a) (x≤3)





 

a: Ta có: \(\sqrt{x-1}=2\)

\(\Leftrightarrow x-1=4\)

hay x=5

b: Ta có: \(\sqrt{3-x}=4\)

\(\Leftrightarrow3-x=16\)

hay x=-13

c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)

\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)

\(\Leftrightarrow-2x=-\dfrac{47}{16}\)

hay \(x=\dfrac{47}{32}\)

d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)

\(\Leftrightarrow x-1=\dfrac{49}{4}\)

hay \(x=\dfrac{53}{4}\)

e: Ta có: \(\sqrt{x-1}-3=1\)

\(\Leftrightarrow\sqrt{x-1}=4\)

\(\Leftrightarrow x-1=16\)

hay x=17

f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)

\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)

\(\Leftrightarrow x+2=\dfrac{1}{64}\)

hay \(x=-\dfrac{127}{64}\)

a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\sqrt{2^2\cdot7}-\sqrt{3^2\cdot7}+\dfrac{\sqrt{7}\cdot\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)

\(=2\sqrt{7}-3\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1\)

\(=-\sqrt{7}\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

\(=\left[\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)

\(=\dfrac{2\cdot4}{\sqrt{x}-3}\)

\(=\dfrac{8}{\sqrt{x}-3}\)

b) \(A>B\) khi 

\(\dfrac{8}{\sqrt{x}-3}< -\sqrt{7}\)

\(\Leftrightarrow8< -\sqrt{7x}+3\sqrt{7}\)

\(\Leftrightarrow x< \dfrac{\left(3\sqrt{7}-8\right)^2}{7}\)

\(a,A=\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{x-2-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

a: \(A=3+\left(-2\right)\cdot\sqrt{3}+3\cdot\sqrt{3}-2-\sqrt{3}\)

\(=3-2=1\)

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

b: B<A

=>B-1<0

=>\(\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}}< 0\)

=>-1/căn x<0

=>căn x>0

=>x>0 và x<>1

\(a,A=2\sqrt{20}-\dfrac{2}{\sqrt{3}+1}-\sqrt{80}+\sqrt{4+2\sqrt{3}}\\ =2.2\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{\sqrt{3^2}-1}-4\sqrt{5}+\sqrt{\left(\sqrt{3}+1\right)^2}\\ =-\dfrac{2\left(\sqrt{3}-1\right)}{2}+\left|\sqrt{3}+1\right|\\ =-\sqrt{3}+1+\sqrt{3}+1\\ =2\)

\(B=\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\left(dk:x\ge0,x\ne1\right)\\ =\left(1+\dfrac{\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\right)\left(1-\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\\ =1-x\)

\(b,A=4\sqrt{B}\Leftrightarrow4\sqrt{1-x}=2\\ \Leftrightarrow\sqrt{1-x}=\dfrac{1}{2}\\ \Leftrightarrow\left|1-x\right|=\dfrac{1}{4}\)

\(\Leftrightarrow1-x=\dfrac{1}{4}\\ \Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)

Vậy \(x=\dfrac{3}{4}\) thì \(A=4\sqrt{B}\).

a) \(A=2\sqrt{20}-\dfrac{2}{\sqrt{3}+1}-\sqrt{80}+\sqrt{4+2\sqrt{3}}\)

\(A=2\cdot2\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}-4\sqrt{5}+\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot1+1^2}\)

\(A=4\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-4\sqrt{5}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(A=-\left(\sqrt{3}-1\right)+\sqrt{3}+1\)

\(A=-\sqrt{3}+1+\sqrt{3}+1\)

\(A=2\)

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

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

\(B=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)

\(B=1^2-\left(\sqrt{x}\right)^2\)

\(B=1-x\)

b) Ta có: \(A=4\sqrt{B}\)

\(\Rightarrow2=4\sqrt{1-x}\)

\(\Leftrightarrow\sqrt{1-x}=\dfrac{1}{2}\)

\(\Leftrightarrow1-x=\dfrac{1}{4}\)

\(\Leftrightarrow x=1-\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)

a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\)

b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

\(\sqrt{x}=a,\sqrt{y}=b\)

Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)

\(\Rightarrow B=x+\sqrt{xy}+y\)

Vậy...

c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\)

d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\)

a:

=\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\)

= 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn)

b:

=\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\)

=\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức )

= (x+\(\sqrt{xy}\)+y)

c:

d:

e:2x +

= 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức)

=2x+\(\dfrac{3x-1}{3x-1}\)

=2x+1

\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)

a: =>2sin(x+pi/3)=-1

=>sin(x+pi/3)=-1/2

=>x+pi/3=-pi/6+k2pi hoặc x+pi/3=7/6pi+k2pi

=>x=-1/2pi+k2pi hoặc x=2/3pi+k2pi

b: =>2sin(x-30 độ)=-1

=>sin(x-30 độ)=-1/2

=>x-30 độ=-30 độ+k*360 độ hoặc x-30 độ=180 độ+30 độ+k*360 độ

=>x=k*360 độ hoặc x=240 độ+k*360 độ

c: =>2sin(x-pi/6)=-căn 3

=>sin(x-pi/6)=-căn 3/2

=>x-pi/6=-pi/3+k2pi hoặc x-pi/6=4/3pi+k2pi

=>x=-1/6pi+k2pi hoặc x=3/2pi+k2pi

d: =>2sin(x+10 độ)=-căn 3

=>sin(x+10 độ)=-căn 3/2

=>x+10 độ=-60 độ+k*360 độ hoặc x+10 độ=240 độ+k*360 độ

=>x=-70 độ+k*360 độ hoặc x=230 độ+k*360 độ

e: \(\Leftrightarrow2\cdot sin\left(x-15^0\right)=-\sqrt{2}\)

=>\(sin\left(x-15^0\right)=-\dfrac{\sqrt{2}}{2}\)

=>x-15 độ=-45 độ+k*360 độ hoặc x-15 độ=225 độ+k*360 độ

=>x=-30 độ+k*360 độ hoặc x=240 độ+k*360 độ

f: \(\Leftrightarrow sin\left(x-\dfrac{pi}{3}\right)=-\dfrac{1}{\sqrt{2}}\)

=>x-pi/3=-pi/4+k2pi hoặc x-pi/3=5/4pi+k2pi

=>x=pi/12+k2pi hoặc x=19/12pi+k2pi

g) \(3+\sqrt[]{5}sin\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=-\dfrac{3}{\sqrt[]{5}}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=sin\left[arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)\right]\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\\x+\dfrac{\pi}{3}=\pi-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)

h) \(1+sin\left(x-30^o\right)=0\)

\(\Leftrightarrow sin\left(x-30^o\right)=-1\)

\(\Leftrightarrow sin\left(x-30^o\right)=sin\left(-90^o\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-30^o=-90^0+k360^o\\x-30^o=180^o+90^0+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-60^0+k360^o\\x=300^0+k360^o\end{matrix}\right.\)

\(\Leftrightarrow x=-60^0+k360^o\)