5/-x^2+5x-6+x+3/2-x=0
Giúp mk với
Bài 1:Tìm x biết:
a.,2x(x-3)-15+5x=0
b,x^3-7x
c,(2x-3)^2-(x+5)^2=0
d,x^3-x^2-4x^2+8x-4=0
giúp mk với mk cầm gấp lắm TvT
a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)
b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)
c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Xem lại đề câu d
tìm x bt
a)x(x-5)-4x+20=0
b)x(x+6)-7x-42=0
c)x^3-5x^2-x+5=0
d)4x^2-25-(2x-5)(3x+7)=0
e)x^3+27+(x+3)(x-9)=0
giúp mk vs ah!!1
a) x(x - 5) - 4x + 20 = 0
\(\Leftrightarrow\) x(x - 5) - (4x + 20)
\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x - 4)
Khi x - 5 = 0 hoặc x - 4 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 4
Vậy S = \(\left\{5;4\right\}\)
b) x(x + 6) - 7x - 42 = 0
\(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0
\(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0
\(\Leftrightarrow\) (x + 6)(x - 7) = 0
Khi x - 6 = 0 hoặc x - 7 = 0
\(\Leftrightarrow\) x = 6 \(\Leftrightarrow\) x = 7
Vậy S = \(\left\{6;7\right\}\)
c) x3 - 5x2 - x + 5 = 0
\(\Leftrightarrow\) (x3 - 5x2) - (x + 5) = 0
\(\Leftrightarrow\) x2 (x - 5) - (x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x2 - 1) = 0
\(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0
Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 1 \(\Leftrightarrow\) x = -1
Vậy S = \(\left\{5;1;-1\right\}\)
d) 4x2 - 25 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x)2 - 52 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0
\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0
Khi 2x - 5 = 0 hoặc -x + 12 = 0
\(\Leftrightarrow\) 2x = 5 \(\Leftrightarrow\) -x = -12
\(\Leftrightarrow\) x = \(\dfrac{5}{2}\) \(\Leftrightarrow\) x = 12
Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)
e) x3 + 27 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) x3 - 33 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 3x + 9) + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3) \(\left[\left(x^2-3x+9\right)+\left(x-9\right)\right]\) = 0
\(\Leftrightarrow\) (x - 3) ( x2 - 3x + 9 + x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 2x) = 0
\(\Leftrightarrow\) (x - 3)x(x - 2)
Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x = 3 \(\Leftrightarrow\) x = 2
Vậy S = \(\left\{3;0;2\right\}\)
Chúc bạn học tốt
a) Ta có: \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
b) Ta có: \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
c) Ta có: \(x^3-5x^2-x+5=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)
d) Ta có: \(4x^2-25-\left(2x-5\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-3x-7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)
Tìm x, biết
a. (5x-7)2-25x2=32
b.(x-3)2+(x+1)2-5=0
giúp mình nhé
\(a,\Leftrightarrow25x^2-70x+49-25x^2=32\\ \Leftrightarrow-70x=-17\Leftrightarrow x=\dfrac{17}{70}\\ b,\Leftrightarrow x^2-6x+9+x^2+2x+1-5=0\\ \Leftrightarrow2x^2-4x+5=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+3=0\\ \Leftrightarrow2\left(x-1\right)^2=-3\Leftrightarrow\left(x-1\right)^2=-\dfrac{3}{2}\left(\text{vô lí}\right)\\ \Leftrightarrow x\in\varnothing\)
Bài1: giải các pt sau:
a, 3-4x+24+6x= x+27+3x
b, 5-(6-x)=4(3-2x)
c, x-(x+1)/3 = (2x+1)/5
d,(2x-1)/3 - (5x+2)/7 = x+13
Bài 2:
a, (x-1)(3x+1)=0
b, (x-5)(7-x)=0
c, ( x-1)(x+5)(-3x+8)=0
d, x(x^2 - 1 )=0
Giúp mình 2 bài này với , mình đang cần gấp , CẢM ƠN M.N ạ><
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
a)Giải bất phương trình
3(x+7)-2x+5>0 x=2/18 - x+3/8 < x-1/9 - x-4/24
b)Giải pt
3x+2+|x+5|=0
giúp mk với ;-;
a: 3(x+7)-2x+5>0
=>3x+21-2x+5>0
=>x+26>0
=>x>-26
Sửa đề: \(\dfrac{x+2}{18}-\dfrac{x+3}{8}< \dfrac{x-1}{9}-\dfrac{x-4}{24}\)
=>\(\dfrac{4\left(x+2\right)}{72}-\dfrac{9\left(x+3\right)}{72}< \dfrac{8\left(x-1\right)}{72}< \dfrac{3\left(x-4\right)}{72}\)
=>\(4\left(x+2\right)-9\left(x+3\right)< 8\left(x-1\right)-3\left(x-4\right)\)
=>\(4x+8-9x-27< 8x-8-3x+12\)
=>-5x-19<5x+4
=>-10x<23
=>\(x>-\dfrac{23}{10}\)
b: \(3x+2+\left|x+5\right|=0\left(1\right)\)
TH1: x>=-5
(1) trở thành: 3x+2+x+5=0
=>4x+7=0
=>\(x=-\dfrac{7}{4}\left(nhận\right)\)
TH2: x<-5
=>x+5<0
=>|x+5|=-x-5
Phương trình (1) sẽ trở thành:
\(3x+2-x-5=0\)
=>2x-3=0
=>2x=3
=>\(x=\dfrac{3}{2}\)
Cho A= x^3 -2x+n B=x-2 Tìm n để A chia hết cho B
(x-3)^2-2x+6=0
x^2-5x+6=0
Giúp mình nhanh vs ạ
\(A=x^3-2x+n\)
\(B=n-2\)
\(A\text{⋮}B\) ⇒ \(\left(x^3-2x+n\right)\text{⋮}\left(n-2\right)\)
⇒ \(\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(2x-4\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)
⇒ \(\left[x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)
⇒ \(\left[\left(x-2\right)\left(x^2+2x+2\right)+\left(n+4\right)\right]\text{⋮}\left(x-2\right)\)
Vì \(\left(x-2\right)\left(x^2+2x+2\right)\text{⋮}\left(n-2\right)\)
Để \(A\text{⋮}B\)
⇒ \(n+4=0\)
⇒ \(n=-4\)
c) (x2 + 1). (x + 2) > 0
d) (x + 4). (x – 3) < 0
e) ( x2 + 2). (x2 + 5) < 0
f) (x + 6). (x – 4) > 0
Giúp mình với, mình cần gấp ạ
Hai bài bị trùng nhau nên các bạn nhìn ảnh hay văn bản đều như nhau ạ
c: =>x+2>0
hay x>-2
d: =>-4<=x<=3
e: =>\(x\in\varnothing\)
f: \(\Leftrightarrow\left[{}\begin{matrix}x>4\\x< -6\end{matrix}\right.\)
Tìm x biết:
a) x(x - 2) - x2 + 5x = 6
b) (x - 3)2 - x + 9 =0
giúp mình câu này nha
\(a,\Leftrightarrow x^2-2x-x^2+5x=6\\ \Leftrightarrow3x=6\\ \Leftrightarrow x=2\)
\(b,\Leftrightarrow x^2-6x+9-x+9=0\\ \Leftrightarrow x^2-7x+18=0\\ \Leftrightarrow\left(x^2-7x+\dfrac{49}{4}\right)+\dfrac{23}{4}=0\\ \Leftrightarrow\left(x-\dfrac{7}{2}\right)^2+\dfrac{23}{4}=0\left(vôlí\right)\)
x^2 - (m-2)*x -6 = 0. Tìm m để pt có 2 nghiệm phân biệt x1 x2 thỏa mãn x1 + x2 - 3 x1x2=0
Giúp mk vs
\(ac=-6< 0\Rightarrow\) phương trình đã cho luôn luôn có 2 nghiệm pb (trái dấu)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m-2\\x_1x_2=-6\end{matrix}\right.\)
Thế vào đề bài:
\(m-2-3\left(-6\right)=0\)
\(\Leftrightarrow m+16=0\Leftrightarrow m=-16\)
\(x^2-\left(m-2\right)x-6=0\left(1\right)\)
\(\Rightarrow\Delta=b^2-4ac=\left[-\left(m-2\right)\right]^2-4.\left(-6\right)\)
\(=m^2-4m+4+24=m^2-4m+28\)
\(=\left(m-2\right)^2+24\)
Thấy \(\left(m-2\right)^2\ge0\)\(\Rightarrow\left(m-2\right)^2+24>0\forall m\)
Vậy phương trình luân có 2 nghiệm phân biệt \(x_1,x_2\)
Áp dụng \(Vi-ét \) ta có :
\(S=x_1+x_2=\dfrac{-b}{a}=m-2\)
\(P=x_1.x_2=\dfrac{c}{a}=-6\)
Ta có \(x_1+x_2-3.x_1.x_2=0\)
\(\Leftrightarrow m-2-3.\left(-6\right)=0\Rightarrow m=-16\)