2x(3x+2)-3x(2x+3)
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NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2021
a.
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos2x=\dfrac{1}{2}-\dfrac{1}{2}cos6x\)
\(\Leftrightarrow cos2x=cos6x\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=2x+k2\pi\\6x=-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=k2\pi\\8x=k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{k\pi}{4}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2021
b.
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos2x+\dfrac{1}{2}-\dfrac{1}{2}cos4x+\dfrac{1}{2}-\dfrac{1}{2}cos6x=\dfrac{3}{2}\)
\(\Leftrightarrow cos2x+cos6x+cos4x=0\)
\(\Leftrightarrow2cos4x.cos2x+cos4x=0\)
\(\Leftrightarrow cos4x\left(2cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\2x=\dfrac{2\pi}{3}+k2\pi\\2x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{3}+k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
KT
20 tháng 8 2018
d) \(2x^3+3x^2+3x+1=2x^3+x^2+2x^2+x+2x+1\)
\(=x^2\left(2x+1\right)+x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(2x^3-5x^2+5x-3=2x^3-3x^2-2x^2+3x+2x-3\)
\(=x^2\left(2x-3\right)-x\left(2x-3\right)+\left(2x-3\right)=\left(2x-3\right)\left(x^2-x+1\right)\)
H9
HT.Phong (9A5)
CTVHS
16 tháng 6 2023
a) Ta có: \(\sqrt{2x-1}\)
Biểu thức này có nghĩa là: \(2x-1\ge0\Leftrightarrow x\ge\dfrac{1}{2}\)
b) Ta có: \(\sqrt{4-x}\)
Biểu thức này có nghĩa là: \(4-x\ge0\Leftrightarrow x\le4\)
c) Ta có: \(\sqrt{\dfrac{3x}{2}}\)
Biểu thức này có nghĩa là: \(\dfrac{3x}{2}\ge0\Leftrightarrow x\ge0\)
d) Ta có: \(\sqrt{2x^2}\)
Biểu thức có nghĩa là: \(2x^2\ge0\Leftrightarrow x^2\ge0\) với mọi x
6 tháng 3 2022
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
KT
20 tháng 8 2018
c) \(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
d) \(2x^3+3x^2+3x+1=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(2x^3-5x^2+5x-3=\left(2x-3\right)\left(x^2-x+1\right)\)
10 tháng 7 2023
a: P(x)=2x^5-2x^5+4x^4-3x^4+5=x^4+5
Q(x)=-5x^4+2x^4-x^3+3x^2-10x+2
=-3x^4-x^3+3x^2-10x+2
b: P(x)+Q(x)
=x^4+5-3x^4-x^3+3x^2-10x+2
=-2x^4-x^3+3x^2-10x+7
Q(x)-P(x)
=-3x^4-x^3+3x^2-10x+2-x^4-5
=-4x^4-x^3+3x^2-10x-3
P(x)-Q(x)=-(Q(x)-P(x))
=4x^4+x^3-3x^2+10x+3
KN
11 tháng 9 2019
\(\left(-3x-2\right)^2+\left(3x+5\right)\left(5-3x\right)=-7\)
\(\Leftrightarrow9x^2+12x+4+15x-9x^2+25-15x=-7\)
\(\Leftrightarrow12x+36=0\Leftrightarrow x=-3\)
KN
11 tháng 9 2019
\(\left(x+2\right)\left(x^2+2x+2\right)-x\left(x-8\right)^2=\left(4x-3\right)\left(4x+3\right)\)
\(\Leftrightarrow x^3+2x^2+2x+2x^2+4x+4-x\left(x^2-16x+64\right)=16x^2-9\)
\(\Leftrightarrow x^3+4x^2+6x+4-x^3+16x^2-64=16x^2-9\)
\(\Leftrightarrow4x^2+6x-51=0\)
\(\cdot\Delta=6^2-4.4.\left(-51\right)=852\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-6+\sqrt{852}}{8}\);\(x_2=\frac{-6-\sqrt{852}}{8}\)
= 6x2+4x2- 6x2-9x
= 6x2 (4x2-9x)