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

Bài 3:

1: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(AB=\sqrt{12^2-5^2}=\sqrt{144-25}=\sqrt{119}\simeq10,9\)

Xét ΔABC vuông tại A có \(sinB=\dfrac{AC}{BC}=\dfrac{5}{12}\)

nên \(\widehat{B}\simeq24^037'\)

=>\(\widehat{C}=90^0-\widehat{B}\simeq65^023'\)

2: ΔABC vuông tại A 

=>\(\widehat{B}+\widehat{C}=90^0\)

=>\(\widehat{B}=90^0-47^0=43^0\)

Xét ΔABC vuông tại A có \(tanC=\dfrac{AB}{AC}\)

=>\(AB=AC\cdot tanC=5\cdot tan47\simeq5,4\)

ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC\simeq\sqrt{5,4^2+5^2}\simeq7,4\)

3: ΔABC vuông tại A

=>\(\widehat{B}+\widehat{C}=90^0\)

=>\(\widehat{C}=90^0-74^0=16^0\)

Xét ΔABC vuông tại A có \(tanB=\dfrac{AC}{AB}\)

=>\(AC=7\cdot tan74\simeq24,4\)

ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC\simeq\sqrt{24,4^2+7^2}\simeq25,4\)

4: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC=\sqrt{12^2+13^2}=\sqrt{313}\simeq17,7\)

Xét ΔABC vuông tại A có \(tanC=\dfrac{AB}{AC}=\dfrac{13}{12}\)

nên \(\widehat{C}\simeq47^017'\)

=>\(\widehat{B}=90^0-\widehat{C}\simeq42^043'\)

\(n_{H_2SO_4}=0,1.0,1=0,01\left(mol\right)\)

\(Na_2O+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

0,1<-----0,1

\(a=m_{Na_2O}=0,1.62=6,2\left(g\right)\)

a: ĐKXĐ: \(x\notin\left\{1;-1;\dfrac{1}{2}\right\}\)

\(A=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

\(=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}-\dfrac{x-5}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{-2x+1}\)

\(=\dfrac{-\left(x+1\right)+2\left(x-1\right)-x+5}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{-2x+1}\)

\(=\dfrac{-x-1+2x-2-x+5}{-2x+1}=\dfrac{2}{-2x+1}\)

b: Để A>0 thì \(\dfrac{2}{-2x+1}>0\)

mà 2>0

nên -2x+1>0

=>-2x>-1

=>\(x< \dfrac{1}{2}\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x\ne-1\end{matrix}\right.\)

a: \(x^2-3x+1>2\left(x-1\right)-x\left(3-x\right)\)

=>\(x^2-3x+1>2x-2-3x+x^2\)

=>-3x+1>-x-2

=>-2x>-3

=>\(x< \dfrac{3}{2}\)

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

=>\(x^2-2x+1+x^2< =x^2+2x+1+x^2+4x+4\)

=>-2x+1<=6x+5

=>-7x<=4

=>\(x>=-\dfrac{4}{7}\)

c: 

\(\left(x^2+1\right)\left(x-6\right)< =\left(x-2\right)^3\)

=>\(x^3-6x^2+x-6< =x^3-6x^2+12x-8\)

=>x-6<=12x-8

=>-11x<=-8+6=-2

=>\(x>=\dfrac{2}{11}\)

Vì \(\dfrac{1}{3}\ne\dfrac{2}{2}\)

nên hệ luôn có nghiệm duy nhất

\(\left\{{}\begin{matrix}x+2y=7\\3x+2y=2m+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x+2y-x-2y=2m+1-7\\x+2y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2m-6\\2y=7-x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=m-3\\2y=7-m+3=-m+10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m-3\\y=-0,5m+5\end{matrix}\right.\)

x+2=y

=>-0,5m+5=m-3+2=m-1

=>-1,5m=-6

=>m=4

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

=>\(\dfrac{3x+5-2x}{2}>=\dfrac{3+x+2}{3}\)

=>\(\dfrac{x+5}{2}-\dfrac{x+5}{3}>=0\)

=>\(\dfrac{3\left(x+5\right)-2\left(x+5\right)}{6}>=0\)

=>\(\dfrac{x+5}{6}>=0\)

=>x+5>=0

=>x>=-5

b: \(\dfrac{x-2}{3}-x-2< =\dfrac{x-17}{2}\)

=>\(\dfrac{2\left(x-2\right)}{6}+\dfrac{6\left(-x-2\right)}{6}< =\dfrac{3\left(x-17\right)}{6}\)

=>\(2\left(x-2\right)+6\left(-x-2\right)< =3\left(x-17\right)\)

=>\(2x-4-6x-12< =3x-51\)

=>-4x-16<=3x-51

=>-7x<=-35

=>x>=5

c: \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}< =\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)

=>\(\dfrac{4\left(2x+1\right)-3\left(x-4\right)}{12}< =\dfrac{2\left(3x+1\right)-x+4}{12}\)

=>4(2x+1)-3(x-4)<=2(3x+1)-x+4

=>8x+4-3x+12<=6x+2-x+4

=>5x+16<=5x+6

=>16<=6(sai)

Vậy: BPT vô nghiệm

a: \(\dfrac{3\left(2x+1\right)}{20}+1>\dfrac{3x+52}{10}\)

=>\(\dfrac{6x+3}{20}+\dfrac{20}{20}>\dfrac{6x+104}{20}\)

=>6x+23>6x+104

=>23>104(sai)

vậy: \(x\in\varnothing\)

b: \(\dfrac{4x-1}{2}+\dfrac{6x-19}{6}< =\dfrac{9x-11}{3}\)

=>\(\dfrac{3\left(4x-1\right)+6x-19}{6}< =\dfrac{2\left(9x-11\right)}{6}\)

=>12x-3+6x-19<=18x-22

=>-22<=-22(luôn đúng)

Vậy: \(x\in R\)

0