2x + 10 ⋮ x + 3
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26 tháng 2 2023
a: =>14x+20+5=6x-9-9x
=>14x+25=-3x-9
=>17x=-34
=>x=-2
b: =>\(2x^2-30x+2x-30=2x^2+10x-10x-50\)
=>-28x-30=-50
=>-28x=-20
=>x=20/28=5/7
c: =>2x+x^3-x=x^3+1
=>x=1
d: =>x^3-3x^2+3x-1-x(x^2+2x+1)=10x-2x^2-11x-22
=>x^3-3x^2+3x-1-x^3-2x^2-x=-2x^2-x-22
=>-5x^2+2x-1+2x^2+x+22=0
=>-3x^2+3x+21=0
=>x^2-x-7=0
=>\(x=\dfrac{1\pm\sqrt{29}}{2}\)
14 tháng 10 2021
a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)
\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)
\(\Leftrightarrow-12x=12\)
hay x=-1
P
0
24 tháng 9 2023
`#3107`
b)
`2.3^x = 162`
`\Rightarrow 3^x = 162 \div 2`
`\Rightarrow 3^x = 81`
`\Rightarrow 3^x = 3^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
c)
`(2x - 15)^5 = (2 - 15)^3`
\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`
\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)
24 tháng 9 2023
`d)`
\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!
`e)`
\(7\cdot4^{x-1}+4^{x-1}=23?\)
\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)
Bạn xem lại đề!
`f)`
\(2\cdot2^{2x}+4^3\cdot4^x=1056\)
\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
_____
\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)
\(\Rightarrow\left(x\div3+17\right)\div10=2\)
\(\Rightarrow x\div3+17=20\)
\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)
Vậy, `x = 9.`
22 tháng 5 2022
\(\dfrac{3}{2x+10}+\dfrac{3}{x-5}-\dfrac{2x}{x^2-25}\)
\(=\dfrac{3\left(x-5\right)}{2\left(x+5\right)\left(x-5\right)}+\dfrac{6\left(x+5\right)}{2\left(x+5\right)\left(x-5\right)}-\dfrac{4x}{2\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3x-15+6x+30-4x}{2\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{5x+15}{2\left(x+5\right)\left(x-5\right)}\)
TB
2
18 tháng 5 2021
Máy bài này khá dễ chỉ cần suy nghĩ tí là làm được.
Đặt `2x^2-x=a(a>=-1/8)`
`pt<=>1/(a+1)+3/(a+3)=10/(a+7)`
`<=>(a+3)(a+7)+3(a+1)(a+7)=10(a+1)(a+3)`
`<=>a^2+10a+21+3(a^2+8a+7)=10(a^2+4a+3)`
`<=>a^2+3a^2+10a+24a+21+21=10a^2+40a+30`
`<=>4a^2+34a+42=10a^2+40a+30`
`<=>6a^2+6a-12=0`
`<=>a^2+a-2=0`
`a+b+c=0`
`=>a_1=1,a_2=-2(l)`
`a=1=>2x^2-x=1`
`=>2x^2-x-1=0`
`a+b+c=0`
`=>x_1=1,x_1=-1/2`
Vậy `S={1,-1/2}`
18 tháng 5 2021
Đặt \(2x^2-x+1=a\left(a\ge\dfrac{7}{8}\right)\)
PTTT : \(\dfrac{1}{a}+\dfrac{3}{a+2}=\dfrac{10}{a+6}\)
\(\Leftrightarrow\left(a+2\right)\left(a+6\right)+3a\left(a+6\right)=10a\left(a+2\right)\)
\(\Leftrightarrow a^2+2a+6a+12+3a^2+18a=10a^2+20a\)
\(\Leftrightarrow-6a^2+6a+12=0\)
\(\Leftrightarrow\left(a+1\right)\left(a-2\right)=0\)
\(\Leftrightarrow a=2\)
-Thay lại a = 2 ta được : \(2x^2-x-1=0\)
<=> \(\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
L
0
\(\frac{2x+10}{x+3}=\frac{2\left(x+3\right)+4}{x+3}=2+\frac{4}{x+3}\)
Vậy để \(2x+10⋮x+3\) thì \(x+3\inƯ\left(4\right)\)
Mà Ư(4)={1;-1;2;-2;4;-4}
=>x+3={1;-1;2;-2;4;-4}
Ta có bẳng sau:
Vậy x={-2;-4;-1;-5;1;-7}
\(2x+10⋮x+3\\ \Rightarrow2\left(x+3\right)+4⋮x+3\\ \Rightarrow4⋮x+3\\ \Rightarrow x+3\in\text{Ư}\left(4\right)=\left\{1;2;4;-1;-2;-4\right\}\\ x\in\left\{-2;-1;1;-4;-5;-7\right\}\)