Tìm x để (x+5)(3x-12)>0
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DA
2
19 tháng 10 2017
\(a,x^3+3x^2=4x+12\)
\(x^2\left(x+3\right)=4\left(x+3\right)\)
\(\Rightarrow\left(x+3\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm2\end{cases}}\)
\(b,49x^2=\left(3x+2\right)^2\)
\(7x=3x+2\)
\(\Rightarrow7x-3x=2\)
\(\Rightarrow4x=2\)
\(\Rightarrow x=\frac{1}{2}\)
các câu còn lại tương tự nha
19 tháng 10 2017
\(a,x^3+3x^2=4x+12\)
\(x^3+3x^2-4x-12=0\)
\(\Rightarrow x^2\left(x+3\right)-4\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\\left(x+2\right)\left(x-2\right)=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm2\end{cases}}\)
\(b,49x^2=\left(3x+2\right)^2\)
\(\Rightarrow\left(7x\right)^2=\left(3x+2\right)^2\)
\(\Rightarrow7x=3x+2\)
\(\Rightarrow7x-3x=2\)
\(\Rightarrow4x=2\)
\(\Rightarrow x=\frac{1}{2}\)
\(c,3x^2\left(x-5\right)+12\left(5-x\right)=0\)
\(3x^2\left(x-5\right)-12\left(x-5\right)=0\)
\(\left(x-5\right)\left(3x^2-12\right)=0\)
\(\Rightarrow3.\left(x-5\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\pm2\end{cases}}}\)
\(d,x^2\left(x-5\right)+45-9x=0\)
\(x^2\left(x-5\right)+9\left(5-x\right)=0\)
\(x^2\left(x-5\right)-9\left(x-5\right)=0\)
\(\left(x-5\right)\left(x^2-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2-9=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=\pm3\end{cases}}\)
TT
2
2T
17 tháng 8 2019
\(x\left(3x-5\right)-9x+15=0\)
\(\Leftrightarrow x\left(3x-5\right)-3\left(3x-5\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\3x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{5}{3}\end{cases}}\)
2T
17 tháng 8 2019
\(3x\left(x-5\right)-2\left(5-x\right)=0\)
\(\Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+2=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=5\end{cases}}\)
14 tháng 4 2019
b
\(\left|6+x\right|\ge0;\left(3+y\right)^2\ge0\Rightarrow\left|6+x\right|+\left(3+y\right)^2\ge0\)
Suy ra \(\left|6+x\right|+\left(3+y\right)^2=0\)\(\Leftrightarrow\hept{\begin{cases}6+x=0\\3+y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-6\\y=-3\end{cases}}\)
14 tháng 4 2019
a
Ta có:\(\left|3x-12\right|=3x-12\Leftrightarrow3x-12\ge0\Leftrightarrow3x\ge12\Leftrightarrow x\ge4\)
\(\left|3x-12\right|=12-3x\Leftrightarrow3x-12< 0\Leftrightarrow3x< 12\Leftrightarrow x< 4\)
Với \(x\ge4\) ta có:
\(3x-12+4x=2x-2\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\left(KTMĐK\right)\)
Với \(x< 4\) ta có:
\(12-3x+4x=2x-2\)
\(\Rightarrow10=x\left(KTMĐK\right)\)
11 tháng 3 2020
ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)
\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left(\frac{2}{x-2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)
Đề sai à ??
K
2
B
8 tháng 12 2016
Sửa: a)\(3x^2-12=0\)
\(\Rightarrow3x^2=12\)
\(\Rightarrow x^2=\frac{12}{3}=4\)
\(\Rightarrow x=\sqrt{4}=2\)
Vậy: x=2
b)\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+5=0\\2-x=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=-5\\x=2\end{array}\right.\)
Vậy: \(x=-5;2\)
c)\(\Rightarrow2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-1=0\\2x+5=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=1\\x=-\frac{5}{2}\end{array}\right.\)
Vậy: \(x=1;-\frac{5}{2}\)
x lớn hơn hoặc bằng 5