Ôn tập chương 1: Căn bậc hai. Căn bậc ba

5:

a: \(=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{x-25}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)

b: Khi x=9 thì A=(3-5)/(3+5)=-2/8=-1/2

c: A<0

=>căn x-5<0

=>0<x<25

\(1+\sqrt{2}>0\)

\(1-\sqrt{2}+1-\sqrt{2}< 0\)

\(1+\sqrt{2}+\sqrt{35}>0\)

OF/sinE=OE/sinF

=>10/sin30=OE/sin80

=>OE=19,7(cm)

IK=AK*cosAKI

=>AK=591,18m

góc B=90-50-15=25 độ

AB/sin15=AK/sinB

=>AB=362,05(m)

đk a >= 0 

\(\left(2\sqrt{a}\right)^2-2.2\sqrt{a}.1+1=\left(2\sqrt{a}-1\right)^2\)

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

ĐKXĐ: \(x\ge0\)

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

\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}+10\)

\(\Leftrightarrow3\sqrt{x}=12\)

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

\(\Leftrightarrow x=16\left(tm\right)\)

\(b,=\dfrac{4\left(\sqrt{3}+\sqrt{7}\right)}{3-7}+\dfrac{\sqrt{3}+\sqrt{2}}{3-2}\\ =-\sqrt{3}-\sqrt{7}+\sqrt{3}+\sqrt{2}=\sqrt{2}-\sqrt{7}\)

\(=\dfrac{4\left(\sqrt{3}+\sqrt{7}\right)}{-4}+\dfrac{\sqrt{3}+\sqrt{2}}{1}=-\sqrt{3}-\sqrt{7}+\sqrt{3}+\sqrt{2}=-\sqrt{7}+\sqrt{2}\)

a: ĐKXĐ: x<>-2; x<>3

\(P=\dfrac{x^2-9-x^2+4-x+8}{\left(x-3\right)\left(x+2\right)}:\dfrac{2x+4-x-6}{2\left(x+2\right)}\)

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

\(=\dfrac{-2}{x-2}\)

b: |2x-3|+1=x

=>|2x-3|=x-1

\(\Leftrightarrow\left\{{}\begin{matrix}x>=1\\\left(2x-3-x+1\right)\left(2x-3+x-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=1\\\left(x-2\right)\left(3x-4\right)=0\end{matrix}\right.\)

=>x=4/3

Khi x=4/3 thì \(P=\dfrac{-2}{\dfrac{4}{3}-2}=-2:\dfrac{-2}{3}=3\)

Bài 3: 

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)

Bài 6:

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

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{x+4}{\sqrt{x}+2}\)

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

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

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

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

`a)`\(M=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

\(M=\left[\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{1}{\sqrt{a}-1}\right]:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)

\(M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)

\(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

`b)`\(M=-1\)

\(\Leftrightarrow\dfrac{\sqrt{a}-1}{\sqrt{a}}=-1\)

\(\Leftrightarrow\sqrt{a}-1=-\sqrt{a}\)

\(\Leftrightarrow a=\dfrac{1}{4}\) ( tm )

`c)`\(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

\(M=1-\dfrac{1}{\sqrt{a}}\)

Ta có: \(\sqrt{a}>0\)

\(\Rightarrow\dfrac{1}{\sqrt{a}}>0\)

\(\Rightarrow M< 1-0=1\)

`d)`\(M< 0\)

\(\Leftrightarrow1-\dfrac{1}{\sqrt{a}}< 0\)

\(\Leftrightarrow\dfrac{1}{\sqrt{a}}>1\)

\(\Leftrightarrow1>\sqrt{a}\)

\(\Leftrightarrow a< 1\)

Vậy \(S=\left\{a|0< a< 1\right\}\)