Tìm x thuộc Z
a,|x+2|-x=2
b,|x-3|+x-3=0
c,|x+1|+|x+2|=1
d,|x-5|\+x-8=6
Tìm các giới hạn sau :
a, lim\(\dfrac{2x^2+x-6}{x^3+8}\) khi x→-2
b, lim\(\dfrac{x^4-x^2-72}{x^2-2x-3}\) khi x→3
c, lim\(\dfrac{x^5+1}{x^3+1}\) khi x→-1
d, lim \(\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)\) khi x→1
a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).
b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).
c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).
d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).
Giải phương trình:
b) (x+2) (2x^2-5) - (x^3-8) = 0
c) x-5/x-1 + 2/x-3 = 1
d) 2x+5/x+3 + 1 = 4/x^2+2x-3 - 3x-1/1-x
b, bạn xem lại đề
c, đk : x khác 1 ; 3
\(\Rightarrow x^2-8x+15+2x-2=x^2-4x+3\Leftrightarrow-2x=-10\Leftrightarrow x=5\left(tm\right)\)
d, đk: x khác -3 ; x khác 1
\(\Rightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2+3x-5+x^2+2x-3=4+3x^2+8x-3\)
\(\Leftrightarrow-3x=5\Leftrightarrow x=-\dfrac{5}{3}\left(tm\right)\)
tìm x , biết
a) 17/6- x( x-7/6)= 7/4
b) 3/35 - ( 3/5-x)= 2/7
tìm x thuộc Z , biết
3/4-5/6 < x/12 < 1 -( 2/3-1/4)
tìm x biết
a ) 2x-3=x + 1/2
b) 4x- ( x+ 1/2) = 2x - ( 1/2 - 5 )
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
Bài 3:
a) Ta có: \(2x-3=x+\dfrac{1}{2}\)
\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)
\(\Leftrightarrow x=\dfrac{7}{2}\)
b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)
\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)
\(\Leftrightarrow x=5\)
Tìm x :
a) (x + 2) - x(x + 3) = 2
b) (x + 2)(x -2) - (x + 1)2 = 7
c) 6x2 - (2x + 1)(3x - 2) = 1
d) (x + 2)(x + 3) - (x - 2)(x + 1) = 2
e) 6(x - 1)( x + 1) - (2x - 1)(3x + 2) + 3 = 0
a (x + 2) - x(x + 3) = 2
x + 2 - x(x + 3) - 2 = 0
x + x(x + 3) = 0
x(1 + x + 3) = 0
x(x + 4) = 0
x = 0 hoặc x + 4 = 0
*) x + 4 = 0
x = -4
Vậy x = -4; x = 0
b) (x + 2)(x - 2) - (x + 1)² = 7
x² - 4 - x² - 2x - 1 = 7
-2x - 5 = 7
-2x = 7 + 5
-2x = 12
x = 12 : (-2)
x = -6
c) 6x² - (2x + 1)(3x - 2) = 1
6x² - 6x² + 4x - 3x + 2 = 1
x + 2 = 1
x = 1 - 2
x = -1
d) (x + 2)(x + 3) - (x - 2)(x + 1) = 2
x² + 3x + 2x + 6 - x² - x + 2x + 2 = 2
6x + 8 = 2
6x = 2 - 8
6x = -6
x = -6 : 6
x = -1
e) 6(x - 1)(x + 1) - (2x - 1)(3x + 2) + 3 = 0
6x² - 6 - 6x² - 4x + 3x + 2 + 3 = 0
-x - 1 = 0
x = -1
1. tìm x biết
a, (2x - 3)\(^2\) = |3 - 2x|
b, (x - 1)\(^2\) + (2x - 1)\(^2\) = 0
c, 5 - x\(^2\) = 1
d, x - 2\(\sqrt{x}\) = 0
g, (x - 1) + \(\dfrac{1}{7}\) = 0
`#3107.101107`
`1.`
`a,`
`(2x - 3)^2 = |3 - 2x|`
`=> (2x - 3)^2 = |2x - 3|`
`=>`\(\left[{}\begin{matrix}2x-3=\left(2x-3\right)^2\\2x-3=-\left(2x-3\right)^2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3-\left(2x-3\right)^2=0\\2x-3+\left(2x-3\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\left(2x-3\right)\left(1-2x+3\right)=0\\\left(2x-3\right)\left(1+2x-3\right)=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3=0\\4-2x=0\\2x-2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)
Vậy, `x \in {3/2; 2; 1}`
`b,`
`(x - 1)^2 + (2x - 1)^2 = 0`
`=>`\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(2x-1\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x \in {1; 1/2}`
`c,`
`5 - x^2 = 1`
`=> x^2 = 4`
`=> x^2 = (+-2)^2`
`=> x = +-2`
Vậy, `x \in {-2; 2}`
`d,`
`x - 2\sqrt{x} = 0`
`=> x^2 - (2\sqrt{x})^2 = 0`
`=> x^2 - 4x = 0`
`=> x(x - 4) = 0`
`=>`\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy, `x \in {0; 4}`
`g,`
`(x - 1) + 1/7 = 0`
`=> x - 1 + 1/7 = 0`
`=> x - 6/7 = 0`
`=> x = 6/7`
Vậy, `x = 6/7.`
1.tìm x,y biết
a, x.(y-3)≥0
b, (2.x-1).(y-1)≤0
c,(x-1).(2.k+1)≥0
2. tìm x,y ϵ Z biết
a, x(x+3)=0
b,(x-2).(5-x)=0
c,(x-1).(x^2+1)=0
d, x.y+3.x-7.y=21
e,x.y+3.x-2y=11
Bài 2:
a: =>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
Tìm x∈Z, biết:
a)x.(x-6)=0
b)(-7-x).(-x+5)=0
c)(x+3).(x-7)=0
d)(x-3).(x2+12)=0
e)(x+1).(2-x) ≥0
f)(x-3).(x-5) ≤0
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
c) => \(\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
a) (2x+1)^2-2x-1=2b) (x^2-3x)^2+5(x^2-3x)+6=0c) (x^2-x-1)(x^2-x)-2=0d) (5-2x)^2+4x-10=8e) (x^2+2x+3)(x^2+2x+1)=3f) x(x-1)(x^2-x+1)-6=0
sửa lại chút: a) (2x+1)^2-2x-1=2 b) (x^2-3x)^2+5(x^2-3x)+6=0 c) (x^2-x-1)(x^2-x)-2=0 d) (5-2x)^2+4x-10=8 e) (x^2+2x+3)(x^2+2x+1)=3 f) x(x-1)(x^2-x+1)-6=0
a) Ta có: \(\left(2x+1\right)^2-2x-1=2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{1}{2};-1\right\}\)
tìm x
a) 3.(x-3)-4x+12=0
b)(x+2)^2-(x+2).(x-2) =0
c)x^3+3x=3x^2+1
d)2/3x.(x^2-4)=0
e)(2x-3)^2-(+5)^2=0
\(a,=3x-9-4x+12=-x+3=0\)
\(\Leftrightarrow x=3\)
Vậy ..
\(b,=\left(x+2\right)\left(x+2-x+2\right)=4\left(x+2\right)=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy ..
\(c,=x^3-3x^2+3x-1=\left(x-1\right)^3=0\)
\(\Leftrightarrow x=1\)
Vậy ..
\(d,\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy ..
\(e,=\left(2x-3-5\right)\left(2x-3+5\right)=\left(2x-8\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}=4\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\)
Vậy ...
a) Ta có: 3(x-3)-4x+12=0
\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\)
hay x=3
Vậy: S={3}
b) Ta có: \(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow x^2+4x+4-x^2+4=0\)
\(\Leftrightarrow4x=-8\)
hay x=-2
Vậy: S={-2}
c) Ta có: \(x^3+3x=3x^2+1\)
\(\Leftrightarrow x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
Vậy: S={1}
d) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: S={0;2;-2}
a) 3.(x-3)-4x+12=0
=> 3x - 9 - 4x + 12 = 0
=> -x + 3 = 0
=> x = 3
b) (x+2)^2-(x+2).(x-2) =0
\(\Rightarrow\left(x+2\right)^2-x^2+4=0\)
\(\Rightarrow x^2+4x+4-x^2+4=0\)
=> 4x + 8 = 0
=> x = -2
c) x^3+3x=3x^2+1
\(\Rightarrow x^3+3x-3x^2-1=0\)
\(\Rightarrow\left(x-1\right)^3=0\)
=> x = 1
d) \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Rightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
=> x = 0 hoặc x = 2 hoặc x = -2
e) \(\left(2x-3\right)^2-5^2=0\)
\(\Rightarrow\left(2x-8\right)\left(2x+2\right)=0\)
=> x = 4 hoăc x = -1
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