Câu 1:

\(A=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{x+9\sqrt{x}}{x-9}\left(x\ge0;x\ne9\right)\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

Câu 2:

\(V\left(3\right)=12000000-1400000.3=7800000\)

Có: \(V\left(t\right)=6400000\) \(\Leftrightarrow12000000-1400000t=6400000\)

\(\Leftrightarrow t=4\) => Sau 4 năm thì gtri chiếc máy tính này còn 6400000 đ

b,\(\left\{{}\begin{matrix}2x+y=5\\mx+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{4-mx}{3}=5\\y=\dfrac{4-mx}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\left(6-m\right)=11\left(1\right)\\y=\dfrac{4-mx}{3}\end{matrix}\right.\)

Xét \(m=6\) thay vào pt ta đc \(\left\{{}\begin{matrix}2x+y=5\\6x+3y=4\end{matrix}\right.\) (vô nghiệm)

\(\Rightarrow m\ne6\)

Từ (1) \(\Rightarrow x=\dfrac{11}{6-m}\)

 \(\Rightarrow y=\dfrac{4-\dfrac{11m}{6-m}}{3}\)\(=\dfrac{24-15m}{3\left(6-m\right)}\)

\(xy>0\Leftrightarrow\dfrac{11}{6-m}.\dfrac{24-15m}{3\left(6-m\right)}>0\)

\(\Leftrightarrow\dfrac{11\left(24-15m\right)}{3\left(6-m\right)^2}>0\) 

\(\Leftrightarrow24-15m>0\Leftrightarrow m< \dfrac{24}{15}\)

A=\(\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\) (đk: \(x\ge0;x\ne1\))

\(=\left[\dfrac{1}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right]:\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\left[\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{\left(\sqrt{x}+1\right)^2}\right]:\dfrac{1}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

b, \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

Để \(A\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\in Z\) mà với mọi \(x\in Z\),\(x\ge0\) thì \(\left\{{}\begin{matrix}\dfrac{2}{\sqrt{x}+1}>0\\\dfrac{2}{\sqrt{x}+1}\le\dfrac{2}{\sqrt{0}+1}=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{\sqrt{x}+1}=1\\\dfrac{2}{\sqrt{x}+1}=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(L\right)\\x=0\left(TM\right)\end{matrix}\right.\)

Vậy x=0 thì A=-1

c, Vì \(\dfrac{2}{\sqrt{x}+1}>0\Rightarrow\) \(A=1-\dfrac{2}{\sqrt{x}+1}< 1< 2\)

Vậy A<2 
d, \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\le2\)

\(\Rightarrow1-\dfrac{2}{\sqrt{x}+1}\ge1-2\)

Hay \(A\ge-1\)

Dấu = xảy ra khi x=0

Đk: x+2y và x+y+1 khác 0

Từ pt (1) \(\Rightarrow\)\(\left(x+y+1\right)^2+\left(x+2y\right)^2=2\left(x+2y\right)\left(x+y+1\right)\)

\(\Leftrightarrow\left(x+y+1\right)^2-2\left(x+y+1\right)\left(x+2y\right)+\left(x+2y\right)^2=0\)

\(\Leftrightarrow\left(x+y+1-2y-x\right)^2=0\)

\(\Leftrightarrow\left(1-y\right)^2=0\)

\(\Leftrightarrow1=y\)

Từ pt (2) \(\Rightarrow x=\dfrac{4-y}{3}=\dfrac{4-1}{3}=1\) (thỏa mãn)

Vậy (x;y)=(1;1)

a,Có \(\Delta=4\left(m+2\right)^2-4.-\left(4m+12\right)=4m^2+32m+64=4\left(m+4\right)^2\ge0\forall m\)

=> Phương trình luôn có nghiệm với mọi m

b,Phương trình có nghiệm \(\left[{}\begin{matrix}x=\dfrac{-2\left(m+2\right)+2\left(m+4\right)}{2}=2\\x=\dfrac{-2\left(m+2\right)-2\left(m+4\right)}{2}=-2m-6\end{matrix}\right.\) (ở đây không cần chia trường hợp của m bởi khi chia trường hợp thì x chỉ đổi giá trị cho nhau)

TH1: \(x_1=x_2^2\Leftrightarrow4=\left(-2m-6\right)^2\)\(\Leftrightarrow\left[{}\begin{matrix}m=-2\\m=-4\end{matrix}\right.\) (Thay vào pt thấy không thỏa mãn)

TH2:\(x_1=x_2^2\Leftrightarrow-2m-6=2^2\)\(\Leftrightarrow m=-5\) (Thay vào pt thấy thỏa mãn)

Vậy ...

\(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{3}\)

\(\Leftrightarrow3y-3x-xy=0\) \(\Leftrightarrow y\left(3-x\right)+3\left(3-x\right)=9\) \(\Leftrightarrow\left(y+3\right)\left(3-x\right)=9\)

Do \(y\in Z^+\Rightarrow y+3\inƯ\left(9\right),y+3>3\) \(\Rightarrow y+3=9\) \(\Rightarrow3-x=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)

\(A=cos2x+sin4x-cos6x\)

\(=\left(cos2x-cos6x\right)+sin4x=-2.sin4x.sin\left(-2x\right)+sin4x\)

\(=2sin4x.sin2x+sin4x=sin4x\left(2sin2x+1\right)\)

\(B=sinx-sin2x+sin5x+sin8x\)

\(=\left(sin5x+sinx\right)+\left(sin8x-sin2x\right)\)

\(=2.sin3x.cos2x+2.sin3x.cos5x\)

\(=2sin3x\left(cos2x+cos5x\right)\)

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

\(\Leftrightarrow\left(x^2-x\right)^2+\left(x^2-x\right)-6=0\)

\(\Leftrightarrow\left[\left(x^2-x\right)-2\right]\left[\left(x^2-x\right)+3\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x=2\\x^2-x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-2=0\\x^2-x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)\left(x+1\right)=0\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0\left(vn\right)\end{matrix}\right.\) 

\(\Rightarrow\)\(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Câu 7: 

Đặt A=\(\sqrt{a^2+abc}+\sqrt{b^2+abc}+\sqrt{c^2+abc}\)

\(=\sqrt{a}\sqrt{a+bc}+\sqrt{b}\sqrt{b+ac}+\sqrt{c}\sqrt{c+ab}\)\(\le\sqrt{\left(a+b+c\right)\left(a+b+c+ab+bc+ac\right)}\) (theo bđt bunhia)

\(\Rightarrow A\le\sqrt{1+ab+bc+ac}\)

mà  \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\) (bạn tự chứng minh được) 

\(\Rightarrow A\le\sqrt{1+\dfrac{\left(a+b+c\right)^2}{3}}=\sqrt{1+\dfrac{1}{3}}=\dfrac{2\sqrt{3}}{3}\)

Áp dụng bđt cosi có:

\(1=a+b+c\ge\sqrt[3]{abc}\) \(\Leftrightarrow abc\le\dfrac{1}{27}\)

Có \(M=A+9\sqrt{abc}\le\dfrac{2\sqrt{3}}{3}+9\sqrt{\dfrac{1}{27}}=\dfrac{5\sqrt{3}}{3}\)

=> maxM\(=\dfrac{5\sqrt{3}}{3}\) \(\Leftrightarrow a=b=c=\dfrac{1}{3}\)

\(\overrightarrow{OA'}+\overrightarrow{OB'}+\overrightarrow{OC'}\)

\(=\overrightarrow{OA}+\overrightarrow{AA'}+\overrightarrow{OB}+\overrightarrow{BB'}+\overrightarrow{OC}+\overrightarrow{CC'}\)

\(=\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)+\left(\overrightarrow{BA}+\overrightarrow{CB}+\overrightarrow{AC}\right)\)

\(=\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)+\overrightarrow{CC}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\) (đpcm)

đk: \(-x^4+3x-1\ge0\)

Có \(-\left(x^4+1\right)\le-2x^2\)

 \(\Rightarrow\sqrt{-x^4+3x-1}+\sqrt{2x^2-3x+2}\le\sqrt{3x-2x^2}+\sqrt{2x^2-3x+2}\) 

Áp dụng bunhia có: \(\sqrt{3x-2x^2}+\sqrt{2x^2-3x+2}\le\sqrt{\left(1+1\right)\left(3x-2x^{^2}+2x^2-3x+2\right)}=2\)

\(\Rightarrow\sqrt{-x^4+3x-1}+\sqrt{2x^2-3x+2}\le2\)  (*)

Có: \(x^4-x^2-2x+4=\left(x^4+1\right)-x^2-2x+3\ge2x^2-x^2-2x+3=\left(x-1\right)^2+2\ge2\) (2*)

Từ (*) (2*) dấu = xảy ra khi x=1 (TM)

Vậy x=1