Ta có: ΔABC đều

mà BP,CM là các đường trung tuyến

nên BP,CM là các đường cao

Xét tứ giác BMPC có 

\(\widehat{BMC}=\widehat{BPC}=90^0\)

nên BMPC là tứ giác nội tiếp

hay B,M,P,C cùng thuộc 1 đường tròn

16 mờ quá

\(17,D\\ 18,C\\ 19,C\)

a,\(P\left(x\right)=x^3-2x^4+x^5-\dfrac{1}{2}x^2+=x^5-2x^4+x^3-\dfrac{1}{2}x^2+1\)

bậc :5

b,\(Q\left(x\right)=-x^3+3x^2-5x^4-x^2+3x^3-\dfrac{1}{2}=-5x^4+2x^3+2x^2-\dfrac{1}{2}\)

bậc :4

b,\(P\left(x\right)-Q\left(x\right)=\left(x^5-2x^4+x^3-\dfrac{1}{2}x^2+1\right)-\left(-5x^4+2x^3+2x^2-\dfrac{1}{2}\right)\)

\(=\text{​​}\text{​​}\text{​​}\text{​​}x^5+3x^4-x^3-\dfrac{5}{2}x^2+\dfrac{3}{2}\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 2\\x>\dfrac{9}{2}\end{matrix}\right.\\-\dfrac{1}{3}< x< 7\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{3}< x< 2\\\dfrac{9}{2}< x< 7\end{matrix}\right.\)

Hay \(S=\left(-\dfrac{1}{3};2\right);\left(\dfrac{9}{2};7\right)\)

d.

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le-\dfrac{11}{5}\\x\ge7\end{matrix}\right.\\-\dfrac{1}{2}< x< 3\end{matrix}\right.\) \(\Rightarrow x\in\varnothing\) hay BPT vô nghiệm

a, Theo tc 2 tt cắt nhau: \(AE=EC;BF=CF\)

Vậy \(AE+BF=EC+CF=EF\)

b, Vì \(\left\{{}\begin{matrix}AE=EC\\\widehat{EAO}=\widehat{ECO}=90^0\\OE.chung\end{matrix}\right.\) nên \(\Delta AOE=\Delta COE\)

\(\Rightarrow\widehat{AOE}=\widehat{EOC}\) hay OE là p/g \(\widehat{AOC}\)

Cmtt: \(\Delta BOF=\Delta COF\Rightarrow\widehat{BOF}=\widehat{COF}\) hay OF là p/g \(\widehat{BOC}\)

Vậy \(\widehat{EOF}=\widehat{COF}+\widehat{COE}=\dfrac{1}{2}\left(\widehat{AOC}+\widehat{BOC}\right)=90^0\) hay OE⊥OF

a: Xét ΔHAO vuông tại A và ΔHIO vuông tại I có

OH chung

góc AOH=góc IOH

=>ΔHAO=ΔHIO

b: H là tâm đường tròn nội tiếp ΔKOB

=>d(H,BK)=HA=4cm

Bài 6: 

\(\Leftrightarrow6n+4⋮2n-1\)

\(\Leftrightarrow2n-1\in\left\{1;-1;7;-7\right\}\)

hay \(n\in\left\{1;0;4;-3\right\}\)

bn lm chi tiết hơn giúp mk đc hông bn?

Bài 1:

1: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

2: Thay x=1/9 vào A, ta được:

\(A=\left(-5\cdot\sqrt{\frac19}+2\right):\left(\sqrt{\frac19}+3\right)=\left(-5\cdot\frac13+2\right):\left(\frac13+3\right)\)

\(=\frac13:\frac{10}{3}=\frac{1}{10}\)

Thay \(x=4-2\sqrt3=\left(\sqrt3-1\right)^2\) vào A, ta được:

\(A=\frac{-5\cdot\sqrt{\left(\sqrt3-1\right)^2}+2}{\sqrt{\left(\sqrt3-1\right)^2}+3}=\frac{-5\left(\sqrt3-1\right)+2}{\sqrt3-1+3}\)

\(=\frac{-5\sqrt3+7}{2+\sqrt3}=\left(7-5\sqrt3\right)\left(2-\sqrt3\right)=14-7\sqrt3-10\sqrt3+15=29-17\sqrt3\)

3: \(A=\frac12\)

=>\(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac12\)

=>\(-10\sqrt{x}+4=\sqrt{x}+3\)

=>\(-11\sqrt{x}=3-4=-1\)

=>\(\sqrt{x}=\frac{1}{11}\)

=>\(x=\frac{1}{121}\) (nhận)

4: A>-1

=>A+1>0

=>\(\frac{-5\sqrt{x}+2+\sqrt{x}+3}{\sqrt{x}+3}>0\)

=>\(-4\sqrt{x}+5>0\)

=>\(-4\sqrt{x}>-5\)

=>\(\sqrt{x}<\frac54\)

=>0<=x<25/16

Kết hợp DKXĐ, ta được: 0<=x<25/16 và x<>1

5: A nguyên khi \(-5\sqrt{x}+2\) ⋮\(\sqrt{x}+3\)

=>\(-5\sqrt{x}-15+17\) ⋮\(\sqrt{x}+3\)

=>17⋮\(\sqrt{x}+3\)

mà \(\sqrt{x}+3\ge3\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}+3=17\)

=>\(\sqrt{x}=14\)

=>x=196(nhận)

6: \(A+5=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}+5=\frac{-5\sqrt{x}+2+5\sqrt{x}+15}{\sqrt{x}+3}=\frac{17}{\sqrt{x}+3}>0\)

=>A>-5

Bài 2:

ĐKXĐ: x>=0; x<>4

1: \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{x+5}{x-\sqrt{x}-2}\)

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

\(=\frac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\frac{-x-7\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{-\left(\sqrt{x}+6\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\frac{-\sqrt{x}-6}{\sqrt{x}-2}\)

2: Khi x=9 thì \(B=\frac{-3-6}{3-2}=-9\)

Khi \(x=9-4\sqrt5=\left(\sqrt5-2\right)^2\) vào B, ta được:

\(B=\frac{-\sqrt{\left(\sqrt5-2\right)^2}-6}{\sqrt{\left(\sqrt5-2\right)^2}-2}=\frac{-\sqrt5+2-6}{\sqrt5-2-2}=\frac{-\sqrt5-4}{\sqrt5-4}\)

\(=\frac{4+\sqrt5}{4-\sqrt5}=\frac{\left(4+\sqrt5\right)^2}{\left(4-\sqrt5\right)\left(4+\sqrt5\right)}=\frac{16+8\sqrt5+5}{16-5}=\frac{21+8\sqrt5}{11}\)

3: B=-1

=>\(-\sqrt{x}-6=-1\left(\sqrt{x}-2\right)=-\sqrt{x}+2\)

=>-6=2(loại)

4: B<-1

=>B+1<0

=>\(\frac{-\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}-2}<0\)

=>\(-\frac{8}{\sqrt{x}-2}<0\)

=>\(\sqrt{x}-2>0\)

=>\(\sqrt{x}>2\)

=>x>4

câu hỏi đâu zậy ??????????

nó bị lỗi nên mình làm câu khác rồi
câu hỏi đây nè bạn
Tìm GTLN của \(A=\dfrac{\left(x+2\right)^2}{2}\times\left(1-\dfrac{x^2}{x+2}\right)-\dfrac{x^2+6x+4}{x}\)