a, Căn (x-3)^2=3-x b, căn (4x^2 -20x+25) +2x=5 c, căn (1-12x+36x^2)=5 Mk cần gấp các bạn làm hộ mk với 26/08/2021 Bởi Kaylee a, Căn (x-3)^2=3-x b, căn (4x^2 -20x+25) +2x=5 c, căn (1-12x+36x^2)=5 Mk cần gấp các bạn làm hộ mk với
a)ĐKXD:`x≤3` `\sqrt{(x-3)^2}=3-x` `⇔|x-3|=3-x` `⇔`\(\left[ \begin{array}{l}x-3=3-x\\x-3=x-3\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=3\\x∈R\end{array} \right.\) Kết hợp ĐKXD `⇒x≤3` b)ĐKXD:`x≤5/2` `\sqrt{4x^2-20x+25}+2x=5` `⇔\sqrt{(2x-5)^2}=5-2x` `⇔|2x-5|=5-2x ` `⇔`\(\left[ \begin{array}{l}2x-5=5-2x\\2x-5=2x-5\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=\frac{5}{2}\\x∈R\end{array} \right.\) Kết hợp ĐKXD `⇒x≤5/2` c) `\sqrt{1-12x+36x^2}=5` `⇔\sqrt{(1-6x)^2}=5` `⇔|1-6x|=5` `⇔`\(\left[ \begin{array}{l}1-6x=5\\1-6x=-5\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}6x=-4\\6x=6\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=\frac{-2}{3}\\x=1\end{array} \right.\) Bình luận
Giải thích các bước giải: $a) \sqrt{(x-3)^{2}}=3-x\\\Leftrightarrow \left | x-3 \right |=3-x\\\Leftrightarrow \left | x-3 \right |+x=3\\\Leftrightarrow \left[ \begin{array}{l}x-3+x=3,x-3\geq 0\\-(x-3)+x=3,x-3<0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=3,x\geq 3\\x\in \mathbb{R},x<3\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=3\\x\leq 3\end{array} \right.\\b) \sqrt{4x^{2}-20x+25}+2x=5\\\Leftrightarrow \sqrt{(2x-5)^{2}}+2x=5\\\Leftrightarrow \left | 2x-5 \right |+2x=5\\\Leftrightarrow \left[ \begin{array}{l}2x-5+2x=5,2x-5\geq 0\\-(2x-5)+2x=5,2x-5<0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\frac{5}{2},x\geq \frac{5}{2}\\x\in \mathbb{R},x<\frac{5}{2}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{5}{2}\\x\leq \dfrac{5}{2}\end{array} \right.\\c) \sqrt{1-12x+36x^{2}}=5\\\Leftrightarrow \sqrt{(1-6x)^{2}}=5\\\Leftrightarrow \left[ \begin{array}{l}1-6x=5\\-(1-6x)=5\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=-\dfrac{2}{3}\\x=1\end{array} \right.$ Bình luận
a)ĐKXD:`x≤3`
`\sqrt{(x-3)^2}=3-x`
`⇔|x-3|=3-x`
`⇔`\(\left[ \begin{array}{l}x-3=3-x\\x-3=x-3\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=3\\x∈R\end{array} \right.\)
Kết hợp ĐKXD
`⇒x≤3`
b)ĐKXD:`x≤5/2`
`\sqrt{4x^2-20x+25}+2x=5`
`⇔\sqrt{(2x-5)^2}=5-2x`
`⇔|2x-5|=5-2x `
`⇔`\(\left[ \begin{array}{l}2x-5=5-2x\\2x-5=2x-5\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\frac{5}{2}\\x∈R\end{array} \right.\)
Kết hợp ĐKXD
`⇒x≤5/2`
c)
`\sqrt{1-12x+36x^2}=5`
`⇔\sqrt{(1-6x)^2}=5`
`⇔|1-6x|=5`
`⇔`\(\left[ \begin{array}{l}1-6x=5\\1-6x=-5\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}6x=-4\\6x=6\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\frac{-2}{3}\\x=1\end{array} \right.\)
Giải thích các bước giải:
$a) \sqrt{(x-3)^{2}}=3-x\\\Leftrightarrow \left | x-3 \right |=3-x\\\Leftrightarrow \left | x-3 \right |+x=3\\\Leftrightarrow \left[ \begin{array}{l}x-3+x=3,x-3\geq 0\\-(x-3)+x=3,x-3<0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=3,x\geq 3\\x\in \mathbb{R},x<3\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=3\\x\leq 3\end{array} \right.\\b) \sqrt{4x^{2}-20x+25}+2x=5\\\Leftrightarrow \sqrt{(2x-5)^{2}}+2x=5\\\Leftrightarrow \left | 2x-5 \right |+2x=5\\\Leftrightarrow \left[ \begin{array}{l}2x-5+2x=5,2x-5\geq 0\\-(2x-5)+2x=5,2x-5<0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\frac{5}{2},x\geq \frac{5}{2}\\x\in \mathbb{R},x<\frac{5}{2}\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{5}{2}\\x\leq \dfrac{5}{2}\end{array} \right.\\c) \sqrt{1-12x+36x^{2}}=5\\\Leftrightarrow \sqrt{(1-6x)^{2}}=5\\\Leftrightarrow \left[ \begin{array}{l}1-6x=5\\-(1-6x)=5\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=-\dfrac{2}{3}\\x=1\end{array} \right.$