tìm GTNN
A=\(\dfrac{3x^2-2x+3}{x^2+1}\)
B=\(\dfrac{x^2-2011}{4\left(x^2+1\right)}\)
C=\(\dfrac{3x^2-8x+13}{x^2+5}\)
ai giúp mk với
tìm GTNN
A=\(\dfrac{3x^2-2x+3}{x^2+1}\)
B=\(\dfrac{x^2-2011}{4\left(x^2+1\right)}\)
C=\(\dfrac{3x^2-8x+13}{x^2+5}\)
C=\(\dfrac{3x^2-2x+3}{x^2+1}\)
E=\(\dfrac{3x^2-8x+13}{x^2+5}\)
D=\(\dfrac{x^2-2011}{4\left(x^2+1\right)}\)
Giải các phương trình sau :
a)\(\dfrac{5x+2}{6}\)\(-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
b)\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
c)\(2x^3 +6x^2=x^2+3x\)
d)\(\left|x-4\right|+3x=5\)
`a,` \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
`<=> (5(5x+2))/30 - (10(8x-1))/30 = (6(4x+2))/30 - (5.30)/30`
`<=> 5(5x+2) - 10(8x-1) =6(4x+2) - 5.30`
`<=> 25x + 10 - 80x + 10 = 24x+12 - 150`
`<=> -55x +20 = 24x-138`
`<=> -55x -24x=-138-20`
`<=>-79x=-158`
`<=> x=2`
Vậy pt có nghiệm `x=2`
`b,` \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)
Ta có : `(x+2)/(x-2) -1/x = 2/(x(x-2))`
`<=> (x(x+2))/(x(x-2)) - (x-2)/(x(x-2)) = 2/(x(x-2))`
`=> x^2 +2x - x +2 = 2`
`<=> x^2 + x =0`
`<=>x(x+1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-1\end{matrix}\right.\)
Vậy pt có nghiệm `x=-1`
`c,2x^3 + 6x^2 =x^2 +3x`
`<=> 2x^3 + 6x^2 -x^2 -3x=0`
`<=> 2x^3 + 5x^2 -3x=0`
`->` Đề có sai ko ạ ?
`d,` \(\left|x-4\right|+3x=5\) `(1)`
Thường hợp `1` : `x-4 >= 0<=> x >=0` thì phương trình `(1)` thở thành :
`x-4 = 5-3x`
`<=> x+3x=5+4`
`<=> 4x=9`
`<=> x= 9/4 (t//m)`
Trường hợp `2` : `x-4< 0<=> x<0` thì phương trình `(1)` trở thành :
`-(x-4) =5-3x`
`<=> -x +4=5-3x`
`<=> -x+3x=5-4`
`<=> 2x =1`
`<=>x=1/2 ( kt//m)`
Vậy phương trình có nghiệm `x=9/4`
đây là phương trình mà đâu phải bất phương trình đâu
Tìm GTNN:
a) \(\dfrac{1}{-x^2+2x-4}\)
b) \(\dfrac{12}{12x-4x^2-13}\)
c) \(\dfrac{x^2-4x-4}{x^2-4x+5}\)
d) \(\dfrac{15}{-6x^2-5y^2+10xy-4x+10y-19}\)
e)\(\dfrac{x^2-2011}{4.\left(x^2+1\right)}\)
\(b,\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=-\left(\dfrac{7}{5}+\dfrac{7}{10}.x\right)\)
\(c,\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
\(d,\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\left(\dfrac{7}{x}-2\right)\)
\(e,2\left(x-1\right)=\left(x-1\right)^2\)
\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Giải các phương trình sau:
a) \(\dfrac{15-x}{2000}+\dfrac{14-x}{2001}=\dfrac{13-x}{2002}+\dfrac{12-x}{2003}\)
b) \(\dfrac{x-5}{2010}+\dfrac{x-4}{2011}=\dfrac{x-2010}{5}+\dfrac{x-2011}{4}\)
c) \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
giúp mk vs
a) \(\dfrac{15-x}{2000}+\dfrac{14-x}{2001}=\dfrac{13-x}{2002}+\dfrac{12-x}{2003}\)
\(\Leftrightarrow\dfrac{15-x}{2000}+1+\dfrac{14-x}{2001}+1=\dfrac{13-x}{2002}+1+\dfrac{12-x}{2003}+1\)
\(\Leftrightarrow\dfrac{2015-x}{2000}+\dfrac{2015-x}{2001}=\dfrac{2015-x}{2002}+\dfrac{2015-x}{2003}\)
\(\Rightarrow\dfrac{2015-x}{2000}+\dfrac{2015-x}{2001}-\dfrac{2015-x}{2002}-\dfrac{2015-x}{2003}=0\)
\(\Leftrightarrow\left(2015-x\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow2015-x=0\)
<=> x=2015
Vậy phương trình có nghiệm là x=2015
b) \(\dfrac{x-5}{2010}+\dfrac{x-4}{2011}=\dfrac{x-2010}{5}+\dfrac{x-2011}{4}\)
\(\Leftrightarrow\dfrac{x-5}{2010}-1+\dfrac{x-4}{2011}-1=\dfrac{x-2010}{5}-1+\dfrac{x-2011}{4}-1\)
\(\Leftrightarrow\dfrac{x-2015}{2010}+\dfrac{x-2015}{2011}=\dfrac{x-2015}{5}+\dfrac{x-2015}{4}\)
\(\Rightarrow\dfrac{x-2015}{2010}+\dfrac{x-2015}{2011}-\dfrac{x-2015}{5}-\dfrac{x-2015}{4}=0\)
\(\Leftrightarrow\left(x-2015\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x-2015=0\)
=> x=2015
Vậy phương trình có nghiệm x=2015
Giải các phương trình sau:
\(a.\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(b.\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(c.2\left(x+\dfrac{3}{5}\right)=5-\left(\dfrac{13}{5}+x\right)\)
\(d.\dfrac{7}{8}x-5\left(x-9\right)=\dfrac{20x+1,5}{6}\)
\(e.\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)
\(f.\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
Cần giúp nhanh vs
Bài 1. Tìm x
a) \(\left|x+\dfrac{7}{4}\right|=\dfrac{1}{2}\)
b) \(\left|2x+1\right|-\dfrac{2}{5}=\dfrac{1}{3}\)
c) \(3x.\left(x+\dfrac{2}{3}\right)=0\)
d) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)\)
Bài 2. Tính nhanh
\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
Bài 1:
a.
$|x+\frac{7}{4}|=\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)
b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$
$|2x+1|=\frac{11}{15}$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)
c.
$3x(x+\frac{2}{3})=0$
\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)
d.
$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$
$\Leftrightarrow x=\frac{2}{5}$
Bài 2:
$\frac{1}{100}-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}$
$=\frac{99}{100}$
$\Rightarrow A=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=\frac{-49}{50}$
Bài 1:
a) Ta có: \(\left|x+\dfrac{7}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{7}{4}=\dfrac{1}{2}\\x+\dfrac{7}{4}=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)
b) Ta có: \(\left|2x+1\right|-\dfrac{2}{5}=\dfrac{1}{3}\)
\(\Leftrightarrow\left|2x+1\right|=\dfrac{11}{15}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{11}{15}\\2x+1=\dfrac{-11}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-4}{15}\\2x=\dfrac{-26}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{15}\\x=\dfrac{-13}{15}\end{matrix}\right.\)
c) Ta có: \(3x\left(x+\dfrac{2}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)
giải phương trình
a.\(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
b.\(x\left(2x-9\right)=3x\left(x-5\right)\)
c.\(3x-15=2x\left(x-5\right)\)
d.\(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
e.\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)