a)\(4x-12=0\)
b)\(x\left(x+1\right)-\left(x+2\right)\left(x-3\right)=7\)
c)\(\dfrac{x-3}{x+1}=\dfrac{x^2}{x^2-1}\)
a)\(4x-12=0\)
b)\(x\left(x+1\right)-\left(x+2\right)\left(x-3\right)=7\)
c)\(\dfrac{x-3}{x+1}=\dfrac{x^2}{x^2-1}\)
\(a,4x-12=0\\ \Leftrightarrow x=3\\ b,x\left(x+1\right)-\left(x+2\right)\left(x-3\right)=7\\ \Leftrightarrow\left(x^2+x\right)-\left(x^2-x-6\right)-7=0\\ \Leftrightarrow x^2+x-x^2+x+6-7=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{x-3}{x+1}=\dfrac{x^2}{x^2-1}\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-4x+3-x^2}{\left(x-1\right)\left(x+1\right)}=0\\ \Rightarrow-4x+3=0\\ \Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)
a, \(4x=12\Leftrightarrow x=3\)
b, \(x^2+x-x^2+x+6=7\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)
c, đk : x khác -1 ; 1
\(\Rightarrow\left(x-3\right)\left(x-1\right)=x^2\Leftrightarrow x^2-4x+3=x^2\Leftrightarrow-4x+3=0\Leftrightarrow x=\dfrac{3}{4}\)
a.
`4x-12=0`
`<=> 4x=12`
`<=> x= 3`
b.
`x(x+1)-(x+2)(x-3)=7`
`<=> (x^2+x)-(x^2-x-6)-7=0`
`<=> x^2+x+x^2+x+6-7=0`
`<=> 2x-1=0`
`<=> x= 1/2`
c.
`(x-3)/(x+1)= (x^2)/(x^2-1)` `(đkxđ:` \(\ne\) `+-1`
`<=> ((x-3)(x-1))/((x-1)(x+1)) - (x^2)/((x-1)(x+1)) = 0`
`<=> (x^2-4x+3-x^2)/((x-1)(x+1))=0`
`=> x^2-4x+3-x^2=0`
`<=> -4x+3=0`
`<=> x=3/4` `(tmđk)`
Tìm GTNN của các biểu thức sau:
a. A= 2a2 + 3ab + b22
b. x2 - 4x + y2 - 6y + 1
c. x2 - 4xy + 5y2 -2y + 5
a, xem lại đề
\(b,x^2-4x+y^2-6y+1\\ =\left(x^2-4x+4\right)+\left(y^2-6y+9\right)-12\\ =\left(x-2\right)^2+\left(y-3\right)^2-12\ge-12\)
Dấu "=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy ...
\(c,x^2-4xy+5y^2-2y+5\\ =\left(x^2-4xy+4y^2\right)+\left(y^2-2y+1\right)+4\\ =\left(x-2y\right)^2+\left(y-1\right)^2+4\ge4\)
Dấu "=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy ...
a,
b,x2−4x+y2−6y+1=(x2−4x+4)+(y2−6y+9)−12=(x−2)2+(y−3)2−12≥−12b,x2−4x+y2−6y+1=(x2−4x+4)+(y2−6y+9)−12=(x−2)2+(y−3)2−12≥−12
Dấu "=" xảy ra⇔{x=2y=3⇔{x=2y=3
Vậy ...
c,x2−4xy+5y2−2y+5=(x2−4xy+4y2)+(y2−2y+1)+4=(x−2y)2+(y−1)2+4≥4c,x2−4xy+5y2−2y+5=(x2−4xy+4y2)+(y2−2y+1)+4=(x−2y)2+(y−1)2+4≥4
Dấu "=" xảy ra⇔{x=2y=1⇔{x=2y=1
Vậy ...
giải phương trình:
a)2x(3x-1)=6x^2-13
b)x/3-2x+1/6=x/6-x
c)x+1/x-1-x-1/x+1=x^2+3/x^2-1
a: \(\Leftrightarrow6x^2-2x=6x^2-13\)
=>-2x=-13
hay x=13/2
b: \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-x\)
=>2x-2x-1=x-6x
=>-5x=-1
hay x=1/5
c: \(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=x^2+3\)
\(\Leftrightarrow x^2+3-x^2-2x-1+x^2-2x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
=>x=3
1.\(x^2+2\left(m-1\right)x+m-4=0\)
Khi nghiệm bằng 2 <=> x=2
Khi đó:
\(2^2+2\left(m-1\right).2+m-4=0\)
\(\Leftrightarrow4+4m-4+m-4=0\)
\(\Leftrightarrow5m=4\)
\(\Leftrightarrow m=\dfrac{4}{5}\)
2.\(\left(3m-1\right)x+m-4=2x+5\) ; \(m=-2\)
\(\Leftrightarrow\left(3.-2-1\right)x+-2-4=2x+5\)
\(\Leftrightarrow6x-6-2x-5=0\)
\(\Leftrightarrow4x=11\)
\(\Leftrightarrow x=\dfrac{11}{4}\)
a, \(2-6x+5=0\Leftrightarrow7-6x=0\Leftrightarrow x=\dfrac{7}{6}\)
b, \(3x\left(2x-1\right)=0\Leftrightarrow x=0;x=\dfrac{1}{2}\)
c, \(\Rightarrow10x-5+60x=36-6x\Leftrightarrow76x=41\Leftrightarrow x=\dfrac{41}{76}\)
d, đk : x khác 0 ; 2
\(\Rightarrow x^2+2x-x+2=2\Leftrightarrow x^2+x=0\Leftrightarrow x=0\left(ktm\right);x=-1\left(tm\right)\)
Gọi quãng đường là x ( x > 0 )
Theo bài ra ta có pt \(\dfrac{x}{30}-\dfrac{x}{40}=\dfrac{45}{60}=\dfrac{3}{4}\Rightarrow x=90\left(tm\right)\)
1, Thay x = -3 ta được \(m-15=-4\Leftrightarrow m=11\)
2, Thay m = -1 vào ta được
\(-6x-1=x\Leftrightarrow-7x=1\Leftrightarrow x=-\dfrac{1}{7}\)
a, \(4x-3x=1+3\Leftrightarrow x=4\)
b, \(x\left(x-1\right)+2\left(x-1\right)=0\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\Leftrightarrow x=-2;x=1\)
c, đk x khác -2 ; 2
\(2x-4+3x+6=3x-8\Leftrightarrow2x=-10\Leftrightarrow x=-5\left(tm\right)\)
Giúp mình nhanh với mình đang cần gấp 😄
Bài 4:
\(m^2x-\left(x+1\right)m=2x+1\)
\(\Leftrightarrow m^2x-xm-m=2x+1\)
\(\Leftrightarrow x\left(m^2-m-2\right)=m+1\)
Để phương trình vô nghiệm thì m-2=0
hay m=2
để phương trình có nghiệm duy nhất thì \(\left(m-2\right)\left(m+1\right)< >0\)
hay \(m\notin\left\{2;-1\right\}\)
Để phương trình có vô số nghiệm thì m+1=0
hay m=-1