Chương I - Căn bậc hai. Căn bậc ba

a) \(4\sqrt{2x+1}-\sqrt{8x+4}+\dfrac{1}{2}\sqrt{32x+16}=12\) (ĐK: \(x\ge-\dfrac{1}{2}\)) 

\(\Leftrightarrow4\sqrt{2x+1}-\sqrt{4\left(2x+1\right)}+\dfrac{1}{2}\cdot4\sqrt{2x+1}=12\)

\(\Leftrightarrow4\sqrt{2x+1}-2\sqrt{2x+1}+2\sqrt{2x+1}=12\)

\(\Leftrightarrow4\sqrt{2x+1}=12\)

\(\Leftrightarrow\sqrt{2x+1}=\dfrac{12}{4}\)

\(\Leftrightarrow2x+1=3^2\)

\(\Leftrightarrow2x=9-1\)

\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=\dfrac{8}{2}\)

\(\Leftrightarrow x=4\left(tm\right)\)

b) \(\sqrt{4x^2-4x+1}=5\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\)

\(\Leftrightarrow\left|2x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\left(x\ge\dfrac{1}{2}\right)\\2x-1=-5\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{2}\\x=-\dfrac{4}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

c) \(\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}=-\dfrac{1}{2}\)(ĐK: \(x\ge0;x\ne1\))

\(\Leftrightarrow-\left(\sqrt{x}-1\right)=2\left(2\sqrt{x}-3\right)\)

\(\Leftrightarrow-\sqrt{x}+1=4\sqrt{x}-6\)

\(\Leftrightarrow4\sqrt{x}+\sqrt{x}=1+6\)

\(\Leftrightarrow5\sqrt{x}=7\)

\(\Leftrightarrow\sqrt{x}=\dfrac{7}{5}\)

\(\Leftrightarrow x=\dfrac{49}{25}\left(tm\right)\)

`\sqrt{4x-3}-\sqrt{x^2-3}=0`    `ĐK: x >= \sqrt{3}`

`<=>\sqrt{4x-3}=\sqrt{x^2-3}`

`<=>4x-3=x^2-3`

`<=>x^2-4x=0`

`<=>x(x-4)=0`

`<=>[(x=0(ko t//m)),(x=4(t//m)):}`

Vậy `S={4}`.

l2x-3l=l1-xl

\(\Leftrightarrow\)3x=4   ;     x=2

\(\Leftrightarrow\)x=3/4   ;x=2   

a) \(9\sqrt{x+2}-\dfrac{1}{3}\sqrt{9x+18}=24\) (ĐK: \(x\ge-2\)) 

\(\Leftrightarrow9\sqrt{x+2}-\dfrac{1}{3}\sqrt{9\left(x+2\right)}=24\)

\(\Leftrightarrow9\sqrt{x+2}-\dfrac{1}{3}\cdot3\sqrt{x+2}=24\)

\(\Leftrightarrow9\sqrt{x+2}-\sqrt{x+2}=24\)

\(\Leftrightarrow8\sqrt{x+2}=24\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{24}{8}\)

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

\(\Leftrightarrow x+2=9\)

\(\Leftrightarrow x=9-2\)

\(\Leftrightarrow x=7\left(tm\right)\)

b) \(\sqrt{x^2-6x+9}-2=0\)

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

\(\Leftrightarrow\left|x-3\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=-2\left(x< 3\right)\\x-3=2\left(x\ge3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3-2\\x=3+2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

a) \(A=\sqrt{\left(3-\sqrt{5}\right)^2}+\dfrac{3}{4}\sqrt{80}-6\)

\(A=\left|3-\sqrt{5}\right|+\dfrac{3}{4}\cdot4\sqrt{5}-6\)

\(A=3-\sqrt{5}+3\sqrt{5}-6\)

\(A=\left(3\sqrt{5}-\sqrt{5}\right)+\left(3-6\right)\)

\(A=2\sqrt{5}-3\)

b) \(B=\dfrac{\sqrt{18}}{\sqrt{6}}+\dfrac{4}{\sqrt{5}-1}-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\)

\(B=\dfrac{\sqrt{6}\cdot\sqrt{3}}{\sqrt{6}}+\dfrac{4\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\dfrac{\sqrt{3}\left(1+\sqrt{3}\right)}{1+\sqrt{3}}\)

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

\(B=\sqrt{5}+1\)

ĐKXĐ: x>=0; x<>4

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

M nguyên khi \(x-2\sqrt{x}+4\sqrt{x}-8+12⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(\sqrt{x}\in\left\{3;1;4;0;5;6;8;14\right\}\)

=>\(x\in\left\{9;1;16;0;25;36;64;196\right\}\)

chúc bạn học tốt

a: ΔAKC vuông tại K

=>góc KAC+góc KCA=90 độ

=>góc KAC=90-30=60 độ

Xét ΔAKC vuông tại K có

sin C=AK/AC

=>3/AC=sin30=1/2

=>AC=6(cm)

ΔAKC vuông tại K

=>AK^2+KC^2=AC^2

=>KC^2=6^2-3^2=27

=>KC=3*căn 3(cm)

b: \(\dfrac{BC}{cotB+cotC}=\dfrac{BC}{\dfrac{BK}{AK}+\dfrac{CK}{AK}}\)

\(=BC:\dfrac{BC}{AK}=AK\)

c: góc BAC=180-68-30=82 độ

Xét ΔABC có BC/sin A=AC/sinB=AB/sinC

=>5/sin82=AC/sin68=AB/sin30

=>\(AC\simeq4,68\left(cm\right);AB\simeq2,52\left(cm\right)\)

\(S_{BAC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC=\dfrac{1}{2}\cdot4.68\cdot2.52\cdot sin82\simeq5,84\left(cm^2\right)\)