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31 tháng 3 2018
\(A=\left(\frac{3-x}{x+3}\times\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\) \(\left(ĐKXĐ:x\ne\pm3\right)\)
\(A=\left(\frac{3-x}{x+3}\times\frac{x+3}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{3-x}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left[\frac{\left(3-x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right]:\frac{3x^2}{x+3}\)
\(A=\left(\frac{9-3x}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}\times\frac{x+3}{3x^2}\)
\(A=\frac{-1}{x^2}\)
31 tháng 3 2018
Ta có :\(x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(L\right)\\x=2\left(tm\right)\end{cases}}\)
\(\Rightarrow A=\frac{-1}{2^2}\)
\(A=\frac{-1}{4}\)
TA
19 tháng 1 2018
a/\(\dfrac{8}{x-8}+1+\dfrac{11}{x-11}+1=\dfrac{9}{x-9}+1+\dfrac{10}{x-10}+1\)
=>\(\dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10}\)
=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)
=>x.\(\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}+\dfrac{1}{x-9}+\dfrac{1}{x-10}\right)=0\)
=>x=0
b/\(\dfrac{x}{x-3}-1+\dfrac{x}{x-5}-1=\dfrac{x}{x-4}-1+\dfrac{x}{x-6}-1\)
=>\(\dfrac{x-x+3}{x-3}+\dfrac{x-x+5}{x-5}-\dfrac{x-x+4}{x-4}-\dfrac{x-6+6}{x-6}=0\)
=>\(\dfrac{3}{x-3}+\dfrac{5}{x-5}-\dfrac{4}{x-4}-\dfrac{6}{x-6}=0\)
Đến đây thì bạn giải giống câu a
25 tháng 8 2017
a)
\(4x-10< 0\\ 4x< 10\\ x< \dfrac{10}{4}=\dfrac{5}{2}\)
b)
\(2x+x+12\ge0\\ 3x\ge-12\\ x\ge-\dfrac{12}{3}=-4\)
c)
\(x-5\ge3-x\\ 2x\ge8\\ x\ge4\)
d)
\(7-3x>9-x\\ -2>2x\\ x< -1\)
đ)
\(2x-\left(3-5x\right)\le4\left(x+3\right)\\ 2x-3+5x\le4x+12\\ 3x\le15\\ x\le5\)
e)
\(3x-6+x< 9-x\\ 5x< 15\\ x< 3\)
f)
\(2t-3+5t\ge4t+12\\ 3t\ge15\\ t\ge5\)
g)
\(3y-2\le2y-3\\ y\le-1\)
h)
\(3-4x+24+6x\ge x+27+3x\\ 0\ge2x\\ 0\ge x\)
i)
\(5-\left(6-x\right)\le4\left(3-2x\right)\\ 5-6+x\le12-8x\\ \\ 9x\le13\\ x\le\dfrac{13}{9}\)
k)
\(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\\ 10x-15-20x+28\ge19-2x-22\\ 13-10x\ge-2x-3\\ -8x\ge-16\\ x\le\dfrac{-16}{-8}=2\)
l)
\(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\\ \dfrac{40x-100}{60}-\dfrac{90x-30}{2}< \dfrac{36-12x}{60}-\dfrac{30x-15}{60}\\ \Rightarrow40x-100-90x+30< 36-12x-30x+15\\ 130-50x< 51-42x\\ 92x< -79\\ x< -\dfrac{79}{92}\)
m)
\(5x-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+x\\ \dfrac{10x}{2}-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+\dfrac{2x}{2}\\ \Rightarrow10x-3+2x>7x-5+2x\\ 12x-3>9x-5\\ 3x>-2\\ x>-\dfrac{2}{3}\)
n)
\(\dfrac{7x-2}{3}-2x< 5-\dfrac{x-2}{4}\\ \dfrac{28x-8}{12}-\dfrac{24x}{12}< \dfrac{60}{12}-\dfrac{3x-6}{12}\\ \Rightarrow28x-8-24x< 60-3x+6\\ 4x-8< -3x+66\\ 7x< 74\\ x< \dfrac{74}{7}\)
Q
25 tháng 8 2017
a) \(4x-10< 0\)
\(\Leftrightarrow4x< 10\)
\(\Leftrightarrow x< \dfrac{5}{2}\)
b) ???
c) \(x-5\ge3-x\)
\(\Leftrightarrow2x-5\ge3\)
\(\Leftrightarrow2x\ge8\)
\(\Leftrightarrow x\ge4\)
d) \(7-3x>9-x\)
\(\Leftrightarrow7-2x>9\)
\(\Leftrightarrow-2x>2\)
\(\Leftrightarrow x< -1\)
đ) ???
e) \(3x-6+x< 9-x\)
\(\Leftrightarrow4x-6< 9-x\)
\(\Leftrightarrow5x-6< 9\)
\(\Leftrightarrow5x< 15\)
\(\Leftrightarrow x< 3\)
f) ???
g) ???
h) \(3-4x+24+6x\ge x+27+3x\)
\(\Leftrightarrow2x+27\ge4x+27\)
\(\Leftrightarrow-2x\ge0\)
\(\Leftrightarrow x\le0\)
i) \(5-\left(6-x\right)\le4\left(3-2x\right)\)
\(\Leftrightarrow5-6+x\le12-8x\)
\(\Leftrightarrow x-1\le12-8x\)
\(\Leftrightarrow9x-1\le12\)
\(\Leftrightarrow9x\le13\)
\(\Leftrightarrow x\le\dfrac{13}{9}\)
k) \(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\)
\(\Leftrightarrow10x-15-20x+28\ge19-2x-22\)
\(\Leftrightarrow-10x+23\ge-3-2x\)
\(\Leftrightarrow-8x+13\ge-3\)
\(\Leftrightarrow-8x\ge-16\)
\(\Leftrightarrow x\ge2\)
l) \(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\)
\(\Leftrightarrow-\dfrac{5}{6}x-\dfrac{7}{6}< -\dfrac{7}{10}x+\dfrac{17}{20}\)
\(\Leftrightarrow-\dfrac{2}{15}x-\dfrac{7}{6}< \dfrac{17}{20}\)
\(\Leftrightarrow-\dfrac{2}{15}x< \dfrac{121}{60}\)
\(\Leftrightarrow x>-\dfrac{121}{8}\)
m, n) làm tương tự:
đáp án: m. \(x>-\dfrac{2}{3}\); n. \(x< \dfrac{74}{7}\)
CN
22 tháng 12 2017
a) \(x^3-\dfrac{1}{4}x=0\)
⇔ \(x.\left(x^2-\dfrac{1}{4}\right)=0\)
⇔ \(x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
⇔ x = 0 hoặc \(x=\dfrac{1}{2}\) hoặc \(x=\dfrac{-1}{2}\)
b) (2x - 1)2 - (x + 3)2 = 0
⇔ (2x - 1 - x - 3)(2x - 1 + x + 3) = 0
⇔ (x - 4)(3x +2) = 0
⇔ x = 4 hoặc \(x=\dfrac{-2}{3}\)
c) 2x2 - x - 6 = 0
⇔ 2x2 - 4x + 3x - 6 = 0
⇔ 2x(x - 2) + 3(x - 2) = 0
⇔ (x - 2) (2x + 3) = 0
⇔ x = 2 hoặc \(x=\dfrac{-3}{2}\)
NT
22 tháng 12 2017
2)a.
\(B=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}\\ =\left(\dfrac{x\left(x^2+6x\right)-\left(x-6\right)\left(x^2-36\right)}{\left(x^2-36\right)\left(x^2+6x\right)}\right).\dfrac{x^2+6x}{2x-6}\\ =\dfrac{x^2\left(x+6\right)-\left(x-6\right)^2.\left(x+6\right)}{x^2-36}.\dfrac{1}{2x-6}\\ =\dfrac{\left(x+6\right)\left(x^2-\left(x-6\right)^2\right)}{x^2-36}.\dfrac{1}{2x-6}\\ =\dfrac{\left(x-x+6\right)\left(x+x-6\right)}{x-6}.\dfrac{1}{2x-6}\\ =\dfrac{6.\left(2x-6\right)}{x-6}.\dfrac{1}{2x-6}\\ =\dfrac{6}{x-6}\)
b)
\(x=2\Leftrightarrow B=\dfrac{6}{x-6}=\dfrac{6}{2-6}=\dfrac{6}{-4}=-\dfrac{3}{2}\)
MN
1
4 tháng 12 2017
Ta có : x^2+y^2/xy=12/25
=>12(x^2+y^2)=25xy
=>12(x^2+2xy+y^2)=49xy
=>12(x+y)^2=49xy
=>(x+y)^2=49xy/12 (1)
Ta có : x^2+y^2/xy=12/25
=>12(x^2+y^2)=25xy
=>12(x^2-2xy+y^2)=xy
=>12(x-y)^2=xy
=>(x-y)^2=xy/12 (2)
Từ (1) và (2) suy ra :
(x-y)^2/(x+y)^2=1/49
Vì x<y<0 nên x-y/x=y=-1/7
Tick cho mik nhé 
KH
10 tháng 12 2018
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
10 tháng 12 2018
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
\(\dfrac{x^2+4+6}{5-6x}< 0\)
Nhận xét:
\(x^2\ge0,\forall x\)
\(\Rightarrow x^2+4+6\ge10,\forall x\)
Do đó \(\dfrac{x^2+4+6}{5-6x}< 0\) khi và chỉ khi:
\(5-6x< 0\)
\(\Leftrightarrow6x>5\)
\(\Leftrightarrow x>\dfrac{5}{6}\)
Vậy \(x>\dfrac{5}{6}\)
Sửa đề:
(x + 2)² + 2 > 0
5 - 6x < 0
-6x < -5