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\(2x+4⋮x-1\Rightarrow2\left(x-1\right)+6⋮x-1\)
\(\Rightarrow6⋮x-1\Rightarrow x-1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Rightarrow x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)
Vậy...........................................
\(2x^2+\left(-3\right)^2=41\)
\(\Rightarrow2x^2=41-9=32\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=\pm4\)
\(2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Rightarrow2x-10-3x-21=14\)
\(\Rightarrow2x-3x=14+21+10\)
\(\Rightarrow-x=45\Rightarrow x=-45\)
\(-7\left(5-x\right)-2\left(x-10\right)=15\)
\(\Rightarrow-35+x-2x+20=15\)
\(\Rightarrow x-2x=15-20+35\)
\(\Rightarrow-x=30\Rightarrow x=-30\)
\(\left(x+1\right)\left(x+7\right)< 0\)
thì \(x+1;x+7\)khác dấu
th1\(\hept{\begin{cases}x+1< 0\\x+7>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x>-7\end{cases}\Rightarrow}-7< x< -1\left(tm\right)}\)
th2\(\hept{\begin{cases}x+1>0\\x+7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< -7\end{cases}\Rightarrow}-1< x< -7\left(vl\right)}\)
vậy với\(-7< x< -1\)thì \(\left(x+1\right)\left(x+7\right)< 0\)
a) (2x - 3) = 5
<=> 2x - 3 = 5
<=> 2x = 5 + 3
<=> 2x = 8
<=> x = 4
=> x = 4
b) (5x - 3) = 1/2
<=> 5x - 3 = 1/2
<=> 5x = 1/2 + 3
<=> 5x = 7/2
<=> x = 7/10
=> x = 7/10
c) (x + 1)(x + 7) < 0
<=> x = -1; -7
<=> x < -7 <=> x = -8 <=> (-8 + 1)(-8 + 7) < 0 <=> 7 < 0 (loại)
<=> -7 < x < -1 <=> x = -6 <=> (-6 + 1)(-6 + 7) < 0 <=> -5 < 0 (nhận)
<=> x > -1 <=> x = 0 <=> (x + 1)(x + 7) < 0 <=> 7 < 0 (loại)
Vậy: -7 < x < -1
Lí luận chung cho cả 4 câu :
Để tích này bé hơn 0 thì các thừa số phải trái dấu với nhau
a) Dễ thấy \(x-2>x-7\)
\(\Rightarrow\hept{\begin{cases}x-2>0\\x-7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>2\\x< 7\end{cases}\Leftrightarrow}2< x< 7}\)
b) tương tự
c) \(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-10\right)< 0\)
\(\Leftrightarrow\left(x^4-11x^2+10\right)\left(x^4-11x^2+28\right)< 0\)
Dễ thấy \(x^4-11x^2+10< x^4-11x^2+28\)
\(\Rightarrow\hept{\begin{cases}x^4-11x^2+10< 0\\x^4+11x^2+10>0\end{cases}}\)
Tự giải nốt nha bạn mình bận rồi
a, \(x^2-9=0\Rightarrow x^2=9\Rightarrow x\pm3\)
b, \(\left(x-3\right)^2-25=0\Rightarrow\left(x-3\right)^2=25\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
c, \(\left(x-3\right)\left(2x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
d, \(\left(x-3\right)x-2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
e, \(3x\left(x-1\right)-5\left(1-x\right)=0\)
\(\Rightarrow3x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
g, \(x^2+6x-7=0\)
\(\Rightarrow x^2-x+7x-7=0\)
\(\Rightarrow x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
h,\(2x^2+5x-7=0\)
\(\Rightarrow2x^2-2x+7x-7=0\)
\(\Rightarrow2x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Chúc bạn học tốt!!!
a) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) vậy \(x=3;x=-3\)
b) \(\left(x-3\right)^2-25=0\Leftrightarrow\left(x-3\right)^2=25\Leftrightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
vậy \(x=8;x=-2\)
c) \(\left(x-3\right)\left(2x-5\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
vậy \(x=3;x=\dfrac{5}{2}\)
d)\(\left(x-3\right).x-2\left(x-3\right)=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) vậy \(x=2;x=3\)
e) \(3x\left(x-1\right)-5\left(1-x\right)=0\Leftrightarrow\left(3x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-5}{3};x=1\)
câu e t thấy sai sai nhưng vẫn làm ; bn coi lại đề nha
g) \(x^2+6x-7=0\Leftrightarrow x^2-x+7x-7=0\)
\(\Leftrightarrow x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\) vậy \(x=-7;x=1\)
h) \(2x^2+5x-7=0\Leftrightarrow2x^2-2x+7x-7=0\)
\(\Leftrightarrow2x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(2x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+7=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-7}{2};x=1\)
Bài 2:
a: =>x/7=1/21
=>x=1/3
c: =>x(3x-2)=0
=>x=0 hoặc x=2/3
Bài1:
a: \(=\left(-\dfrac{7}{3}\right)^{3-2}=\dfrac{-7}{3}\)
b: \(=\left(-\dfrac{4}{9}\right)^{1-3}=\left(-\dfrac{4}{9}\right)^{-2}=\dfrac{81}{16}\)
c: \(=\left(\dfrac{1}{5}\right)^{10-7}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}\)
a) 3x - 2 = 0 => 3x = 2 => x = 2/3
b) 2x - 1 = 0 => 2x = 1 => x = 1/2
c) 5 ( 4+2x) = 8+5x
<=> 20 + 10x = 8 + 5x
<=> 10x - 5x = 8 - 20
<=> 5x = -12
x = -12/5
d) \(\frac{1}{2}+\frac{3}{4}x=6-\frac{4}{5}x\)
\(\frac{3}{4}x+\frac{4}{5}x=6-\frac{1}{2}\)
\(\frac{31}{20}x=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{31}{20}=\frac{110}{31}\)
e) 3 + 2x = 4 - 8x
<=> 2x + 8x = 4 - 3
10 x = 1
x = 1/10
f \(5+\frac{1}{2}\left(x+5\right)=3\)
\(\frac{1}{2}\left(x+5\right)=3-5=-2\)
\(x+5=-2:\frac{1}{2}=-4\)
\(x=-4-5=1\)
Vậy ......
1.
a) (—7/3)3:(—7/3)2=(—7/3)3–2=—7/3
b) (—4/9):(—4/9)3= (—4/9)1–3=(—4/9)—2=81/16
c) (1/5)10:(1/5)7=(1/5)10–7=(1/5)3=1/125
2.
a) —x/7 =1/—21
==> —x.(—21)=7.1
==> —x.(—21)=7
==> —x=7:(—21)
==> —x=—1/3
==> x=1/3
b) 4 2/5 . 0,5–1 3/7= 22/5 . 1/2 —10/7= 22.1/5.2–10/7= 11/5 —10/7= 77/35 — 50/35= 27/35
c) 3x2–2x=0
==> x3(3–2)=0
x3.1=0
x3=0:1
x3=0
==> x=0
c) 9x2–1=0
9x2=0+1
9x2=1
x2=1:9
x2=1/9
x2=12/32 hoặc x2=(—1/3)2
Vậy x=1/3 hoặc x=—1/3
3) tổng có số ước la (10 +1)(1 + 1) = 11.2 = 22 ước dó
2) ta có x( x - 3) < 0 nên x và x -3 trái dấu nhau mặt khác x > x-3 nên :
x > 0 và x - 3 < 0 => x < 3 vạy chung lại ta có 0 < x < 3 do x nguyên nên x = 1, x = 2
a) l 3x + 1l = 15
=>\(\hept{\begin{cases}3x+1=15\\3x+1=-15\end{cases}}\Rightarrow\hept{\begin{cases}x=\left(15-1\right):3=\frac{14}{3}\\x=\left(-15-1\right):3=\frac{-16}{3}\end{cases}}\)
a, |3x+1|=15
=>3x+1=15 hoặc -15
- Với 3x+1=15
=>3x=14
=>x=14/3
- Với 3x+1=-15
=>3x=-16
=>x=-16/3
b, (6x+12).(x-2)=0
=>6x+12=0 hoặc x-2=0
=>x=-2 hoặc x=2
c, 5.(x-3)+4=2(x+1)+7
=>5x-15+4=2x+2+7
=>5x-11=2x+9
=>3x=20
=>x=20/3
d chịu
1) (3x - 15)(5x - 10) = 0
TH1: 3x - 15 = 0
⇒ 3x = 15
⇒ x = 5
TH2: 5x - 10 = 0
⇒ 5x = 10
⇒ x = 2
Vậy x ∈ {5; 2}
2) (2x - 4)(x - 3) = 0
TH1: 2x = 0
⇒ 2x = 4
⇒ x = 2
TH2: x - 3 = 0
⇒ x = 3
Vậy x ∈ {2; 3}
3) (9x - 3)(x - 7) = 0
TH1: 9x - 3 = 0
⇒ 9x = 3
⇒ x = \(\frac13\)
TH2: x - 7 = 0
⇒ x = 7
Vậy x ∈ {\(\frac13\); 7}
cẩn thận nhé Mạo Danh, 1) TH2 phải là 5x+10=0
=> 5x=-10
=> x=-2
2) vt nhầm thành 2x=0( ko vấn đề gì lắm nếu người đăng bài ko có ngốc để mà tự bt:))
\(1)\left(3x-15\right)\left(5x+10\right)=0\)
TH1: \(3x-15=0\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
TH2: \(5x+10=0\)
\(\Leftrightarrow5x=-10\)
\(\Leftrightarrow x=-2\)
Vậy \(x\in\left\lbrace5;-2\right\rbrace\)
\(2)\left(2x-4\right)\left(x-3\right)=0\)
TH1: \(2x-4=0\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
TH2: \(x-3=0\)
\(\Leftrightarrow x=3\)
Vậy \(x\in\left\lbrace2;3\right\rbrace\)
\(3)\left(9x-3\right)\left(x-7\right)=0\)
TH1: \(9x-3=0\)
\(\Leftrightarrow9x=3\)
\(\Leftrightarrow x=\frac39\)
\(\Leftrightarrow x=\frac13\)
TH2: \(x-7=0\)
\(\Leftrightarrow x=7\)
Vậy \(x\in\left\lbrace\frac13;7\right\rbrace\)
- Trường hợp 1:
- Trường hợp 2:
Vậy \(x \in \{5; -2\}\).\(3x-15=0\)
\(3x=15\)
\(x=\frac{15}{3}\)
\(x=5\)
\(5x+10=0\)
\(5x=-10\)
\(x=\frac{-10}{5}\)
\(x=-2\)
Bài 2: \((2x - 4) \cdot (x - 3) = 0\)
- Trường hợp 1:
- Trường hợp 2:
Vậy \(x \in \{2; 3\}\).\(2x-4=0\)
\(2x=4\)
\(x=\frac{4}{2}\)
\(x=2\)
\(x-3=0\)
\(x=3\)
Bài 3: \((9x - 3) \cdot (x - 7) = 0\)
- Trường hợp 1:
- Trường hợp 2:
Vậy \(x \in \left\{\frac{1}{3}; 7\right\}\).\(9x-3=0\)
\(9x=3\)
\(x=\frac{3}{9}\)
\(x=\frac{1}{3}\)
\(x-7=0\)
\(x=7\)