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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
c: (3x-2)(x+3)<0
=>x+3>0 và 3x-2<0
=>-3<x<2/3
d: \(\dfrac{x-2}{x-10}>=0\)
=>x-10>0 hoặc x-2<=0
=>x>10 hoặc x<=2
e: \(3x^2+7x+4< 0\)
\(\Leftrightarrow3x^2+3x+4x+4< 0\)
=>(x+1)(3x+4)<0
=>-4/3<x<-1
a) Ta có: (2x-3)(x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{2};-2\right\}\)
b) Ta có: (3x-1)(2x-5)=(3x-1)(x+2)
⇔\(\left(3x-1\right)\left(2x-5\right)-\left(3x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left[\left(2x-5\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(3x-1\right)\left(2x-5-x-2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=7\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{3};7\right\}\)
c) Ta có: \(\left(x^2-25\right)+\left(x-5\right)\left(2x-11\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+\left(x-5\right)\left(2x-11\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5+2x-11\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\cdot3\cdot\left(x-2\right)=0\)
mà 3≠0
nên \(\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
Vậy: x∈{5;2}
d) Ta có: \(\left(x^2-6x+9\right)-4=0\)
\(\Leftrightarrow\left(x-3\right)^2-2^2=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy: x∈{5;1}
e) Ta có: \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;\frac{3}{2}\right\}\)
Bài 1:
a) (5x-4)(4x+6)=0
\(\Leftrightarrow\orbr{\begin{cases}5x-4=0\\4x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=4\\4x=-6\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{4}{5}\\y=\frac{-3}{2}\end{cases}}}\)
b) (x-5)(3-2x)(3x+4)=0
<=> x-5=0 hoặc 3-2x=0 hoặc 3x+4=0
<=> x=5 hoặc x=\(\frac{3}{2}\)hoặc x=\(\frac{-4}{3}\)
c) (2x+1)(x2+2)=0
=> 2x+1=0 (vì x2+2>0)
=> x=\(\frac{-1}{2}\)
bài 1:
a) (5x - 4)(4x + 6) = 0
<=> 5x - 4 = 0 hoặc 4x + 6 = 0
<=> 5x = 0 + 4 hoặc 4x = 0 - 6
<=> 5x = 4 hoặc 4x = -6
<=> x = 4/5 hoặc x = -6/4 = -3/2
b) (x - 5)(3 - 2x)(3x + 4) = 0
<=> x - 5 = 0 hoặc 3 - 2x = 0 hoặc 3x + 4 = 0
<=> x = 0 + 5 hoặc -2x = 0 - 3 hoặc 3x = 0 - 4
<=> x = 5 hoặc -2x = -3 hoặc 3x = -4
<=> x = 5 hoặc x = 3/2 hoặc x = 4/3
c) (2x + 1)(x^2 + 2) = 0
vì x^2 + 2 > 0 nên:
<=> 2x + 1 = 0
<=> 2x = 0 - 1
<=> 2x = -1
<=> x = -1/2
bài 2:
a) (2x + 7)^2 = 9(x + 2)^2
<=> 4x^2 + 28x + 49 = 9x^2 + 36x + 36
<=> 4x^2 + 28x + 49 - 9x^2 - 36x - 36 = 0
<=> -5x^2 - 8x + 13 = 0
<=> (-5x - 13)(x - 1) = 0
<=> 5x + 13 = 0 hoặc x - 1 = 0
<=> 5x = 0 - 13 hoặc x = 0 + 1
<=> 5x = -13 hoặc x = 1
<=> x = -13/5 hoặc x = 1
b) (x^2 - 1)(x + 2)(x - 3) = (x - 1)(x^2 - 4)(x + 5)
<=> x^4 - x^3 - 7x^2 + x + 6 = x^4 + 4x^3 - 9x^2 - 16x + 20
<=> x^4 - x^3 - 7x^2 + x + 6 - x^4 - 4x^3 + 9x^2 + 16x - 20 = 0
<=> -5x^3 - 2x^2 + 17x - 14 = 0
<=> (-x + 1)(x + 2)(5x - 7) = 0
<=> x - 1 = 0 hoặc x + 2 = 0 hoặc 5x - 7 = 0
<=> x = 0 + 1 hoặc x = 0 - 2 hoặc 5x = 0 + 7
<=> x = 1 hoặc x = -2 hoặc 5x = 7
<=> x = 1 hoặc x = -2 hoặc x = 7/5
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\\ \left(3x+2\right)\left(x^2-1\right)-\left(9x^2-4\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left[\left(x-1\right)-\left(3x-2\right)\right]=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(1-2x\right)=0\\ \left[{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
\(b.x\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\\ x\left(x^2-9\right)-\left(x^3+8\right)=0\\ x^3-9x-x^3-8=0\\ -9x-8=0\\ -9x=8\\ x=\frac{-8}{9}\)
\(c.2x\left(x-3\right)+5\left(x-3\right)=0\\ \left(x-3\right)\left(2x+5\right)=0\\ \left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-5}{2}\end{matrix}\right.\)
\(d.\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\\ \left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\\ \left(3x-1\right)\left[\left(x^2+2\right)-\left(7x-10\right)\right]=0\\ \left(3x-1\right)\left(x^2+2-7x+10\right)=0\\ \left(3x-1\right)\left(x^2-7x+12\right)=0\\ \left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \left(3x-1\right)\left[\left(x^2-4x\right)+\left(-3x+12\right)\right]=0\\ \left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}3x-1=0\\x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\\x=3\end{matrix}\right.\)
\(e.\left(x+2\right)\left(3-4x\right)=x^2+4x+4\\ \left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\\ \left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\\ \left(x+2\right)\left[\left(3-4x\right)-\left(x+2\right)\right]=0\\ \left(x+2\right)\left(3-4x-x-2\right)=0\\ \left(x+2\right)\left(1-5x\right)=0\left[{}\begin{matrix}x+2=0\\1-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{1}{5}\end{matrix}\right.\)
\(f.x\left(2x-7\right)-4x+14=0\\ x\left(2x-7\right)-2\left(2x-7\right)=0\\ \left(2x-7\right)\left(x-2\right)=0\\ \left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)
\(g.3x-15=2x\left(x-5\right)\\ 3\left(x-5\right)=2x\left(x-5\right)\\ 3\left(x-5\right)-2x\left(x-5\right)=0\\ \left(x-5\right)\left(3-2x\right)=0\\ \left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)
\(h.\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \left(2x+1\right)\left[\left(3x-2\right)-\left(5x-8\right)\right]=0\\ \left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \left(2x+1\right)\left(6-2x\right)=0\\ \left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=3\end{matrix}\right.\)
c,
<=> \(\left[\begin{matrix}x-1=0\\x^2+5x+2=0\\x^3-1=0\end{matrix}\right.\)
+/ x - 1 = 0 <=> x = 1
+/x2 + 5x + 2 =0 <=> (x + \(\frac{5}{2}\))2 - \(\frac{17}{4}\)= 0 <=> (x + \(\frac{5}{2}\))2 = \(\frac{17}{4}\)<=> x + \(\frac{5}{2}\)= \(\pm\)\(\sqrt{\frac{17}{4}}\)
<=> x = \(\pm\)\(\sqrt{\frac{17}{4}}\) - \(\frac{5}{2}\)
+/ x3 - 1 = 0 <=.> ( x - 1 )(x2 + x + 1 ) = 0
<=> x = 1
Vậy phương trình có Nghiệm là x = 1 và x = \(\pm\)\(\sqrt{\frac{17}{4}}\) - \(\frac{5}{2}\)
d,
x2 + (x + 3)(10 -2x ) = 9
<=> x2 + 10x - 2x2 + 30 - 6x -9 = 0
<=> x2 + 4x + 21 = 0
<=> 7x - x2 + 21 -3x = 0
<=> (x +3)(7-x) =0
<=> \(\left[\begin{matrix}7-x=0\\x+3=0\end{matrix}\right.\) <=> \(\left[\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy pt có nghiệm là x = -3 và x = 7