y}-\dfrac{1}{x-y}=-\dfrac{3}{8}\end{matrix}\right.\)

Đặt \(\dfrac{1}{x+y}=a;\dfrac{1}{x-y}=b\)

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}a+b=\dfrac{5}{8}\\a-b=-\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b+a-b=\dfrac{5}{8}-\dfrac{3}{8}\\a-b=-\dfrac{3}{8}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2a=\dfrac{2}{8}=\dfrac{1}{4}\\b=a+\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{8}\\b=\dfrac{1}{8}+\dfrac{3}{8}=\dfrac{4}{8}=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x+y}=\dfrac{1}{8}\\\dfrac{1}{x-y}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\x-y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y+x-y=8+2\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=10\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=5-2=3\end{matrix}\right.\left(nhận\right)\)

d: ĐKXĐ: \(\left\{{}\begin{matrix}y\ne3x\\y\ne\dfrac{2}{3}x\end{matrix}\right.\)

Đặt \(\dfrac{1}{3x+y}=a;\dfrac{1}{2x-3y}=b\)

Hệ phương trình sẽ trở thành:

\(\left\{{}\begin{matrix}5a+4b=-2\\3a-5b=21\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}15a+12b=-6\\15a-25b=105\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}37b=-111\\5a+4b=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b=-3\\5a=-2-4b=-2-4\cdot\left(-3\right)=-2+12=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b=-3\\a=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9x+3y=\dfrac{3}{2}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{7}{6}\\3x+y=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{7}{66}\\y=\dfrac{1}{2}-3x=\dfrac{1}{2}-\dfrac{7}{22}=\dfrac{4}{22}=\dfrac{2}{11}\end{matrix}\right.\left(nhận\right)\)

e: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne y-2\\x\ne-y+1\end{matrix}\right.\)

Đặt \(\dfrac{1}{x-y+2}=a;\dfrac{1}{x+y-1}=b\)

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}7a-5b=4,5\\3a+2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}21a-15b=13,5\\21a+14b=28\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-29b=13,5-28=-14,5\\3a+2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{2}\\3a=4-2b=4-1=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=1\\b=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y+2=1\\x+y-1=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\x+y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x=2\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3-x=3-1=2\end{matrix}\right.\left(nhận\right)\)

a: \(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2xy+5x-6y-15=2xy-2x+7y-7\\12xy-24x+3y-6=12xy+18x-2y-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x-6y-15=-2x+7y-7\\-24x+3y-6=18x-2y-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7x-13y=-7+15=8\\-42x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42x-78y=48\\-42x+5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-73y=51\\7x-13y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{53}{71}\\7x=13y+8=13\cdot\dfrac{-53}{71}+8=-\dfrac{121}{71}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{53}{71}\\x=-\dfrac{121}{497}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x-xy-y+2xy\\y^2+y-xy-x=y^2-2y+xy-2x-2xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x-y=x-y\\y-x=-2y-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x=0\\3y=-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}x\left(x-3y\right)=4\left(y^2+2\right)\left(1\right)\\\left(xy-4\right)\left(x+y\right)=8\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Rightarrow xy-4;x+y\ne0\)

\(\left(1\right)\Leftrightarrow x^2-3xy-4y^2=8\) (*)

Từ (*) và (2) \(\Rightarrow x^2-3xy-4y^2=\left(xy-4\right)\left(x+y\right)\)

\(\Leftrightarrow x\left(x-4y\right)+y\left(x-4y\right)=\left(xy-4\right)\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)\left(x-4y\right)=\left(x+y\right)\left(xy-4\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\left(L\right)\\x-4y=xy-4\end{matrix}\right.\) \(\Leftrightarrow x\left(1-y\right)+4\left(1-y\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(1-y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\y=1\end{matrix}\right.\)

x = -4 thay vào (*), ta được: \(16-3.\left(-4\right)y-4y^2=8\)

\(\Leftrightarrow8+12y-4y^2=0\) \(\Leftrightarrow y^2-3y-2=0\) 

\(\Leftrightarrow y=\dfrac{3\pm\sqrt{17}}{2}\) ( dùng \(\Delta\) )

y=1 thay vào (*), ta được: \(x^2-3x-4=8\)

\(\Leftrightarrow x^2-3x-12=0\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt{57}}{2}\)

Vậy ...