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\(3x^3-48x=0\)
\(3x\cdot\left(x^2-16\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\left\{\pm4\right\}\end{cases}}\)
Vậy,............
\(-22x^3-\left(-21x^3+19x^2+23^0\right)-\left(-x^3-18x^2\right)+\left(x^2-23^1\right)\)
\(=-22x^3+21x^3-19x^2-1+x^3+18x^2+x^2-23\)
\(=\left(-22x^3+21x^3+x^3\right)+\left(-19x^2+18x^2+x^2\right)+\left(-1-23\right)\)
\(=0x^3+0x^2-24\)
\(=-24\)
Vậy biểu thức trên có giá trị không phụ thuộc vào biến.
a)x=-2
b)x=1
c)x=1/2
f)x=1 hoặc x=-1
h)x=0 hoặc x=6
i)x=2
hok tốt!
_Lan Lan_
Áp dụng hằng đẳng thức:\(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)
\(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\)
Áp dụng vào từng bài là được:
\(VD1:x^3+3x^2+3x+1=-1\)
\(\Rightarrow\left(x+1\right)^3=-1\)
\(\Rightarrow x=-2\)
\(VD2:x^3-9x^2+27x-27=-8\)
\(\Rightarrow\left(x-3\right)^3=-8\)
\(\Rightarrow x=1\)
a) \(2x^2+x-6=0\Leftrightarrow2x^2+4x-3x-6=0\)
\(\Leftrightarrow2x\left(x+2\right)-3\left(x+2\right)=0\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=3\\x=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
vậy \(x=\dfrac{3}{2};x=-2\)
b) \(-5x^2+17x-6=0\Leftrightarrow-5x^2+15x+2x-6=0\)
\(\Leftrightarrow-5x\left(x-3\right)+2\left(x-3\right)\Leftrightarrow\left(-5x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}-5x+2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\)
vậy \(x=\dfrac{2}{5};x=3\)
c) \(3x^2+22x-16=0\Leftrightarrow3x^2+24x-2x-16=0\)
\(\Leftrightarrow3x\left(x+8\right)-2\left(x+8\right)=0\Leftrightarrow\left(3x-2\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x=2\\x=-8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)
vậy \(x=\dfrac{2}{3};x=-8\)
d) \(2x^3+3x^2-8x+3=0\Leftrightarrow2x^3-3x^2+x+6x^2-9x+3=0\)
\(\Leftrightarrow x\left(2x^2-3x+1\right)+3\left(2x^2-3x+1\right)=0\Leftrightarrow\left(x+3\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x^2-2x-x+1\right)=0\Leftrightarrow\left(x+3\right)\left(2x\left(x-1\right)-\left(x-1\right)\right)=0\)
\(\left(x+3\right)\left(2x-1\right)\left(x-1\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\2x-1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\2x=1\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\) vậy \(x=-3;x=\dfrac{1}{2};x=1\)
(22x+3)2=121
<=>(22x+3)2=112
<=>22x+3=11
<=>22x=11-3
<=>22x=8
<=>x=8:22
<=>x=\(\frac{4}{11}\)
vậy x=\(\frac{4}{11}\)
k mk nha nếu thấy đúng
3 − 2 2 x − 3 = 2 5 + 2 9 − 6 x − 3 2 3 − 2 2 x − 3 = 2 5 − 2 6 x − 9 − 3 2 3 − 2 2 x − 3 = 2 5 + 2 3 2 x − 3 − 3 2 2 3 2 x − 3 − 2 2 x − 3 = 2 5 − 3 2 − 3 2 − 6 3 2 x − 3 = 4 − 15 − 30 10 − 4 3 2 x − 3 = − 41 10 4 3 2 x − 3 = 41 10 4.10 = 41.3. 2 x − 3 40 = 123. 2 x − 3 2 x − 3 = 40 123 2 x = 40 123 + 3 2 x = 40 + 369 123 2 x = 409 123 x = 409 246