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

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

a, ĐKXĐ: \(x\ge0,x\ne1\)

Khi a=\(3+\sqrt{8}\)

\(\Rightarrow A=\sqrt{3+\sqrt{8}}-1=\sqrt{2+2\sqrt{2}+1}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1=\left|\sqrt{2}+1\right|-1=\sqrt{2}+1-1=\sqrt{2}\)

c,Để A>0\(\Rightarrow\sqrt{a}-1>0\Leftrightarrow\sqrt{a}>1\Leftrightarrow a>1\)

Để A=0

\(\Rightarrow A=\sqrt{a}-1=0\Leftrightarrow\sqrt{a}=1\Leftrightarrow a=1\)

Lời giải:

Ta có:

\(a(4-b)(4-c)=a(16-4b-4c+bc)=a[16-4(4-a-\sqrt{abc})+bc]\)

\(=a(4a+4\sqrt{abc}+bc)=4a^2+4a\sqrt{abc}+abc\)

\(=(2a+\sqrt{abc})^2\)

\(\Rightarrow \sqrt{a(4-b)(4-c)}=2a+\sqrt{abc}\)

Hoàn toàn tương tự với các biểu thức còn lại, suy ra:

\(M=2a+\sqrt{abc}+2b+\sqrt{abc}+2c+\sqrt{abc}-\sqrt{abc}\)

\(=2(a+b+c+\sqrt{abc})=2.4=8\)

@ngonhuminh

a: \(B-2=\dfrac{4\sqrt{a}-4a-1}{2a+1}=\dfrac{-\left(2\sqrt{a}-1\right)^2}{2a+1}< 0\)

=>B<2

b: Để 1/D là số nguyên thì \(\sqrt{x}+2⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1+3⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1\in\left\{1;-1;3\right\}\)

hay \(x\in\left\{4;0;16\right\}\)

a: =>căn 3x+19=x+5

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+10x+25=3x+19\\x>=-\dfrac{19}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+7x+6=0\\x>=-\dfrac{19}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(x+6\right)=0\\x>=-\dfrac{19}{3}\end{matrix}\right.\Leftrightarrow x\in\left\{-6;-1\right\}\)

b: =>\(\sqrt{4x}=\sqrt{9x-3}\)

=>9x-3=4x

=>5x=3

=>x=3/5

a:

Sửa đề: \(\left(\dfrac{7-\sqrt{7}}{\sqrt{7}}-2\right)\left(\dfrac{6}{\sqrt{7}+1}+4\right)\)

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

\(=\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)=7-9=-2\)

b: \(=\sqrt{\dfrac{5-2\sqrt{6}}{12}}+\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}+\dfrac{1}{\sqrt{6}}\)

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

\(x+2\sqrt{x+2}+3\)

\(=x+2\sqrt{x+2}+2+1\)

\(=\left(x+2\right)+2\sqrt{x+2}+1\)

\(=\left(\sqrt{x+2}\right)^2+2\sqrt{x+2}+1^2\) ( hằng đẳng thức thứ nhất )

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

chúc bạn học tốt.....

\(a,\sqrt{25-10x+x^2}+7=x\)(\(x\ge5\))

\(\Leftrightarrow\sqrt{\left(5-x\right)^2}=7-x\)

\(\Leftrightarrow5-x=7-x\)

\(\Leftrightarrow0x=2\) (vô nghiệm)

Vậy PT đã cho vô nghiệm

\(b,\sqrt{x+6\sqrt{x-9}}=4\left(x\ge9\right)\)

\(\Leftrightarrow\sqrt{x-9+6\sqrt{x-9}+9}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-9}+3\right)^2}=4\)

\(\Leftrightarrow\sqrt{x-9}+3=4\)

\(\Leftrightarrow\sqrt{x-9}=1\)

\(\Leftrightarrow x-9=1\)

\(\Leftrightarrow x=10\left(TM\right)\)

Vậy PT có nghiệm là \(x=10\)