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Lời giải:
a) $|4x^2-25|=0$
$\Leftrightarrow 4x^2-25=0$
$\Leftrightarrow (2x-5)(2x+5)=0$
$\Rightarrow x=\pm \frac{5}{2}$
b)
$|x-2|=3$
\(\Rightarrow \left[\begin{matrix} x-2=-3\\ x-2=3\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-1\\ x=5\end{matrix}\right.\)
c)
\(|x-3|=2x-1\Rightarrow \left\{\begin{matrix} 2x-1\geq 0\\ \left[\begin{matrix} x-3=2x-1\\ x-3=1-2x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\end{matrix}\right.\Rightarrow x=\frac{4}{3}\)
d)
$|x-5|=|3x-2|$
\(\Rightarrow \left[\begin{matrix} x-5=3x-2\\ x-5=2-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{7}{4}\end{matrix}\right.\)
a: =>|5x-2|=|2x-3|
=>5x-2=2x-3 hoặc 5x-2=-2x+3
=>3x=-1 hoặc 7x=5
=>x=5/7 hoặc x=-1/3
b: =>|5x-2|-|2x+2|=3x+5
TH1 x<-1
PT sẽ là 2-5x+2x+2=3x+5
=>-3x+4=3x+5
=>-6x=1
=>x=-1/6(loại)
TH2: -1<=x<2/5
Pt sẽ là 2-5x-2x-2=3x+5
=>-7x=3x+5
=>-4x=5
=>x=-5/4(loại)
Th3: x>=2/5
PT sẽ là 5x-2-2x-2=3x+5
=>3x-4=3x+5
=>0x=9(loại)
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
a, \(x<2\)
\(2-x+2x=7\)
\(x=5(\)ko \(t/m)\)
\(x>2\)
\(-x=5\)
\(x=-5(ko\) \(t/m)\)
a: |x-2|+2x=7
=>|x-2|=-2x+7
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{7}{2}\\\left(-2x+7\right)^2=\left(x-2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{7}{2}\\\left(2x-7-x+2\right)\left(2x-7+x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{7}{2}\\\left(x-5\right)\left(3x-9\right)=0\end{matrix}\right.\Leftrightarrow x=3\)
b: |x-3|-4x=5
=>|x-3|=4x+5
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{5}{4}\\\left(4x+5-x+3\right)\left(4x+5+x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{5}{4}\\\left(3x+8\right)\left(5x+2\right)=0\end{matrix}\right.\Leftrightarrow x=-\dfrac{2}{5}\)
c: |2x+3|+x=2x+3
=>|2x+3|=x+3
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\\left(2x+3-x-3\right)\left(2x+3+x+3\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;-2\right\}\)
a) x(4x + 2) = 4x2 - 14
⇔ 4x2 + 2x = 4x2 - 14
⇔ 4x2 - 4x2 + 2x = -14
⇔ 2x = -14
⇔ x = -7
Vậy tập nghiệm S = ......
b) (x2 - 9)(2x - 1) = 0
⇔ x2 - 9 = 0 hoặc 2x - 1 = 0
⇔ x2 = 9 hoặc 2x = 1
⇔ x = 3 hoặc -3 hoặc x = \(\dfrac{1}{2}\)
Vậy .......
c) \(\dfrac{3}{x-2}\) + \(\dfrac{4}{x+2}\) = \(\dfrac{x-12}{x^2-4}\)
⇔ \(\dfrac{3}{x-2}\) + \(\dfrac{4}{x+2}\) = \(\dfrac{x-12}{\left(x-2\right)\left(x+2\right)}\)
ĐKXĐ: x - 2 ≠ 0 và x + 2 ≠ 0
⇔ x ≠ 2 và x ≠ -2MSC (mẫu số chung): (x - 2)(x + 2)Quy đồng mẫu hai vế và khử mẫu ta được:3x + 6 + 4x - 8 = x - 12⇔ 3x + 4x - x = 8 - 6 - 12⇔ 6x = -10⇔ x = \(-\dfrac{5}{3}\) (nhận)Vậy ........a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
Bài 1: Giải các phương trình sau:
a) 3(2,2-0,3x)=2,6 + (0,1x-4)
<=> 6.6 - 0.9x = 2,6 + 0,1x - 4
<=> - 0.9x - 0,1x = -6.6 -1,4
<=> -x = -8
<=> x = 8
Vậy x = 8
b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)
<=> 3,6 - x - 0,5 = x - 5,5 + x
<=> - x - 3,1 = -5,5
<=> - x = -2.4
<=> x = 2.4
Vậy x = 2.4
a: 3(x-1)+2=2x-1
=>3x-3+2=2x-1
=>3x-1=2x-1
hay x=0
b: (x+1)(x-3)=0
=>x+1=0 hoặc x-3=0
=>x=-1 hoặc x=3
c: \(\Leftrightarrow x\left(x-1\right)-\left(2x-3\right)\left(x+1\right)=x+3\)
\(\Leftrightarrow x^2-x-2x^2-2x+3x+3=x+3\)
\(\Leftrightarrow-x^2-x=0\)
=>x=0(nhận) hoặc x=-1(loại)
a) Ta có: \(\left|x^2-x+2\right|-3x-7=0\)
\(\Leftrightarrow\left|x^2-x+2\right|=3x+7\)
\(\Leftrightarrow x^2-x+2=3x+7\)(Vì \(x^2-x+2>0\forall x\))
\(\Leftrightarrow x^2-x+2-3x-7=0\)
\(\Leftrightarrow x^2-4x-5=0\)
\(\Leftrightarrow x^2-5x+x-5=0\)
\(\Leftrightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
Vậy: S={5;-1}
bạn giải giúp mình câu b nữa với
mai mình phải nộp bài rồi!!!
a) x4 + 4x2 = 5
<=>x4+4x2-5=0
<=>x4+4x2+4-9=0
<=>(x2+2)2-9=0
<=>(x2+2-3)(x2+2+3)=0
<=>(x2-1)(x2+5)=0
<=>(x-1)(x+1)(x2+5)=0
<=>x-1=0 hoặc x+1=0 hoặc x2+5=0
<=>x=1 hoặc x=-1 hoặc x=\(-\sqrt{5}\)hoặc x= \(\sqrt{5}\)
Vậy S={1;-1;\(\sqrt{5};-\sqrt{5}\)}
c)c) |x + 2| + |-1| = 3 (1)
Nếu x+2\(\ge0\)<=>x\(\ge\)-2
(1) trở thành :
x+2+|-1|=3
<=>x+2+1=3
<=>x+3=3
<=>x=0 (loại)
Nếu x+2<0 <=>x<-2
(1) trở thành :
-x-2+|-1|=3
<=>-x-2+1=3
<=>-x-1=3
<=>-x=4
<=>x=-4(thỏa mãn )
Vậy S={-4}
a, x^4 + 4x^2 + 4 - 9 = 0 => (x^ 2 + 2 - 3) (x^2+2+3) = 0 => x = +- 1 hoặc x^2 + 5 = 0 (loại )
Vậy x = +- 1
b, Bạn xết ba trường hợp nhé x<=1 ; 1<x<=3/2 và x> 3 / 2
C, |x + 2| + |-1| = 3 => |x + 2 | = 2 => x + 2 = 2 hoặc x+ 2 = -2 => x = 0 hoặc x = -4