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27 tháng 7 2017
x2 - 4 = 0
x2 = 4
\(\orbr{\begin{cases}x^2=2^2\\x^2=\left(-2\right)^2\end{cases}}\)
\(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
3x2 - 75 = 0
3x2 = 75
x2 = 25
\(\orbr{\begin{cases}x^2=5^2\\x^2=\left(-5\right)^2\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
( x + 2 )2 = 25
\(\orbr{\begin{cases}\left(x+2\right)^2=5^2\\\left(x+2\right)^2=\left(-5\right)^2\end{cases}}\)
\(\orbr{\begin{cases}x+2=5\\x+2=-5\end{cases}}\)
\(\orbr{\begin{cases}x=3\\x=-7\end{cases}}\)
6 tháng 8 2016
\(x^4+1\)
\(=x^4+2x^2+1-2x^2\)
\(=\left(x^2+1\right)^2-2x^2\)
\(=\left(x^2+1\right)^2-\left(\sqrt{2}x\right)^2\)
\(=\left(x^2+1-\sqrt{2}x\right)\left(x^2+1+\sqrt{2}x\right)\)
28 tháng 9 2019
a) \(\left(x^2+2x-2\right)\left(x^2+2x+3\right)=6\)
Đặt \(x^2+2x=a\)
\(pt\Leftrightarrow\left(a-2\right)\left(a+3\right)=6\)
\(\Leftrightarrow a^2+a-6=6\)
\(\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow a^2+3a-4a-12=0\)
\(\Leftrightarrow a\left(a+3\right)-4\left(a+3\right)=0\)
\(\Leftrightarrow\left(a-4\right)\left(a+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a-4=0\\a+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=4\\a=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+2x=4\\x^2+2x=-3\end{cases}}\)
\(Th1:x^2+2x=4\Leftrightarrow x^2+2x-4=0\)
\(\cdot\Delta=2+4.4=18\)
pt có 2 nghiệm \(x_1=\frac{-2+\sqrt{18}}{2}\);\(x_2=\frac{-2-\sqrt{18}}{2}\)
\(Th1:x^2+2x=-3\Leftrightarrow x^2+2x+3=0\)
\(\cdot\Delta=2-4.3=-10< 0\)
Vậy pt này không có nghiệm
Vậy \(x_1=\frac{-2+\sqrt{18}}{2}\);\(x_2=\frac{-2-\sqrt{18}}{2}\)
28 tháng 9 2019
b) \(\left(x^2-4x+6\right)\left(x^2-4x+8\right)=8\)
Đặt \(x^2-4x=t\)
\(pt\Leftrightarrow\left(t+6\right)\left(t+8\right)=8\)
\(\Leftrightarrow t^2+14x+48=8\)
\(\Leftrightarrow t^2+14x+40=0\)
\(\Delta=14^2-4.40=36,\sqrt{\Delta}=6\)
pt có 2 nghiệm: \(t_1=\frac{-14+6}{2}=-4\);\(t_2=\frac{-14-6}{2}=-10\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4x=-4\\x^2-4x=-10\end{cases}}\)
\(TH1:x^2-4x=-4\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
\(TH2:x^2-4x=-10\Leftrightarrow x^2-4x+10=0\)
\(\Delta=\left(-4\right)^2-4.10=-24< 0\)
Vậy pt này không có nghiệm
Vậy x = 2
YK
18 tháng 8 2019
1) Đặt \(x^2\)+2x=t
Ta có \(\left(x^2+2x-2\right)\left(x^2+2x+3\right)-6\)=(t-2)(t+3)-6=\(t^2+t-6-6\)\(=t^2+t-12\)\(\left(t-3\right)\left(t+4\right)\)\(=\left(x^2+2x-3\right)\left(x^2+2x+4\right)=\left(x+3\right)\left(x-1\right)\left(x^2+2x+4\right)\)
2) Đặt \(x^2-4x+7=t\)
Ta có : (\(\left(x^2-4x+6\right)\left(x^2-4x+8\right)\)\(-8\)\(=\left(t-1\right)\left(t+1\right)-8=t^2-1-8=t^2-9=\left(t-3\right)\left(t+3\right)\)\(=\left(x^2-4x+4\right)\left(x^2-4x+10\right)\)\(=\left(x-2\right)^2\left(x^2-4x+10\right)\)
31 tháng 12 2019
a) (2x - 1)(3x + 5) - 2(-4x + 1)2 = 6x2 + 10x - 3x - 5 - 2(16x2 - 8x + 1) = 6x2 - 3x - 5 - 32x2 + 16x - 2 = -26x2 + 13x - 7
b) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=-\frac{x+4}{x}\)
c) \(\frac{2x-9}{x^2-5x+6}+\frac{2x+1}{x-3}+\frac{x+3}{2-x}\)
= \(\frac{2x-9}{x^2-2x-3x+6}+\frac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{2x-9+2x^2-3x-2-x^2+9}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{x^2-2x+x-2}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{\left(x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\frac{x+1}{x-3}\)
d) (x - 1)3 - (x + 1)3 + 6(x + 1)(x - 1)
= (x - 1 - x - 1)[(x - 1)2 + (x - 1)(x + 1) + (x + 1)2] + 6(x2 - 1)
= -2(x2 - 2x + 1 + x2 - 1 + x2 + 2x + 1) + 6x2 - 6
= -2(3x2 + 1) + 6x2 - 6
= -6x2 - 2 + 6x2 - 6
= -8
e) (2x + 7)2 - (4x + 14)(2x - 8) + (8 - 2x)2
= (2x + 7)2 - 2(2x + 7)(2x - 8) + (2x - 8)2
= (2x + 7 - 2x + 8)2
= 152 = 225
11 tháng 10 2020
a) \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{-2;3\right\}\)
b) \(\left(x+8\right)^2=121\)
\(\Leftrightarrow\left(x+8\right)^2-121=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+19\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-19\end{matrix}\right.\)
Vậy \(x\in\left\{-19;3\right\}\)
c) \(x^3-4x^2+4x=0\)
\(\Leftrightarrow x\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)^2=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{0;2\right\}\)
d) \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
14 tháng 6 2022
a: \(B=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{10}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)
b: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)
\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)
\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)
c: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=\dfrac{-5}{2}\)
d: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)
\(=\dfrac{1-4x^2}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x+4}\cdot\dfrac{3}{2\left(1-2x\right)}=\dfrac{3\left(2x+1\right)}{x+4}\)
\(=6-100x+8x^2-8x^2+600+x=-99x+606\)