Tính và tìm x:
a) | x - 1,7| = 2,3
b) |x + 3/4| - 1/3 = 0
c) | x+1/4| -3/4 = 5
d) 2 - | 3/2.x - 1/4| = 5/4
e) |4 + 2x| +4x = 0
Tìm x:
a) 2x - 2/3 = -3/4
b) x : 3/4 + 1/4 = -2/3
c) 1/4x + 2/5 = 7/5
d) /x//5 - 2 = -3/5
MÌNH CẦN GẤP. CẢM ƠN Ạ
`c)1/4x+2/5=7/5`
`=>1/4x=7/5-1/5=1`
`=>x=1:1/4=4`
Vậy `x=4`
`a)2x-2/3=-3/4`
`=>2x=-3/4+2/3=-1/12`
`=>x=-1/24`
Vậy `x=-1/24`
`b)x:3/4+1/4=-2/3`
`=>x:3/4=-2/3-1/4=-11/4`
`=>x=-11/4 xx 3/4=-33/16`
Vậy `x=-33/16`
TÌM X
a) /x-17/=2,3
b) /x+3/4/=0
c)/x+3/4/+1/3=0
a) \(\left|x-17\right|=2,3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-17=2,3\\x-17=-2,3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=19,3\\x=14,7\end{matrix}\right.\)
b) \(\left|x+\dfrac{3}{4}\right|=0\)
\(\Leftrightarrow x+\dfrac{3}{4}=0\Leftrightarrow x=-\dfrac{3}{4}\)
c) \(\left|x+\dfrac{3}{4}\right|+\dfrac{1}{3}=0\)
\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=-\dfrac{1}{3}\)( vô lý do \(\left|x+\dfrac{3}{4}\right|\ge0\forall x\))
Vậy \(S=\varnothing\)
Tìm x:
a) (3-2x)2-(3+2x)2=8
b) 9x5-72x2=0
c) 5x4-8x2-4=0
d) (x3+x2-4x-4) : (x-2)=0
Lời giải:
a. PT $\Leftrightarrow (3-2x-3-2x)(3-2x+3+2x)=8$
$\Leftrightarrow -4x.6=8$
$\Leftrightarrow -24x=8\Leftrightarrow x=\frac{-1}{3}$
b.
$9x^5-72x^2=0$
$\Leftrightarrow 9x^2(x^3-8)=0$
$\Leftrightarrow x^2=0$ hoặc $x^3=8$
$\Leftrightarrow x=0$ hoặc $x=2$
c.
$5x^4-8x^2-4=0$
$\Leftrightarrow 5x^4-10x^2+2x^2-4=0$
$\Leftrightarrow 5x^2(x^2-2)+2(x^2-2)=0$
$\Leftrightarrow (5x^2+2)(x^2-2)=0$
$\Leftrightarrow 5x^2+2=0$ (loại) hoặc $x^2-2=0$ (chọn)
$\Leftrightarrow x=\pm \sqrt{2}$
d.
PT $\Leftrightarrow [x^2(x+1)-4(x+1)]:(x-2)=0$
$\Leftrightarrow (x^2-4)(x+1):(x-2)=0$
$\Leftrightarrow (x-2)(x+2)(x+1):(x-2)=0$
$\Leftrightarrow (x+2)(x+1)=0$
$\Leftrightarrow x+2=0$ hoặc $x+1=0$
$\Leftrightarrow x=-2$ hoặc $x=-1$
a: Ta có: \(\left(3-2x\right)^2-\left(3+2x\right)^2=8\)
\(\Leftrightarrow9-12x+4x^2-9-12x-4x^2=8\)
\(\Leftrightarrow-24x=8\)
hay \(x=-\dfrac{1}{3}\)
b: Ta có: \(9x^5-72x^2=0\)
\(\Leftrightarrow9x^2\left(x^3-8\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c: Ta có: \(5x^4-8x^2-4=0\)
\(\Leftrightarrow5x^4-10x^2+2x^2-4=0\)
\(\Leftrightarrow x^2-2=0\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
Tìm x:
a)(x+2)^2-2(x+2)(x-5)=0
b)2x^2+3x-5=0
c)x+2√2x^2+2x^3=0
d)(3x-1)^2-4(x+5)^2=0
a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)
Tìm x:
a) 4.(2-x)+x.(x+6)=x2
b) x.(x-7)-(x-2).(x+5)=0
c) (2x+3).(3-2x)+(2x-1)2=2
a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)
\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)
\(\Leftrightarrow2x=-8\)
hay x=-4
b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)
\(\Leftrightarrow-10x=-10\)
hay x=1
c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)
\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)
\(\Leftrightarrow-4x=-8\)
hay x=2
a)x-3/x-2 + x-2/x-4 = -1
b)3x + 12 = 0
c)5 + 2x = x - 5
d)2x(x - 2) + 5(x - 2) = 0
e)3x-4/2 = 4x+1/3
f)2x/x-1 - x/x+1 =1
g)2x/x-1 + 3-2x/x+2 = 6/(x-1)(x+2)
\(a)\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1.\left(x\ne2;4\right).\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1.\\ \Rightarrow x^2-4x-3x+12+x^2-4x+4+x^2-4x-2x+8=0.\\ \Leftrightarrow3x^2-17x+24=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}.\\x=3.\end{matrix}\right.\) (TM).
\(b)3x+12=0.\\ \Leftrightarrow3x=-12.\\ \Leftrightarrow x=-4.\)
\(c)5+2x=x-5.\\ \Leftrightarrow2x-x=-5-5.\\ \Leftrightarrow x=-10.\)
\(d)2x\left(x-2\right)+5\left(x-2\right)=0.\\ \Leftrightarrow\left(2x+5\right)\left(x-2\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}.\\x=2.\end{matrix}\right.\)
\(e)\dfrac{3x-4}{2}=\dfrac{4x+1}{3}.\\ \Rightarrow3\left(3x-4\right)-2\left(4x+1\right)=0.\\ \Leftrightarrow9x-12-8x-2=0.\\ \Leftrightarrow x=14.\)
\(f)\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1.\left(x\ne\pm1\right).\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x}{x^2-1}=1.\\ \Leftrightarrow x^2+3x-x^2+1=0.\\ \Leftrightarrow3x+1=0.\\ \Leftrightarrow x=\dfrac{-1}{3}.\)
\(g)\dfrac{2x}{x-1}+\dfrac{3-2x}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\left(x\ne1;-2\right).\\ \Leftrightarrow\dfrac{2x^2+4x+\left(3-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\\ \Rightarrow2x^2+4x+3x-3-2x^2+2x-6=0.\\ \Leftrightarrow9x=9.\)
\(\Leftrightarrow x=1\left(koTM\right).\)
Tìm x:
a) (2x - 1) (x^2 - x + 1) = 2x^3 - 3x^2 + 2
b) (x + 1) (x^2 + 2x + 4) - x^3 - 3x^2 + 16 = 0
c) (x + 1) (x + 2) (x + 5) - x^3 - 8x^2 = 27
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
Giúp với mn ơi ;-;a) 9(2x+1)^2-4(x+1)^2=0b) (x+1)^2+2(x+1)+1=0c) (x-1)(x^2-9)+x+3=0d) (7-x)^2__________ -(x+5)^2=0 4e) 4x^2+(x-1)^2-(2x+1)^2=0f) x^3+1=(x+1)(2-x)
Rối quá bn ơi
Bn ko dùng dấu enter để xuống dòng à
Mk sẽ giúp bn 3 câu đầu thôi còn 3 câu sau thì..... chịu vì ko biết làm sao
a) \(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\)
(=)\(\left[9\left(2x+1\right)-4\left(x+1\right)\right]\left[9\left(2x+1\right)+4\left(x+1\right)\right]=0\)
(=)\(\left(18x+9-4x-4\right)\left(18x+9+4x+4\right)=0\)
(=)\(\left(14x+5\right)\left(22x+12\right)=0\)
(=)\(\left(14x+5\right)=0\) hoặc \(\left(22x+12\right)=0\)
1) \(14x+5=0\Rightarrow x=-\dfrac{5}{14}\)
2) \(22x+12=0\Rightarrow x=-\dfrac{6}{11}\)
Vậy........
b)\(\left(x+1\right)^2+2\left(x+1\right)+1=0\)
(=)\(\left(x+1\right)^2+2\left(x+1\right).1+1^2=0\)
(=) \(\left(x+2\right)^2=0\)
(=) \(x+2=0\Rightarrow x=-2\)
Vậy........
c) \(\left(x-1\right)\left(x^2-9\right)+x+3=0\)
(=) \(\left(x-1\right)\left(x-3\right)\left(x+3\right)+x+3=0\)
(=) \(\left(x+3\right)\left(x-3+1\right)\left(x-1\right)=0\)
(=) \(\left(x+3\right)\left(x-2\right)\left(x-1\right)=0\)
(=) \(x+3=0\) hoặc \(x-2=0\) hoặc \(x-1=0\)
1)\(x+3=0\Rightarrow x=-3\)
2)\(x-2=0\Rightarrow x=2\)
3) \(x-1=0\Rightarrow x=1\)
Vậy.........
P=( 3/2x+4 + x/2-x + 2x^2+3/x^2-4) : 2x-1/4x-8
a) Rút gọn P
b) Tính P khi 4x^2-1=0
c) Tìm x để P < 2
a: \(P=\left(\dfrac{3}{2\left(x+2\right)}-\dfrac{x}{x-2}+\dfrac{2x^2+3}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{4\left(x-2\right)}{2x-1}\)
\(=\left(\dfrac{3\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}-\dfrac{2x\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{4x^2+6}{2\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{4\left(x-2\right)}{2x-1}\)
\(=\dfrac{3x-6-2x^2-4x+4x^2+6}{2\left(x+2\right)\left(x-2\right)}\cdot\dfrac{4\left(x-2\right)}{2x-1}\)
\(=\dfrac{2x^2-x}{x+2}\cdot\dfrac{2}{2x-1}=\dfrac{2x}{x+2}\)
b: Khi 4x2-1=0 thì (2x-1)(2x+1)=0
=>x=1/2(loại) và x=-1/2(nhận)
Khi x=-1/2 thì \(P=\left(2\cdot\dfrac{-1}{2}\right):\left(-\dfrac{1}{2}+2\right)=-1:\dfrac{3}{2}=-\dfrac{2}{3}\)