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Phạm Thị Thùy Linh

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Những câu trả lời của Phạm Thị Thùy Linh:

Vào lúc: 2019-10-24 18:59:33 Xem câu hỏi

\(3^{x+2}-5.3^x=36\)

\(\Rightarrow3^x\left(3^2-5\right)=36\)

\(\Rightarrow3^x.\left(9-5\right)=36\)

\(\Rightarrow3^x.4=36\)

\(\Rightarrow3^x=9\)

\(\Rightarrow x=2\)

Vào lúc: 2019-10-20 20:20:53 Xem câu hỏi

\(3^{x+1}-3^x=54\)

\(\Rightarrow3^x\left(3-1\right)=54\)

\(\Rightarrow3^x.2=54\)

\(\Rightarrow3^x=27\)

\(\Rightarrow x=3\)

Vào lúc: 2019-10-20 19:22:22 Xem câu hỏi

\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}\)

\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}=\frac{9}{10}\)

\(A< \frac{9}{10}\Rightarrow A< 1\left(đpcm\right)\)

Vào lúc: 2019-10-19 21:18:26 Xem câu hỏi

\(C=x^4+100x^2+99x+100\)

\(=x^4-x+100x^2+100x+100\)

\(=x\left(x^3-1\right)+100\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+100\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+100\right)\)

Câu 2 em khai triển hằng đẳng thức và rút gọn là ra nhé 

Vào lúc: 2019-10-17 19:28:04 Xem câu hỏi

Ừ nhở chị sai từ chỗ \(\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{x^2+2xy+y^2-x^2}{y\left(x+y\right)}=\frac{y^2+2xy}{y\left(x+y\right)}\)em nhé 

Vào lúc: 2019-10-17 13:10:47 Xem câu hỏi

\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)

\(=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y\left(y-x\right)}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x\left(x-y\right)}{x-2y}\right):\frac{y\left(x+y\right)}{2\left(x-2y\right)}\)

\(=\frac{4y\left(y-x\right)}{\left(x-y\right)\left(x-3y\right)}-\frac{\left(x-y\right)y\left(y-x\right)}{y^2\left(x-3y\right)}\)\(+\frac{x.2\left(x-2y\right)}{2.y\left(x+y\right)}-\frac{x\left(x-y\right).2\left(x-2y\right)}{\left(x-2y\right).y\left(x+y\right)}\)

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

\(=\frac{-4y^2+x^2-2xy+y^2}{y\left(x-3y\right)}+\frac{x^2-2xy-2x^2+2xy}{y\left(x+y\right)}\)

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

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

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

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

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

\(=\frac{x^2-2xy+y^2-x^2}{y\left(x+y\right)}=\frac{-2xy+y^2}{y\left(x+y\right)}\)

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

Thay \(x=\frac{1}{2};y=\frac{1}{3}\)vào A ta có :

\(A=\frac{\frac{1}{3}-2.\frac{1}{2}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}-1}{\frac{3}{6}+\frac{2}{6}}=\frac{2}{3}:\frac{5}{6}=\frac{2.6}{3.5}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)tại \(x=\frac{1}{2};y=\frac{1}{3}\)

Vào lúc: 2019-10-13 14:29:55 Xem câu hỏi

\(e,-5x+x^2-14\)

\(=x^2+2x-7x-14\)

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

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

\(f,x^3+8+6x\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+2x+4\right)+6x\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+8x+4\right)\)

\(g,15x^2-7xy-2y^2\)

\(=15x^2+3xy-10xy-2y^2\)

\(=3\left(5x+y\right)-2y\left(5x+y\right)\)

\(=\left(5x+y\right)\left(3-2y\right)\)

\(h,3x^2-16x+5\)

\(=3x^2-x-15x+5\)

\(=x\left(3x-1\right)+5\left(3x-1\right)\)

\(=\left(3x-1\right)\left(x+5\right)\)

Vào lúc: 2019-10-13 14:22:03 Xem câu hỏi

\(a,x^3+2x^2y+xy^2=x\left(x^2+2xy+y^2\right)\)

\(=x\left(x+y\right)^2\)

\(b,4x^2-9y^2+4x-6y\)

\(=4x^2+4x+1-\left(9y^2+6y+1\right)\)

\(=\left(2x+1\right)^2-\left(3y+1\right)^2\)

\(=\left(2x-3y\right)\left(2x+3y+2\right)\)

\(c,-x^2+5x+2xy-5y-y^2\)

\(=-\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)

\(=-\left(x-y\right)^2+5\left(x-y\right)\)

\(=\left(x-y\right)\left(y-x+5\right)\)

\(d,x^2+4x-12\)

\(=x^2-2x+6x-12\)

\(=x\left(x-2\right)+6\left(x-2\right)\)

\(=\left(x-2\right)\left(x+6\right)\)

Vào lúc: 2019-10-13 14:11:21 Xem câu hỏi

Câu b Th4 cho chị sửa cưng nhé :

\(x+2=3\Rightarrow x=1\) nên có 4 trường hợp để P có giá trị nguyên nhé

Vào lúc: 2019-10-13 14:07:48 Xem câu hỏi

\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne\pm2\end{cases}}\)

\(P=\left(\frac{x^2}{x^3-4x}-\frac{10}{5x+10}-\frac{1}{2-x}\right):\)\(\left(x+2+\frac{6-x^2}{x-2}\right)\)

\(=\left(\frac{x^2}{x\left(x^2-4\right)}-\frac{10}{5\left(x+2\right)}+\frac{1}{x-2}\right)\)\(:\left(\frac{\left(x-2\right)\left(x+2\right)}{x-2}+\frac{6-x^2}{x-2}\right)\)

\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)\(:\left(\frac{x^2-4+6-x^2}{x-2}\right)\)

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

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

\(b,P\in Z\Leftrightarrow\frac{3}{x+2}\in Z\Rightarrow3\)\(⋮\)\(x+2\Rightarrow x+2\inƯ_3\)

MÀ \(Ư_3=\left\{\pm1;\pm3\right\}\)

TH1 : \(x+2=-1\Rightarrow x=-3\)

Th2 : \(x+2=1\Rightarrow x=-1\)

Th3 : \(x+2=-3\Rightarrow x=-5\)

Th4 : \(x+3=3\Rightarrow x=0\left(ktm\right)\)

Vậy để P có giá trị nguyên thì x thuộc { - 3 ; - 5 ;- 1 }

\(c,P=-1\Leftrightarrow\frac{3}{x+2}=-1\)

\(\Rightarrow\frac{3}{x+2}=\frac{-1}{1}\Rightarrow3=-1\left(x+2\right)\)

\(\Rightarrow-x-2=3\Rightarrow-x=5\)

\(\Rightarrow x=-5\)

Vậy để P = -1 thì x = - 5

\(d,P>0\Leftrightarrow\frac{3}{x+2}>0\)

Vì \(x+2>0\)nên để \(\frac{3}{x+2}>0\)thì \(x+2>0\)

\(\Rightarrow x>-2\)

Vậy để \(P>0\)thì \(x>2\) và \(\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)

Vào lúc: 2019-10-10 19:37:32 Xem câu hỏi

\(a,1+\sqrt[3]{x-16}=\sqrt[3]{x+3}.\)

Đặt \(\sqrt[3]{x-16}=a\Rightarrow x-16=a^3\)

\(\sqrt[3]{x+3}=b\Rightarrow x+3=b^3\)

\(\Rightarrow a^3-b^3=-19\)

Mà \(1+a=b\)

\(\Rightarrow a-b=-1\)

Ta có hệ phương trình : 

\(\hept{\begin{cases}a^3-b^3=-19\\a-b=-1\end{cases}}\)\(\Rightarrow\left(a-b\right)\left(a^2+ab+b^2\right)=-19\)

\(\Rightarrow a^2+ab+b^2=19\)

\(\Rightarrow\left(a-b\right)^2-ab=19\)

\(\Rightarrow1-ab=19\Rightarrow ab=-18\)

\(\Rightarrow a=-\frac{18}{b}\)( 1) 

\(a-b=-1\)

Thay vào ( 1 ) ta có : \(-\frac{18}{b}-b=-1\)

Thay vào tính ra b, rồi tính a, và tìm x nhé. ( số hơi xấu 1 tí

b tương tự. đặt ẩn rồi giải hệ phương trình nha. có gì khó hiểu hỏi tớ ^^

Vào lúc: 2019-10-08 21:43:41 Xem câu hỏi

\(1,10a^6+20a^5=10a^5\left(a+10\right)\)

\(2,5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)\)

\(=5\left(x-y\right)^2\)

\(3,3ab^3+6ab^2-18ab\)

\(=3ab\left(b^2+2b-6\right)\)

\(4,15x^3y^2+10x^2y^2-20x^2y^3\)

\(=5x^2y^2\left(3x+2-4y\right)\)

\(5,a^2\left(x-1\right)-b\left(1-x\right)\)

\(=a^2\left(x-1\right)+b\left(x-1\right)\)

\(=\left(x-1\right)\left(a^2+b\right)\)

\(6,x\left(x-5\right)-4\left(5-x\right)\)

\(=x\left(x-5\right)+4\left(x-5\right)\)

\(=\left(x+4\right)\left(x-5\right)\)

Vào lúc: 2019-10-08 18:46:17 Xem câu hỏi

\(1,3\left(x+2\right)-x^2+4=0\)

\(\Rightarrow3x+6-x^2+4=0\)

\(\Rightarrow x^2-3x-10=0\)

\(\Rightarrow x^2+2x-5x-10=0\)

\(\Rightarrow x\left(x+2\right)-5\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-2\\x=5\end{cases}}\)

\(2,\left(x+2\right)\left(x-2\right)-x^2\)

\(=x^2-4-x^2=-4\)

Vào lúc: 2019-10-06 19:24:57 Xem câu hỏi

\(a,\)\(M=\frac{3x+3}{x^3+x^2+x+1}=\frac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}\)

\(=\frac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\frac{3}{x^2+1}\)

\(b,M\in Z\Leftrightarrow\frac{3}{x^2+1}\in Z\)

\(\Rightarrow3\)\(⋮\)\(x^2+1\)\(\Rightarrow x^2+1\inƯ_3\)

Ta có \(Ư_3=\left\{\pm1;\pm3\right\}\)

Mà \(x^2+1\ge1\)với mọi x 

\(\Rightarrow\orbr{\begin{cases}x^2+1=1\\x^2+1=3\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{2}\end{cases}}}\)

\(c,\)\(M_{max}\Leftrightarrow x^2+1\)nhỏ nhất \(\Rightarrow x^2\)nhỏ nhất \(\Rightarrow x=0\)

\(\Rightarrow M_{max}=3\Leftrightarrow x=0\)

Vào lúc: 2019-10-05 13:17:06 Xem câu hỏi

Ta có : \(m< 1\Rightarrow m-1< 0\Leftrightarrow\left(\sqrt{m}-1\right)\left(\sqrt{m}+1\right)< 0\)

Vì \(\sqrt{m}+1>0\Rightarrow\sqrt{m}-1< 0\Rightarrow\sqrt{m}< 1\)\(\left(1\right)\)

Ta lại có : \(\sqrt{m}-1< 0\left(cmt\right)\)

Mà \(\sqrt{m}>0\left(m\ne0\right)\Rightarrow\sqrt{m}\left(\sqrt{m}-1\right)< 0\)

\(\Rightarrow m-\sqrt{m}< 0\Leftrightarrow m< \sqrt{m}\left(2\right)\)

Từ ( 1 ) và ( 2 ) \(\Rightarrow m< \sqrt{m}< 1\)khi \(0< m< 1\)\(\left(đpcm\right)\)

Vào lúc: 2019-10-05 12:57:32 Xem câu hỏi

Cách khác ngắn hơn nè :b

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

\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\)

\(\Rightarrow x^2+2x-3=x^2-4\)

\(\Rightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)

Vào lúc: 2019-10-04 21:21:14 Xem câu hỏi

\(a,\)\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne9;x\ne25\end{cases}}\)

\(P=\frac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}\)\(-\frac{\sqrt{x}+15}{\sqrt{x}-3}-\frac{3\sqrt{x}-1}{5-\sqrt{x}}\)

\(=\frac{8\sqrt{x}-x-31}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)\(-\frac{\sqrt{x}+15}{\sqrt{x}-3}+\frac{3\sqrt{x}-1}{\sqrt{x}-5}\)

\(=\frac{8\sqrt{x}-x-31}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}-\)\(\frac{\left(\sqrt{x}+15\right)\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)\(+\frac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)

\(=\frac{8\sqrt{x}-x-31-x-10\sqrt{x}+75+3x-10\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)

\(=\frac{x-12\sqrt{x}+47}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)

\(\Rightarrow\)Sai đề không cậu ưi 

Vào lúc: 2019-10-03 19:59:05 Xem câu hỏi

\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)-24\)

\(=\left(x^2+3x\right)\left(x^2+3x+2\right)-24\)

\(=\left(x^2+3x\right)\left(x^2+3x\right)+2\left(x^2+3x\right)-24\)

\(=\left(x^2+3x\right)+2\left(x^2+3x\right)+1-25\)

\(=\left(x^2+3x+1\right)^2-5^2\)

\(=\left(x^2+3x+6\right)\left(x^2+3x-4\right)\)

Vào lúc: 2019-10-03 19:53:24 Xem câu hỏi

\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}.\)

\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}+1}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)

\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{13+30\sqrt{2}+30}=\sqrt{43+30\sqrt{2}}\)

Vào lúc: 2019-10-03 12:33:35 Xem câu hỏi

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)

\(\Rightarrow x+1+x+2+...+x+100=5750\)

\(\Rightarrow100x+1+2+...+100=5750\)

\(\Rightarrow100x+\frac{\left(1+100\right).100}{2}=5750\)

\(\Rightarrow100x+5050=5750\)

\(\Rightarrow100x=700\Leftrightarrow x=7\)

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