Tìm x, biết
a) 8.x + 20 = 76
b) 10 + 2.(x – 9) = 45 : 43
c) 54 ⋮ x; 270 ⋮ x và 20 ≤ x ≤ 30
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10 tháng 12 2021
Câu 2:
\(a,\Rightarrow8x=56\\ \Rightarrow x=7\\ b,\Rightarrow2\left(x-9\right)=4^2-10=6\\ \Rightarrow x-9=8\\ \Rightarrow x=17\\ c,\Rightarrow x\in BC\left(54,270\right)=B\left(54\right)=\left\{1;2;3;6;9;18;27;54\right\}\\ \text{Mà }20\le x\le30\Rightarrow x=27\)
Câu 3:
\(a,\text{A có }\left(2017-17\right):2+1=1001\left(\text{phần tử}\right)\\ b,P=\left\{2;3;5;7;9\right\}\)
LD
6 tháng 2 2022
\(a,\dfrac{5}{8}=\dfrac{x}{14}\)
\(\Rightarrow x=\dfrac{5.14}{8}=8,75\)
Vậy \(x=8,75\)
\(b,\dfrac{x}{6}=-\dfrac{1}{3}\)
\(\Rightarrow x=-\dfrac{1.6}{3}=-2\)
Vậy \(x=-2\)
\(c,-\dfrac{3}{5}=\dfrac{x}{10}\)
\(\Rightarrow x=-\dfrac{3.10}{5}=-6\)
Vậy \(x=-6\)
câu d đã có đáp án
2 tháng 10 2021
a) \(\sqrt{\left(2x-3\right)^2}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\left(đk:x\ge-2\right)\)
\(\Leftrightarrow8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}=20\)
\(\Leftrightarrow5\sqrt{x+2}=20\)
\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)
c) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
HP
2 tháng 10 2021
a. \(\sqrt{\left(2x-3\right)^2}=7\)
<=> \(\left|2x-3\right|=7\)
<=> \(\left[{}\begin{matrix}2x-3=7\left(x\ge\dfrac{3}{2}\right)\\-2x+3=7\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}2x=10\\-2x=4\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=5\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
b. \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\) ĐK: \(x\ge-2\)
<=> \(\sqrt{64\left(x+2\right)}-\sqrt{25\left(x+2\right)}+\sqrt{4\left(x+2\right)}-20=0\)
<=> \(8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}-20=0\)
<=> \(\sqrt{x+2}.\left(8-5+2\right)-20=0\)
<=> \(5\sqrt{x+2}=20\)
<=> \(\sqrt{x+2}=4\)
<=> \(\left(\sqrt{x+2}\right)^2=4^2\)
<=> \(\left|x+2\right|=16\)
<=> \(\left[{}\begin{matrix}x+2=16\left(x\ge-2\right)\\x+2=-16\left(x< -2\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=14\left(TM\right)\\x=-18\left(TM\right)\end{matrix}\right.\)
c. \(\sqrt{x^2-9}-3\sqrt{x-3}=0\) ĐK: \(x\ge3\)
<=> \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
<=> \(\sqrt{x-3}.\sqrt{x+3}-3\sqrt{x-3}=0\)
<=> \(\left(\sqrt{x+3}-3\right).\sqrt{x-3}=0\)
<=> \(\left[{}\begin{matrix}\sqrt{x+3}-3=0\\\sqrt{x-3}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=6\\x=3\end{matrix}\right.\)
NL
1
27 tháng 9 2021
a: Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)
\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)
\(\Leftrightarrow-12x=24\)
hay x=-2
b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)
\(\Leftrightarrow2x=-40\)
hay x=-20
em không biết