tìm x
x+3-2*(x-1)=3*(2+x)-1
giải giúp mik
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\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
a)Ta có:5/3.x^2-1/2.x^2y
=(5/3-1/2).x^2y
= 7/6.x^2y(Bậc 3)
b)Ta có: 7/6.(-2)^2(-1)
= 7/6.4.(-1)
= 7/6.(-4)
=-28/6
a, - A=\(\dfrac{5}{3}\).x2.y-\(\dfrac{-1}{2}\).x2.y
=\(\dfrac{13}{6}\).x2.y
- Bậc= 3.
b, A=\(\dfrac{13}{6}\).(-2)2.(-1)
=\(\dfrac{13}{6}\).4.(-1)
=\(\dfrac{-26}{3}\)
`[2-x]/x >= 1`
`<=>[2-x-x]/x >= 0`
`<=>[2-2x]/x >= 0`
`<=>0 < x <= 1`
`->\bb B`
\(x^5-x^3+x^2-1=x^3\left(x^2-1\right)+\left(x^2-1\right)=\left(x^2-1\right)\left(x^3+1\right)=\left(x-1\right)\left(x+1\right)^2\left(x^2-x+1\right)\)
1)
a) \(2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b) \(x\times\left(x+2\right)-3\times\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\times\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
c) \(\frac{x-6}{x+1}=\frac{x^2}{x-1}\)
nhân chéo lên, ngại chết đc
Hệ số góc bằng 2 `=> 3a+4 = 2 <=> a=-2/3`
Tung độ gốc bằng 1 `=> -b+3=1 <=> b=2`
`=> y=-2/3 x +2`
DKXD : x khac -1
\(\frac{-x}{x+1}\)+ 3 =\(\frac{2x+3}{x+1}\)
<=> \(\frac{-x}{x+1}\)+\(\frac{3\left(x+1\right)}{x+1}\)= \(\frac{2x+3}{x+1}\)
=> -x + 3x +3 = 2x +3
<=> 2x -2x =3-3
<=> 0x=0
<=> x=0(TMDK)
a) Thay x=-1 vào g(x),ta được:
\(2\cdot\left(-1\right)^2+5a=0\)
\(\Leftrightarrow5a=-2\)
hay \(a=-\dfrac{2}{5}\)
\(\left|x^2-1\right|=2x+1\left(dk:2x+1\ge0\Leftrightarrow2x\ge-1\Leftrightarrow x\le-\dfrac{1}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+1\\x^2-1=-2x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1-2x-1=0\\x^2-1+2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-2=0\\x^2+2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-2+3=3\\x.\left(x+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=3\\\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1^2\right)-\left(\sqrt{3}\right)^2=0\\\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1-\sqrt{3}\right).\left(x-1+\sqrt{3}\right)=0\\\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1-\sqrt{3}\left(loai\right)\\x=1+\sqrt{3\left(loai\right)}\end{matrix}\right.\\\left[{}\begin{matrix}x=0\left(loai\right)\\x=-2\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy x = -2
Cho g(x) = 0
x + 1 = 0
x = -1
Để f(x) chia hết cho g(x) thì x = -1 cũng là nghiệm của f(x)
Hay f(1) = 0
3.1² + 2.1² - 7.1 - m + 2 = 0
-2 - m + 2 = 0
m = 0
Vậy m = 0 thì f(x) chia hết cho g(x)
Giải chi tiết của em đây :
F(x) = 3x2 + 2x2 - 7x - m + 2
F(x) \(⋮\) x + 1 \(\Leftrightarrow\) F(x) \(⋮\) x - (-1)
Theo bezout ta có : F(x) \(⋮\) x - (-1) \(\Leftrightarrow\) F(-1) = 0
\(\Leftrightarrow\) 3(-1)2 + 2(-1)2 - 7.(-1) - m + 2 = 0
3 + 2 + 7 - m + 2 =0
14 - m = 0
m = 14
Kết luận với m = 14 thì F(x) chia hết cho x + 1
\(x+3-2\left(x-1\right)=3\left(2+x\right)-1\)
\(x+3-2x+2=6+3x-1\)
\(x-2x-3x=6-1-3-2\)
\(-4x=0\)
\(x=0\)
cảm ơn b