Câu hỏi của nguyễn thị thanh huyền

Question

3*căn bậc hai(x+3) - căn bậc hai(x-2) = 7

Nguyễn Hoàng Minh · Accepted Answer

$ĐK:x\ge2\
PT\Leftrightarrow3\sqrt{x+3}-\sqrt{x-2}=7\
\Leftrightarrow\left(3\sqrt{x+3}\right)^2-6\sqrt{\left(x+3\right)\left(x-2\right)}+\left(\sqrt{x-2}\right)^2=49\
\Leftrightarrow9x+27-6\sqrt{x^2+x-6}+x-2=49\
\Leftrightarrow10x+25-6\sqrt{x^2+x-6}=49\
\Leftrightarrow6\sqrt{x^2+x-6}=10x-24\
\Leftrightarrow3\sqrt{x^2+x-6}=5x-12\
\Leftrightarrow9\left(x^2+x-6\right)=\left(5x-12\right)^2\
\Leftrightarrow9x^2+9x-54=25x^2-120x+144\
\Leftrightarrow16x^2-129x+198=0\
\Leftrightarrow\left(x-6\right)\left(16x-33\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\x=\dfrac{33}{16}\left(tm\right)\end{matrix}\right.$Câu trả lời: $ĐK:x\ge2\
PT\Leftrightarrow3\sqrt{x+3}-\sqrt{x-2}=7\
\Leftrightarrow\left(3\sqrt{x+3}\right)^2-6\sqrt{\left(x+3\right)\left(x-2\right)}+\left(\sqrt{x-2}\right)^2=49\
\Leftrightarrow9x+27-6\sqrt{x^2+x-6}+x-2=49\
\Leftrightarrow10x+25-6\sqrt{x^2+x-6}=49\
\Leftrightarrow6\sqrt{x^2+x-6}=10x-24\
\Leftrightarrow3\sqrt{x^2+x-6}=5x-12\
\Leftrightarrow9\left(x^2+x-6\right)=\left(5x-12\right)^2\
\Leftrightarrow9x^2+9x-54=25x^2-120x+144\
\Leftrightarrow16x^2-129x+198=0\
\Leftrightarrow\left(x-6\right)\left(16x-33\right)=0\
\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\x=\dfrac{33}{16}\left(tm\right)\end{matrix}\right.$

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