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\(a,ĐK:5x\ge0\Leftrightarrow x\ge0\\ b,ĐK:3x+7\ge0\Leftrightarrow x\ge-\dfrac{7}{3}\\ c,ĐK:5-x\ge0\Leftrightarrow x\le5\\ đ,ĐK:3-2x\ge0\Leftrightarrow x\le\dfrac{3}{2}\\ f,ĐK:\dfrac{-3}{1+4x}\ge0\Leftrightarrow1+4x< 0\left(-3< 0;1+4x\ne0\right)\\ \Leftrightarrow x< -\dfrac{1}{4}\\ h,ĐK:10+x^2\ge0\Leftrightarrow x\in R\left(10+x^2\ge10>0\right)\)
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14) \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)
15) \(\sqrt{8+2\sqrt{15}}=\sqrt{5}+\sqrt{3}\)
16) \(\sqrt{10-2\sqrt{21}}=\sqrt{7}-\sqrt{3}\)
17) \(\sqrt{11+2\sqrt{18}}=3+\sqrt{2}\)
18) \(\sqrt{7+2\sqrt{10}}=\sqrt{5}+\sqrt{2}\)
19) \(\sqrt{7+4\sqrt{3}}=2+\sqrt{3}\)
20) \(\sqrt{12-2\sqrt{35}}=\sqrt{7}-\sqrt{5}\)
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\(\sqrt{10}-\sqrt{\dfrac{5}{3}}-\sqrt{18}\)
\(=\sqrt{10}-\dfrac{\sqrt{5}}{\sqrt{3}}-3\sqrt{2}\)
\(=\sqrt{10}-\dfrac{\sqrt{15}}{3}-3\sqrt{2}\)
\(=\dfrac{3\sqrt{10}-\sqrt{15}-9\sqrt{2}}{3}\)
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`1)(x+2)(x+3)(x-7)(x-8)=144`
`<=>[(x+2)(x-7)][(x+3)(x-8)]=144`
`<=>(x^2-5x-14)(x^2-5x-24)=144`
`<=>(x^2-5x-19)^2-25=144`
`<=>(x^2-5x-19)^2-169=0`
`<=>(x^2-5x-6)(x^2-5x-32)=0`
`+)x^2-5x-6=0`
`<=>` $\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.$
`+)x^2-5x-32=0`
`<=>` $\left[ \begin{array}{l}x=\dfrac{5+3\sqrt{17}}{2}\\x=\dfrac{5-3\sqrt{17}}{2}\end{array} \right.$
Vậy `S={-1,6,\frac{5+3\sqrt{17}}{2},\frac{5-3\sqrt{17}}{2}}`
1: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)=144\)
\(\Leftrightarrow\left(x^2-7x+2x-14\right)\left(x^2-8x+3x-24\right)=144\)
\(\Leftrightarrow\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+336-144=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+192=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2-6\left(x^2-5x\right)-32\left(x^2-5x\right)+192=0\)
\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x-6\right)-32\left(x^2-5x-6\right)=0\)
\(\Leftrightarrow\left(x^2-5x-6\right)\left(x^2-5x-32\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+1\right)\left(x^2-5x-32\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+1=0\\x^2-5x-32=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\\x=\dfrac{5-3\sqrt{17}}{2}\\x=\dfrac{5+3\sqrt{17}}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{6;-1;\dfrac{5-3\sqrt{17}}{2};\dfrac{5+3\sqrt{17}}{2}\right\}\)
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Sửa đè \(\sqrt{x^2+10x+21}=3\sqrt{x+3}+2\sqrt{x+7}-6\)
Đk:\(x\ge-3\)
\(pt\Leftrightarrow\sqrt{\left(x+3\right)\left(x+7\right)}=3\sqrt{x+3}+2\sqrt{x+7}-6\)
Đặt \(\hept{\begin{cases}\sqrt{x+3}=a\\\sqrt{x+7}=b\end{cases}\left(a,b>0\right)}\) thì
\(\Leftrightarrow ab=3a+2b-6\)
\(\Leftrightarrow\left(a-2\right)\left(b-3\right)=0\Rightarrow\orbr{\begin{cases}a=2\\b=3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x+3}=2\\\sqrt{x+7}=3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\) (thỏa)
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\(\frac{90}{x}-\frac{90}{x}+10=\frac{45}{60}\) Á sai đề chắc
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\(-\frac{3}{10}+x=-\frac{7}{18}\)
\(x=-\frac{7}{10}-\left(-\frac{3}{10}\right)\)
\(x=-\frac{7}{10}+\frac{3}{10}\)
\(x=-\frac{2}{5}\)
X=-4/45
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