(2x + 11)(x + 1)
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Tìm x :
a) | x + 12x | = 2x
=> \(\orbr{\begin{cases}13x=2x\\13x=-2x\end{cases}}\)
=> \(\orbr{\begin{cases}11x=0\\15x=0\end{cases}}\)
=> \(x=0\)
b) 3x − |x + 1| = 1
=> |x + 1| = 3x -1
=>\(\orbr{\begin{cases}x+1=3x-1\\x+1=1-3x\end{cases}}\)
=> \(\orbr{\begin{cases}2x=2\\4x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
c) |2x + 3| = x + 1
=> \(\orbr{\begin{cases}2x+3=x+1\\2x+3=-x-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\\3x=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}\)
b) 3x - |x + 1| = 1
<=> |x + 1| = 3x - 1 (1)
ĐK : \(x\ge\frac{1}{3}\)
Khi đó (1) <=> \(\orbr{\begin{cases}x+1=3x-1\\x+1=-3x+1\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=2\\4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(\text{loại}\right)\\x=1\end{cases}}\)
Vậy x = 1
c) ĐK : x + 1\(\ge0\Rightarrow x\ge-1\)
Khi đó |2x + 3| = x + 1
<=> \(\orbr{\begin{cases}2x+3=x+1\\2x+3=-x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}\left(\text{loại}\right)\)
Vậy \(x\in\varnothing\)
d) ||x + 9| + 11| = 2x + 11 (1)
ĐK : \(2x+11\ge0\Rightarrow x\ge-\frac{5}{2}\)
Khi đó (1) <=> \(\orbr{\begin{cases}\left|x+9\right|+11=2x+11\\\left|x+9\right|+11=-2x-11\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left|x+9\right|=2x\\\left|x+9\right|=-2x-22\end{cases}}\)
Khi |x + 9| = 2x (x \(\ge0\))
<=> \(\orbr{\begin{cases}x+9=2x\\x+9=-2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\left(tm\right)\\x=-3\left(\text{loại}\right)\end{cases}}\)
Khi |x + 9| = -2x - 22 ( \(-\frac{5}{2}\le x\le-11\))
<=> \(\orbr{\begin{cases}x+9=-2x-22\\x+9=2x+22\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{31}{3}\\x=-13\end{cases}}\left(\text{loại}\right)}\)
Vậy x = 9
Xét cấp số cộng 1, 6, 11, ..., 96.
Ta có: 96 = 1 + 5(n − 1) ⇒ n = 20
Suy ra
Và 2x.20 + 970 = 1010
Từ đó x = 1
\(4\left(x+1\right)\left(-x+2\right)+\left(2x-1\right)\left(2x+3\right)=-11\)
\(\text{⇔}-4x^2+4x+8+4x^2+4x-3=-11\)
\(\text{⇔}8x+5=-11\)
\(\text{⇔}8x=-16\)
\(\text{⇔}x=-2\)
Vậy: \(x=-2\)
==========
\(\left(2x+4\right)\left(3x+1\right)\left(x-2\right)-\left(-3x^2+1\right)\left(-2x+\dfrac{2}{3}\right)=-\dfrac{26}{3}\)
\(\text{⇔}6x^3+2x^2-24x-8-6x^3-2x^2-2x+\dfrac{2}{3}=-\dfrac{26}{3}\)
\(\text{⇔}-26x-\dfrac{22}{3}=-\dfrac{26}{3}\)
\(\text{⇔}-26x=-\dfrac{4}{3}\)
\(\text{⇔}x=\dfrac{2}{39}\)
b: =>15-x=-10
hay x=25
a: =>-2x+17=9
=>-2x=-8
hay x=4
d: \(\Leftrightarrow9x^2=81\)
hay \(x\in\left\{3;-3\right\}\)
e: \(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\3-x=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
1)
x^3 -16x=0`
`<=>x(x^2 -16)=0`
\(< =>\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
b)
`x^4 -2x^3=0`
`<=>x^3 (x-2)=0`
\(< =>\left[{}\begin{matrix}x^3=0\\x-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
3)
`(2x-11)(x^2 -1)=0`
\(< =>\left[{}\begin{matrix}2x-11=0\\x^2-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=11\\x^2=1\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=1\\x=-1\end{matrix}\right.\)
4)
`x^3 -36x=0`
`<=>x(x^2 -36)=0`
\(< =>\left[{}\begin{matrix}x=0\\x^2-36=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=36\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)
5)
`2x+19=0`
`<=>2x=-19`
`<=>x=-19/2`
e) \(\left(x+3\right)^3=\left(2x\right)^3\)
\(\Rightarrow x+3=2x\)
\(\Rightarrow2x-x=3\)
\(\Rightarrow x=3\)
f) \(\left(5-x\right)^5=32\)
\(\Rightarrow\left(5-x\right)^5=2^5\)
\(\Rightarrow5-x=2\)
\(\Rightarrow x=5-2\)
\(\Rightarrow x=3\)
g) \(\left(5x-6\right)^3=64\)
\(\Rightarrow\left(5x-6\right)^3=4^3\)
\(\Rightarrow5x-6=4\)
\(\Rightarrow5x=4+6\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=\dfrac{10}{2}\)
\(\Rightarrow x=5\)
h) \(5\cdot9^x=405\)
\(\Rightarrow9^x=\dfrac{405}{5}\)
\(\Rightarrow9^x=81\)
\(\Rightarrow9^x=9^2\)
\(\Rightarrow x=2\)
i) \(11^5:11^{n-2}=11^5\)
\(\Rightarrow11^{n-2}=11^5:11^5\)
\(\Rightarrow11^{n-2}=1\)
\(\Rightarrow11^{n-2}=11^0\)
\(\Rightarrow n-2=0\)
\(\Rightarrow n=2\)
k) \(\left(3x\right)^3=\left(2x+1\right)^3\)
\(\Rightarrow3x=2x+1\)
\(\Rightarrow3x-2x=1\)
\(\Rightarrow x=1\)
a) Tìm được x = 2,2
b) Tìm được x = 2073
c) Tìm được x = 4 hoặc x = -2
d) Điều kiện x≠-1 . Tìm được x = 0 hoặc x = 3
a, \(\left(4x-3\right)\left(x-5\right)-2x\left(2x-11\right)\)
\(=4x^2-20x-3x+15-4x^2+22x\)
\(=-x+15\)
(2x + 11) x (x +1) = mấy trời hay so sánh
Tìm x Î N, biết: