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1
a (9+x)=2 ta có (9+x)= 9+x khi 9+x >_0 hoặc >_ -9
(9+x)= -9-x khi 9+x <0 hoặc x <-9
1)pt 9+x=2 với x >_ -9
<=> x = 2-9
<=> x=-7 thỏa mãn điều kiện (TMDK)
2) pt -9-x=2 với x<-9
<=> -x=2+9
<=> -x=11
x= -11 TMDK
vậy pt có tập nghiệm S={-7;-9}
các cau con lai tu lam riêng nhung cau nhan với số âm thi phan điều kiện đổi chiều nha vd
nhu cau o trên mk lam 9+x>_0 hoặc x>_0
với số âm thi -2x>_0 hoặc x <_ 0 nha
\(a,\Leftrightarrow-4+k=-3\Leftrightarrow k=1\\ b,\Leftrightarrow-3\left(2k-18\right)=40\\ \Leftrightarrow2k-18=-\dfrac{40}{3}\Leftrightarrow k=\dfrac{7}{3}\\ c,\Leftrightarrow10+18=9\left(2+k\right)\\ \Leftrightarrow k+2=\dfrac{28}{9}\Leftrightarrow k=\dfrac{10}{9}\)
a) với x=1=> (1)^3+a.1-4.1-4=0<=> 1+a-8=0<=>a=7
b) ta có phương trình
x^3+7x-4x-4 =0<=> x^3+3x-4=0
<=> x^3-x+4x-4=0
<=> x(x^2-1)+4(x-1)=0
<=> x(x-1)(x+1)+4(x-1)=0
<=> (x-1)(x^2+x+4)=0
<=> ..... tự làm tiếp nha
1, 2mx−1x−1=m−2 (x≠1)(x≠1)
⇔ 2mx−1=(m−2)(x−1)
⇔ 2mx−1=x(m−2)−m+2
⇔ x.(m+2)=−m+3x.(m+2)=−m+3
Nếu m+2=0m+2=0 hay m=−2m=−2 thì 0x=5
⇒ PT vô nghiệm
Nếu m+2≠0 hay m≠−2 thì x=3mm+2
2, 2x2x²−5x+3+9x2x²−x−3=6
⇔ 2x(3x−2).(x−1)+9x(3x−2).(x+1)=6
⇔ 2x(x+1)(3x−2).(x−1)(x+1)+9x(x−1)(3x−2).(x+1)(x−1)=6
⇒ 2x(x+1)+9x(x−1)=6(3x−2)(x+1)(x−1)
⇔ 11x²−7x=18x³−12x²−18x+12
⇔ 18x³−13x²−11x+12=0
Bài 1:
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)
\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)
Suy ra: \(-12x-3=8x-2-6x-8\)
\(\Leftrightarrow-12x-3-2x+10=0\)
\(\Leftrightarrow-14x+7=0\)
\(\Leftrightarrow-14x=-7\)
\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)
\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)
\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)
\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)
\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
3.15:
a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)
b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3.16
\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)
\(\Leftrightarrow-14m+35-2m^2+8=0\)
\(\Leftrightarrow-14m-2m^2+43=0\)
\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)
\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)
\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)
\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)
pt vô nghiệm
\(2\left(5x+1\right)-5=\left(x+6\right)\left(2x+k\right)\left(1\right)\)
Thay \(x=3\) vào \(\left(1\right)\)ta có:
\(2.\left(5.3+1\right)-5=\left(3+6\right)\left(2.3+k\right)\)
Thu gọn, tìm được \(k=-3\)