\(a)A\ge\dfrac{x-\sqrt{x}-3}{\sqrt[]{x}}\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}}\ge\dfrac{x-\sqrt{x}-3}{\sqrt{x}}\)

\(\Leftrightarrow\sqrt{x}-4\ge x-\sqrt{x}-3\)

\(\Leftrightarrow x-2\sqrt{x}+1\le0\)

\(\Leftrightarrow(\sqrt{x}-1)^2\le0\)

\(\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(tm\right)\)

\(b)ĐKXĐ:x>0;x\ne4\)

\(B=\dfrac{x+2\sqrt{x}-10}{x-2\sqrt{x}}+\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}}\)

\(=\dfrac{x+2\sqrt{x}-10}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{(\sqrt{x}+2)(\sqrt{x}-2)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+2\sqrt{x}-10+x-\sqrt{x}-x+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

\(=\dfrac{\sqrt{x}+3}{\sqrt{x}}\left(đpcm\right)\)

\(c)\dfrac{A}{B}=\dfrac{\sqrt{x}-4}{\sqrt{x}+3}\Rightarrow\left|\dfrac{A}{B}\right|=\dfrac{\left|\sqrt{x}-4\right|}{\sqrt{x}+3}\left(vì\sqrt{x}+3>0\right)\)

Xét các TH:

\(TH1:\sqrt{x}-4< 0\Leftrightarrow\sqrt{x}< 4\Leftrightarrow x< 16\left(1\right)\)

\(\Rightarrow\left|\dfrac{A}{B}\right|=\dfrac{4-\sqrt{x}}{\sqrt{x}+3}\)

\(\left|\dfrac{A}{B}\right|>\dfrac{A}{B}\Leftrightarrow\dfrac{4-\sqrt{x}}{\sqrt{x}+3}>\dfrac{\sqrt{x}-4}{\sqrt{x}+3}\)

\(\Leftrightarrow4-\sqrt{x}>\sqrt{x}-4\Leftrightarrow2\sqrt{x}< 8\Leftrightarrow\sqrt{x}< 4\)

\(\Leftrightarrow x< 16\left(2\right)\)

Từ (1)(2) suy ra x<16 suy ra x lớn nhất bằng 15 

\(TH2:\sqrt{x}-4\ge0.\) Giai tương tự TH1 suy ra loại

bài này câu hỏi là gì bạn

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

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

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

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

Phân tích đa thức thành nhân tử nha!

2:

a: =(1+căn 3)^2-5

=4+2căn 3-5

=2căn 3-1

b: \(=\sqrt{\dfrac{125}{7}\cdot\dfrac{35}{81}}=\sqrt{\dfrac{625}{81}}=\dfrac{25}{9}\)

c: \(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)-\sqrt{6}+\sqrt{2}\)

=2-căn 6+căn 2

3:

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

b: \(=\dfrac{6\sqrt{2}+7\sqrt{2}-5\sqrt{2}}{\sqrt{2}}=13-5=8\)

c: \(=\dfrac{12-10+8}{2}=5\)

d: \(=\sqrt{\dfrac{1}{5}:5}-\sqrt{\dfrac{9}{5}:5}+\sqrt{5:5}\)

=1/5-3/5+1

=3/5

Gọi \(M\left(x;y\right)\) là 1 điểm bất kì trên (E) \(\Rightarrow\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\) (1)

Gọi \(M'\left(x';y'\right)\) là ảnh của M qua phép tịnh tiến \(\overrightarrow{v}\Rightarrow M'\in\left(E'\right)\) với (E') là ảnh của (E) qua phép tịnh tiến nói trên

\(\left\{{}\begin{matrix}x'=x+3\\y'=y-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=x'-3\\y=y'+2\end{matrix}\right.\)

Thế vào (1):

\(\dfrac{\left(x'-3\right)^2}{16}+\dfrac{\left(y'+2\right)^2}{9}=1\)

Hay pt (E') có dạng: \(\dfrac{\left(x-3\right)^2}{16}+\dfrac{\left(y+2\right)^2}{9}=1\)

d d d a d a a d a b c a d c b a

\(\Leftrightarrow sinx+sinax=\sqrt{3}cosx-\sqrt{3}cosax\)

\(\Leftrightarrow sinax+\sqrt{3}cosax=\sqrt{3}cosx-sinx\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosax+\dfrac{1}{2}sinax=\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx\)

\(\Leftrightarrow cos\left(ax-\dfrac{\pi}{6}\right)=cos\left(x+\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}ax-\dfrac{\pi}{6}=x+\dfrac{\pi}{6}+k2\pi\\ax-\dfrac{\pi}{6}=-x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(a-1\right)x=\dfrac{\pi}{3}+k2\pi\\\left(a+1\right)x=k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3\left(a-1\right)}+\dfrac{k2\pi}{a-1}\left(a\ne1\right)\\x=\dfrac{k2\pi}{a+1}\left(a\ne-1\right)\end{matrix}\right.\)

nhiệt lượng nước tỏa ra \(Q_1=5.4200.55=1155000\left(J\right)\)

nhiệt lượng nước nồi hấp thụ \(Q_2=Q_1-Q_1.30\%=808500\left(J\right)\)

\(\Leftrightarrow m'.880.25=808500\Rightarrow m'=36,75\left(kg\right)\)