1,giải bpt sau
1,\(x^2-4x-5\)< \(0\)
2, \(x^2-2x-1\)< \(0\)
3, \(2x^2-6x+5\)>\(0\)
giải BPT sau
a,(4x-1)(x^2+12)(-x+4)>0
b,(2x-1)(5-2x)(1-x)<0
\(a,\left(4x-1\right)\left(x^2+12\right)\left(-x+4\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1>0\\x^2+12>0\left(LD\forall x\right)\\-x+4>0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x>1\\-x>-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{4}\\x< 4\end{matrix}\right.\)
Vậy \(S=\left\{x|\dfrac{1}{4}< x< 4\right\}\)
\(b,\left(2x-1\right)\left(5-2x\right)\left(1-x\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1< 0\\5-2x< 0\\1-x< 0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{1}{2}\\x>\dfrac{5}{2}\\x< 1\end{matrix}\right.\)
Vậy \(S=\left\{x|1>x>\dfrac{5}{2}\right\}\)
giải pt sau
1, \(\sqrt{5-2x}=6\)
2,\(\sqrt{2-x}-\sqrt{x+1}=0\)
3, \(\sqrt{4x^2+4x+1}=6\)
4,\(\sqrt{x^2-10x+25}=x-2\)
1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)
\(\Leftrightarrow5-2x=36\)
\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)
2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)
\(\Leftrightarrow2-x=x+1\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)
\(\Leftrightarrow\left|x-5\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
Giải bpt:
a)x\(^2\)-4x+3≤0
b)9x\(^2\)-6x≥0
c)\(\frac{4x-3}{3}-\frac{2x+1}{4}< \frac{x-5}{-2}\)
a. \(x^2-4x+3\le0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(3x-3\right)\le0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\le0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\ge0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le1\\x\ge3\end{matrix}\right.\left(Vo.li\right)\\\left\{{}\begin{matrix}x\ge1\\x\le3\end{matrix}\right.\end{matrix}\right.\)
Vậy \(1\le x\le3\)
b. \(9x^2-6x\ge0\)
\(\Leftrightarrow3x\left(3x-2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x\ge0\\3x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}3x\le0\\3x-2\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge\frac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x\le\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(0\le x\le\frac{2}{3}\)
c. Câu c cậu tự làm nha, tớ đang có việc. Quy đồng lên rồi tính bình thường thôi.
Giải phương trình:
1. \(x^4-6x^2-12x-8=0\)
2. \(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
3. \(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
4. \(2x^2.\sqrt{-4x^4+4x^2+3}=4x^4+1\)
5. \(x^2+4x+3=\sqrt{\dfrac{x}{8}+\dfrac{1}{2}}\)
6. \(\left\{{}\begin{matrix}4x^3+xy^2=3x-y\\4xy+y^2=2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}\sqrt{x^2-3y}\left(2x+y+1\right)+2x+y-5=0\\5x^2+y^2+4xy-3y-5=0\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\sqrt{2x^2+2}+\left(x^2+1\right)^2+2y-10=0\\\left(x^2+1\right)^2+x^2y\left(y-4\right)=0\end{matrix}\right.\)
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)
\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)
Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)
\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)
Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:
\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)
\(\Leftrightarrow10b+40=3\left(b+8\right)b\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)
TH1: \(b=2\Leftrightarrow...\)
TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)
I) THỰC HIỆN PHÉP TÍNH a) 2x(x^2-4y) b)3x^2(x+3y) c) -1/2x^2(x-3) d) (x+6)(2x-7)+x e) (x-5)(2x+3)+x II phân tích đa thức thành nhân tử a) 6x^2+3xy b) 8x^2-10xy c) 3x(x-1)-y(1-x) d) x^2-2xy+y^2-64 e) 2x^2+3x-5 f) 16x-5x^2-3 g) x^2-5x-6 IIITÌM X BIẾT a)2x+1=0 b) -3x-5=0 c) -6x+7=0 d)(x+6)(2x+1)=0 e)2x^2+7x+3=0 f) (2x-3)(2x+1)=0 g) 2x(x-5)-x(3+2x)=26 h) 5x(x-1)=x-1 IV TÌM GTNN,GTLN. a) tìm giá trị nhỏ nhất x^2-6x+10 2x^2-6x b) tìm giá trị lớn nhất 4x-x^2-5 4x-x^2+3
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
Giải các phương trình
1. (x-4)2-25=0
2. (2x-1)2+(2-x)(2x-1)=0
3. x2+6x+9=4x2
4. (2x-5)(x+11)=(5-2x)(2x+1)
5. 2x2+5x+3=0
1. \(\left(x-4\right)^2-25=0\)
<=> (x-4+5).(x-4-5) = 0
<=> (x+1)(x-9) = 0
<=> \(\left[\begin{matrix}x+1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = {-1;9}
2. \(\left(2x-1\right)^2+\left(2-x\right)\left(2x-1\right)=0\)
<=> (2x-1)(2x-1+2-x) = 0
<=> (2x-1)(x+1) = 0
<=> \(\left[\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}2x=1\\x=-1\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0.5\\x=-1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = {-1 ; 0,5}
3. \(x^2+6x+9=4x^2\)
<=> \(\left(x+3\right)^2-4x^2=0\)
<=> (x+3+2x)(x+3-2x) = 0
<=> (3x+3)(3-x) = 0
<=> \(\left[\begin{matrix}3x+3=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}3x=-3\\x=3\end{matrix}\right.\Leftrightarrow}\left[\begin{matrix}x=-1\\x=3\end{matrix}\right.\) Vậy phương trình có tập nghiệm S = {-1 ; 3}
4. (2x-5)(x+11) = (5-2x)(2x+1)
<=> (2x-5)(x+11) = - (2x-5)(2x+1)
<=> x + 11 = -2x - 1
<=> x+2x = -12
<=> 3x = -12
<=> x = -4
Vậy phương trình có một nghiệm duy nhất là x = -4
5. \(2x^2+5x+3=0\)
<=> \(2x^2+2x+3x+3=0\)
<=> \(2x\left(x+1\right)+3\left(x+1\right)=0\)
<=> \(\left(x+1\right)\left(2x+3\right)=0\)
<=> \(\left[\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\Leftrightarrow}\left[\begin{matrix}x=-1\\x=\frac{-3}{2}\end{matrix}\right.\) Vậy phương trình có tập nghiệm S = { -1 ; -3/2 }
1) (x-4)^2-25=0
<=> (x-4+5)(x-4-5)=0
\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
2) (2x-1)2+(2-x)(2x-1)=0
<=> (2x-1)(2+2-x)=0
<=> \(\left[\begin{matrix}x=\frac{1}{2}\\x=4\end{matrix}\right.\)
3) x^2+6x+9=4x^2
<=> 3x^2 -6x-9=0
<=> x^2 -2x -3=0
<=> x^2 -3x+x-3=0
<=> x(x-3)+(x-3)=0
<=> (x-3)(x+1)=0
=>\(\left[\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
4) (2x-5)(x+11)=(5-2x)(2x+1)
-(5-2x)(x+11)-(5-2x)(2x+1)=0
(5-2x)(x+11+2x+1)=0
=>\(\left[\begin{matrix}x=\frac{5}{2}\\x=-4\end{matrix}\right.\)
5)2x^2+5x+3=0
2x^2+2x+3x+3=0
2x(x+1)+3(x+1)=0
(x+1)(2x+3)=0
=>\(\left[\begin{matrix}x=-1\\x=\frac{-3}{2}\end{matrix}\right.\)
a) \(\left(x-4\right)^2-25=0\)
=> (x-4)2 = 25
Suy ra \(\left\{\begin{matrix}x-4=5\\x-4=-5\end{matrix}\right.\)
=> \(\left\{\begin{matrix}x=5+4\\x=-5+4\end{matrix}\right.\)
=>\(\left\{\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
2. (2x-1)2 + (2-x)(2x-1)=0
\(\Rightarrow\left(2x-1\right)\left(2x-1+2-x\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left\{\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)
3. x2+6x+9=4x2
3x2-6x-9 =0
3(x2 - 2x -3) =0
\(3\left(x^2-3x+x-3\right)=0\)
(x-3)(x+1) =0
\(\left\{\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\)
\(\left\{\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
4. (2x-5)(x+11)=(5-2x)(2x+1)
(2x-5)(x+11) + (2x-5)(2x+1)=0
(2x-5)(3x+12) =0
\(\left\{\begin{matrix}2x-5=0\\3\left(x+4\right)=0\end{matrix}\right.\)
\(\left\{\begin{matrix}x=2,5\\x=-4\end{matrix}\right.\)
5. 2x2+5x+3=0
(2x2+2x)+(3x+3)=0
2x(x+1) +3(x+1) =0
(x+1)(2x+3) =0
\(\left\{\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\)
\(\left\{\begin{matrix}x=-1\\x=\frac{-3}{2}\end{matrix}\right.\)
giải phương trình :
a, \(\sqrt{x-3}+\sqrt[3]{x^2+1}+x^2+x-2=0\)
b,\(4x^2+\sqrt{2x+3}=8x+1\)
c, \(2x^2-6x+10-5\left(x-2\right)\sqrt{x+1=0}\)
a.
ĐKXĐ: \(x\ge3\)
(Tốt nhất bạn kiểm tra lại đề cái căn đầu tiên của \(\sqrt{x-3}\) là căn bậc 2 hay căn bậc 3). Vì nhìn ĐKXĐ thì thấy căn bậc 2 là không hợp lý rồi đó
Pt tương đương:
\(\sqrt{x-3}+\sqrt[3]{x^2+1}+\left(x+1\right)\left(x-2\right)=0\)
Do \(x\ge3\Rightarrow x-2>0\Rightarrow\left(x+1\right)\left(x-2\right)>0\)
\(\Rightarrow\sqrt{x-3}+\sqrt[3]{x^2+1}+\left(x+1\right)\left(x-2\right)>0\)
Pt vô nghiệm
b.
ĐKXĐ: \(x\ge-\dfrac{3}{2}\)
Pt: \(2x+3-\sqrt{2x+3}-\left(4x^2-6x+2\right)=0\)
Đặt \(\sqrt{2x+3}=t\ge0\) ta được:
\(t^2-t-\left(4x^2-6x+2\right)=0\)
\(\Delta=1+4\left(4x^2-6x+2\right)=\left(4x-3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t_1=\dfrac{1+4x-3}{2}=2x-1\\t_2=\dfrac{1-4x+3}{2}=2-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+3}=2x-1\left(x\ge\dfrac{1}{2}\right)\\\sqrt{2x+3}=2-2x\left(x\le1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4x^2-4x+1\left(x\ge\dfrac{1}{2}\right)\\2x+3=4x^2-8x+4\left(x\le1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{17}}{4}\\x=\dfrac{5-\sqrt{21}}{4}\end{matrix}\right.\)
c.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2\left(x^2-4x+4\right)+2\left(x+1\right)-5\left(x-2\right)\sqrt{x+1}=0\)
\(\Leftrightarrow2\left(x-2\right)^2-5\left(x-2\right)\sqrt{x+1}+2\left(x+1\right)=0\)
Đặt \(\left\{{}\begin{matrix}x-2=a\\\sqrt{x+1}=b\end{matrix}\right.\) ta được:
\(2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=x-2\left(x\ge2\right)\\\sqrt{x+1}=2x-4\left(x\ge2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-4x+4\\x+1=4x^2-16x+16\end{matrix}\right.\) (\(x\ge2\))
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\) (đã loại nghiệm)
Giai phường trình sau:
a, \(3x^2+2x-1=0\) e, \(4x^2-12x+5=0\) i,\(2x^2+5x-3=0\)
b,\(x^2-5x+6=0\) f, \(2x^2+5x+3=0\) j,\(x^2+6x-16=0\)
c,\(x^2-3x+2=0\) g,\(x^2+x-2=0\)
d,\(2x^2-6x+1=0\) h, \(x^2-4x+3=0\)
a) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)
b) Ta có: \(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: S={2;3}
c) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: S={1;2}
d) Ta có: \(2x^2-6x+1=0\)
\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)
mà \(2\ne0\)
nên \(x^2-3x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)
e) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)