Có:

\(x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}\ge0\)

\(x\left(\sqrt{x}-\sqrt{y}\right)-y\left(\sqrt{x}-\sqrt{y}\right)\ge0\)

\(\left(x-y\right)\left(\sqrt{x}-\sqrt{y}\right)\ge0\)

\(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\ge0\)

\(\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)\ge0\)  (luôn đúng)

Dấu = xảy ra khi x=y

\(a,\) A có nghĩa \(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

\(b,A=\dfrac{1}{\sqrt{x}+2}-\dfrac{2}{\sqrt{x}-2}+\dfrac{x}{x-4}\\ =\dfrac{\sqrt{x}-2-2\left(\sqrt{x}+2\right)+x}{x-4}\\ =\dfrac{\sqrt{x}-2-2\sqrt{x}-4+x}{x-4}\\ =\dfrac{-\sqrt{x}+x-6}{x-4}\\ =\dfrac{2\sqrt{x}+x-3\sqrt{x}-6}{x-4}\\ =\dfrac{\sqrt{x}\left(2+\sqrt{x}\right)-3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)

\(x=16\Rightarrow A=\dfrac{\sqrt{16}-3}{\sqrt{16}-2}=\dfrac{1}{2}\)

\(c,A=\dfrac{2}{3}\Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=\dfrac{2}{3}\\ \Rightarrow3\left(\sqrt{x}-3\right)=2\left(\sqrt{x}-2\right)\\ \Rightarrow3\sqrt{x}-9=2\sqrt{x}-4\)

\(\Rightarrow\sqrt{x}=5\\ \Rightarrow x=25\left(tmdk\right)\) 

Vậy ...

e: =>|x-2|=6

=>x-2=6 hoặc x-2=-6

=>x=-4 hoặc x=8

f: =>x^2-6x+25=x^2+10x+25 và x>=-5

=>x=0

d: =>|x-3|=11+5=16

=>x-3=16 hoặc x-3=-16

=>x=-13 hoặc x=19

b: =>3*căn x-2+2/3*căn x-2=3/2

=>căn x-2=3/2:11/3=3/2*3/11=9/22

=>x-2=81/484

=>x=1049/484

d: \(=\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}+\dfrac{2\sqrt{5}+3}{2\sqrt{5}-3}-\dfrac{1}{\sqrt{7}+\sqrt{5}}\)

\(=\left(3+2\sqrt{2}\right)^2+\dfrac{\left(2\sqrt{5}+3\right)^2}{20-9}-\dfrac{\sqrt{7}-\sqrt{5}}{2}\)

\(=17+12\sqrt{2}+\dfrac{29+12\sqrt{5}}{11}-\dfrac{\sqrt{7}-\sqrt{5}}{2}\)

\(=17+12\sqrt{2}+\dfrac{29}{11}+\dfrac{12}{11}\sqrt{5}-\dfrac{\sqrt{7}}{2}+\dfrac{\sqrt{5}}{2}\)

\(=\dfrac{216}{11}+12\sqrt{2}+\dfrac{35}{22}\sqrt{5}-\dfrac{1}{2}\sqrt{7}\)

1: |1-2x|=2

=>|2x-1|=2

=>2x-1=2 hoặc 2x-1=-2

=>x=-1/2 hoặc x=3/2

15: =>x^2-2x-3=0

=>(x-3)(x+1)=0

=>x=3 hoặc x=-1

16: =>(1-5x-2x+3)/(2x-3)>0

=>(-7x+4)/(2x-3)>0

=>(7x-4)/(2x-3)<0

=>4/7<x<3/2

6: =>|x-1|=4

=>x-1=4 hoặc x-1=-4

=>x=5 hoặc x=-3

8: =>|x^2+1|=3x+5

=>x^2+1=3x+5

=>x^2-3x-4=0

=>(x-4)(x+1)=0

=>x=4 hoặc x=-1

a: khi m=2 thì (d): y=4x-2^2+1=4x-3

PTHĐGĐ:

x^2-4x+3=0

=>x=1 hoặc x=3

Khi x=1 thì y=1

Khi x=3 thì y=9

b: PTHĐGĐ là;

x^2-2mx+m^2-1=0

Δ=(-2m)^2-4(m^2-1)=4>0

=>(P) luôn cắt (d) tại hai điểm phân biệt

2y1+4m*x2-2m^2-3<0

=>2(2mx1-m^2+1)+4m*x2-2m^2-3<0

=>4m*x1-2m^2+2+4m*x2-2m^2-3<0

=>-4m^2+4m*(x1+x2)-1<0

=>-4m^2+4m*(2m)-1<0

=>-4m^2+8m-1<0

=>\(\left[{}\begin{matrix}m< \dfrac{2-\sqrt{3}}{2}\\m>\dfrac{2+\sqrt{3}}{2}\end{matrix}\right.\)

Tổng quát:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)\(=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow S=\dfrac{10}{11}\)

Ta có công thức tổng quát như sau:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left[\left(n+1\right)\sqrt{n}+n\sqrt{n+1}\right]\left[\left(n+1\right)\sqrt{n}-n\sqrt{n+1}\right]}\)

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

\(=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\)

\(=\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}\)

Áp dụng vào tổng S ta có:

\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{121\sqrt{120}+120\sqrt{121}}\)

\(S=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{120}}+\dfrac{1}{\sqrt{121}}\)

\(S=1-\dfrac{1}{\sqrt{121}}=1-\dfrac{1}{11}=\dfrac{10}{11}\)

1,\(=\sqrt{5-2.\sqrt{2}.\sqrt{5}.2}-\sqrt{5-2\sqrt{5}+1}\\ =\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)}^2\\ =\left|\sqrt{5}-\sqrt{2}\right|-\left|\sqrt{5}-1\right|\\ =\sqrt{5}-\sqrt{2}-\sqrt{5}+1\\ =1-\sqrt{2}\)

\(2,=\sqrt{24-2.2\sqrt{6}.3+9}-\sqrt{9-2.3.\sqrt{6}+6}\\ =\sqrt{\left(2\sqrt{6}-3\right)^2}-\sqrt{\left(3-\sqrt{6}\right)}\\ =\left|2\sqrt{6}-3\right|-\left|3-\sqrt{6}\right|\\ =2\sqrt{6}-3-3+\sqrt{6}\\ =3\sqrt{6}-6\)

\(3,=\sqrt{4-2.2.\sqrt{5}+5}+\sqrt{7-2.\sqrt{7}.\sqrt{5}+5}\\ =\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|+\left|\sqrt{7}-\sqrt{5}\right|\\ =\sqrt{5}-2+\sqrt{7}-\sqrt{5}\\ =\sqrt{7}-2\)

\(\left\{{}\begin{matrix}x+y=3000\\1,1x+1,12y=3328\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}1,1x+1,1y=3300\\1,1x+1,12y=3328\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=3000\\-0,02y=-28\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1400=3000\\y=\dfrac{-28}{-0,02}=1400\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3000-1400\\y=1400\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1600\\y=1400\end{matrix}\right.\)

Vậy: \(x=1600,y=1400\)

a: góc ACB=180-6-4=170 độ

Xét ΔCAB có AB/sinC=BC/sinA=AC/sinB

=>762/sin170=BC/sin6=AC/sin4

=>BC=458,69m; AC=306,10(m)

S CAB=1/2*CA*CB*sinC

\(=\dfrac{1}{2}\cdot458.69\cdot306.10\cdot sin170\simeq12190,54\left(m^2\right)\)

=>\(CH=2\cdot\dfrac{12190.54}{762}\simeq32\left(m\right)\)

b: An đến trường lúc:

6h+458,69:1000:4+306,10:1000:19\(\simeq6h8p\)