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13 tháng 5 2018
a)<=>
A,=(x+y)(x-y)=x^2-y^2
x=(-1/2)^5:(1/2)^4=-1/2
x^2=1/4
y=8^2/(-2)^5=-2
y^2=4
A=1/4-4=-15/4
10 tháng 9 2020
a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)
\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)
b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)
\(\Leftrightarrow x=\frac{-3}{2}\)
c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)
\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)
d) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x+1\right)\left(x^2-x+1\right)-3\left(-x-2\right)=5\)
\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)
10 tháng 7 2022
1: \(=-\left(x^2+2x+2\right)=-\left(x^2+2x+1+1\right)=-\left(x+1\right)^2-1< =-1\)
Dấu '=' xảy ra khi x=-1
2: \(=-\left(4x^2-12x-10\right)\)
\(=-\left(4x^2-12x+9-19\right)\)
\(=-\left(2x-3\right)^2+19< =19\)
Dấu '=' xảy ra khi x=3/2
3: \(=-\left(x^2+4x+4-4\right)=-\left(x+2\right)^2+4< =4\)
Dấu '=' xảy ra khi x=-2
YN
4 tháng 10 2021
1, \(3x\left(x-7\right)+2x-14=0\)
\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)
2, \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)
3, \(15x-5+6x^2-2x=0\)
\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)
\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)
4, \(5x-2-25x^2+10x=0\)
\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)
\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)
\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)
6 tháng 8 2022
b \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
hay \(x\in\left\{0;2\right\}\)
c: \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
=>(x-8)(3x+2)=0
=>x=8 hoặc x=-2/3
d: \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
=>x=2 hoặc x=1
e: \(\Leftrightarrow x\left(x^2-11x+30\right)=0\)
=>x(x-5)(x-6)=0
hay \(x\in\left\{0;5;6\right\}\)
6 tháng 8 2022
b: \(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
hay \(x\in\left\{0;2\right\}\)
c: \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
=>(x-8)(3x+2)=0
hay \(x\in\left\{8;-\dfrac{2}{3}\right\}\)
d: \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
=>x=1 hoặc x=2
a) Ta có: \(\frac{x+2}{2}-\frac{2x-3}{5}=\frac{10x+13}{10}\)
\(\Leftrightarrow\frac{5\left(x+2\right)}{10}-\frac{2\left(2x-3\right)}{10}-\frac{10x+13}{10}=0\)
Suy ra: \(5x+10-4x+6-10x-13=0\)
\(\Leftrightarrow-9x+3=0\)
\(\Leftrightarrow-9x=-3\)
hay \(x=\frac{1}{3}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{3}\right\}\)
b) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\frac{x-1}{x-2}-\frac{5}{x+2}=\frac{x^2}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x^2}{\left(x+2\right)\left(x-2\right)}=0\)
Suy ra: \(x^2+x-2-5x+10-x^2=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Leftrightarrow-4x=-8\)
hay x=2(ktm)
Vậy: Tập nghiệm \(S=\varnothing\)