Bài 1:

\(a,A=6\sqrt{2}-6\sqrt{2}+2\sqrt{5}=2\sqrt{5}\\ b,B=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{3}+\sqrt{2}\\ c,=2\sqrt{3}-6\sqrt{3}+15\sqrt{3}-4\sqrt{3}=7\sqrt{3}\\ d,=1+6\sqrt{3}-\sqrt{3}-1=5\sqrt{3}\\ e,=4\sqrt{2}+\sqrt{2}-6\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)

Bài 2:

\(a,ĐK:x\ge\dfrac{3}{2}\\ PT\Leftrightarrow\sqrt{2x-3}=5\Leftrightarrow2x-3=25\Leftrightarrow x=14\\ b,PT\Leftrightarrow x^2=\sqrt{\dfrac{98}{2}}=\sqrt{49}=7\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\end{matrix}\right.\\ c,ĐK:x\ge3\\ PT\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+1\right)=0\\ \Leftrightarrow\sqrt{x-3}=0\left(\sqrt{x+3}+1>0\right)\\ \Leftrightarrow x=3\\ d,ĐK:x\ge1\\ PT\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\\ \Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\left(tm\right)\\ e,PT\Leftrightarrow2x-1=16\Leftrightarrow x=\dfrac{17}{2}\\ f,PT\Leftrightarrow\left|2x-1\right|=\sqrt{3}-1\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{3}-1\\2x-1=1-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}}{2}\\x=\dfrac{2-\sqrt{3}}{2}\end{matrix}\right.\)

Bài 3:

\(a,Q=\dfrac{1+5}{3-1}=3\\ b,P=\dfrac{x+\sqrt{x}-6+x-2\sqrt{x}-3-x+4\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ c,M=\dfrac{\sqrt{x}}{\sqrt{x}-3}\cdot\dfrac{3-\sqrt{x}}{\sqrt{x}+5}=\dfrac{-\sqrt{x}}{\sqrt{x}+5}\)

Vì \(-\sqrt{x}\le0;\sqrt{x}+5>0\) nên \(M< 0\)

Do đó \(\left|M\right|>\dfrac{1}{2}\Leftrightarrow M< -\dfrac{1}{2}\Leftrightarrow-\dfrac{\sqrt{x}}{\sqrt{x}+5}+\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-\sqrt{x}-5}{2\left(\sqrt{x}+5\right)}< 0\Leftrightarrow\sqrt{x}-5< 0\left(\sqrt{x}+5>0\right)\\ \Leftrightarrow0\le x< 25\)

Bài 4:

\(a,A=\dfrac{16+2\cdot4+5}{4-3}=29\\ b,B=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ c,P=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}+1}\\ P=\dfrac{\left(\sqrt{x}+1\right)^2+4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}\\ P\ge2\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{4}{\sqrt{x}+1}}=2\sqrt{4}=4\\ P_{min}=4\Leftrightarrow\left(\sqrt{x}+1\right)^2=4\Leftrightarrow\sqrt{x}+1=2\Leftrightarrow x=1\left(tm\right)\)

\(\left(x+2\right)\left(\dfrac{360}{x}-6\right)=360\)

\(ĐK:x\ne0\)

\(\Leftrightarrow\left(x+2\right)\left(\dfrac{360-6x}{x}\right)=360\)

\(\Leftrightarrow360-6x+\dfrac{720-12x}{x}=360\)

\(\Leftrightarrow360x-6x^2+720-12x=360x\)

\(\Leftrightarrow6x^2+12x-720=0\)

\(\Delta=12^2-4.6.\left(-720\right)\)

    \(=17424>0\)

`->` pt có 2 nghiệm

\(\left\{{}\begin{matrix}x_1=\dfrac{-12-\sqrt{17424}}{12}=-12\\x_2=\dfrac{-12+\sqrt{17424}}{12}=10\end{matrix}\right.\) ( tm )

Vậy \(S=\left\{-12;10\right\}\)

`48/[x+4]+48/[x-4]=5`           `ĐK: x \ne +-4`

`<=>[48(x-4)+48(x+4)]/[(x-4)(x+4)]=[5(x+4)(x-4)]/[(x-4)(x+4)]`

   `=>48x-192+48x+192=5x^2-80`

`<=>5x^2-96x-80=0`

`<=>5x^2-100+4x-80=0`

`<=>5x(x-20)+4(x-20)=0`

`<=>(x-20)(5x+4)=0`

`<=>` $\left[\begin{matrix} x=20\\ x=\dfrac{-4}{5}\end{matrix}\right.$   (t/m)

Vậy `S={-4/5;20}`

ĐK : \(x\ne\pm4\)

\(\Leftrightarrow\cdot\dfrac{48\left(x+4\right)+48\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5\left(x+4\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\)

\(\Leftrightarrow48x+192+48x-192==5x^2-80\)

\(\Leftrightarrow96x=5x^2-80\)

\(\Leftrightarrow5x^2-96x-80=0\)

\(\Leftrightarrow5x^2+4x-100-80=0\)

\(\Leftrightarrow4\left(x-20\right)+5x\left(x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-20=0\\5x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-\dfrac{4}{5}\end{matrix}\right.\)

55470 : 86 = 645

xem trong sách đi em=)

a) \(4x^2-6x=2x\left(2x-3\right)\)

b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(x^2+y\right)\)

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

\(\dfrac{x+2}{x-3}< 0\)vì \(x+2>x-3\)

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

Công của người đó là

\(A=P.h=10m.h=10.60.4=2400\left(J\right)\) 

Công suất trong 2p (120s) là

\(P=\dfrac{A}{t}=\dfrac{2400}{20}=120\left(W\right)\)

a: Khi m=2 thì (1) sẽ là:

2x+y=2 và 4x+3y=10

=>x=-2 và y=6

b: 2x+y=m và 4x+3y=10

=>4x+2y=2m và 4x+3y=10

=>4x+3y=10 và 4x+2y=2m

=>y=10-2m và 2x=m-10+2m=3m-10

=>y=10-2m và x=3/2m-5

x>0 và y>0

=>10-2m>0 và 3/2m-5>0

=>m>5:3/2=10/3 và m<5

=>10/3<m<5

\(1,\\ a,=\left[x^3\left(x-2\right)-4x\left(x-2\right)\right]:\left(x^2-4\right)\\ =x\left(x^2-4\right)\left(x-2\right):\left(x^2-4\right)=x\left(x-2\right)\\ b,=\left(2014-14\right)^2=2000^2=4000000\\ 2,\\ A=2015\cdot2013\cdot\left(2014^2+1\right)\\ A=\left(2014^2-1\right)\left(2014^2+1\right)\\ A=2014^4-1< B=2014^4\)