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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2 tháng 8 2023
Tách ?
`a, 70 -5.(x-3) =45`
`=> 5.(x-3)= 70-45`
`=> 5.(x-3)=25`
`=>x-3=25:5`
`=>x-3=5`
`=>x= 5+3`
`=>x=8`
______
`b,10 + 2.x = 4^5:4^3`
`=> 10 + 2.x = 4^(5-3)`
`=> 10 + 2.x =4^2=16`
`=> 2.x=16-10`
`=>2.x=6`
`=>x=6:2`
`=>x=3`
_____
`c,60-3.x-2=51`
`=> 60-3.x= 51+2`
`=> 60-3.x=53`
`=>3.x=60-53`
`=> 3.x= 7`
`=>x= 7/3`
____
`d, 4.x-20=2^5:2^3`
`=> 4.x-20=2^(5-3)`
`=> 4.x-20=2^2`
`=> 4.x= 4+20`
`=>4.x=24`
`=>x=24:4`
`=>x=6`
____
`2^x . 4=16`
`=> 2^x=16:4`
`=>2^x= 4`
`=>2^x=2^2`
`=>x=2`
____
`f, 3^x . 3=243`
`=>3^x=243:3`
`=> 3^x=81`
`=> 3^x=3^3`
`=>x=3`
_____
`g, 64. 4^x =16^8`
`=> 4^3 . 4^x=(4^2)^8`
`=> 4^3 . 4^x = 4^(16)`
`=> 4^x= 4^(16-3)`
`=>4^x=4^(13)`
`=>x=13`
_____
`2^x . 16^2 =1024`
`=> 2^x= 1024 : 16^2`
`=>2^x=4`
`=>2^x=2^2`
`=>x=2`
2 tháng 8 2023
a: =>5(x-3)=25
=>x-3=5
=>x=8
b: =>2x=16-10=6
=>x=3
c: =>58-3x=51
=>3x=7
=>x=7/3
d: =>4x-20=4
=>4x=24
=>x=6
e: =>2^x=4
=>2^x=2^2
=>x=2
f: =>3^x=81
=>3^x=3^4
=>x=4
g: =>4^x*4^3=4^16
=>x+3=16
=>x=13
h: =>2^x=1024/256=4=2^2
=>x=2
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
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