Tìm x :
a) 0 : x = 0
b) 3x = 9
c) 4x = 64
d) 2x = 16
e) 9x-1 = 9
f) x4 = 16
g) 2x : 25 = 1
Giúp mik nha ai nhanh mik tick
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2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
a) \(\left|x-3\right|+\left|2x-6\right|=8\)
\(x-3+2x-6=8\)
\(3x-9=8\)
\(3x=17\)
\(\Rightarrow x=\frac{17}{3}\)
b) Tương tự câu a .
c) \(\left|2x-3\right|=6-\left|3-2x\right|\)
\(2x-3=6-3-2x\)
\(2x-3=x\)
\(-2x=3\)
\(x=\frac{-3}{2}\)
d) \(\left|3x-2\right|-\left|6-9x\right|=-\left|-16\right|\)
\(3x-2-6-9x=-16\)
\(3x-8-9x=-16\)
\(-6x-8=-16\)
\(-6x=-8\)
\(\Rightarrow x=\frac{8}{6}\)
\(\)
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
a: \(3x\left(x-3\right)+4x-12=0\)
=>\(3x\left(x-3\right)+\left(4x-12\right)=0\)
=>\(3x\left(x-3\right)+4\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(3x+4\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b: Sửa đề:\(\left(x+1\right)\left(x^2-x+1\right)-x^3+2x=17\)
\(\Leftrightarrow x^3+1-x^3+2x=17\)
=>2x+1=17
=>2x=17-1=16
=>\(x=\dfrac{16}{2}=8\)
c: \(\left(x-3\right)\left(x+5\right)+\left(x-1\right)^2-6x^4y^2:3x^2y^2=15x\)
=>\(x^2+2x-15+x^2-2x+1-2x^2=15x\)
=>\(15x=-14\)
=>\(x=-\dfrac{14}{15}\)
a) \(\left(2x-1\right)^2-25=0\)
⇔ \(\left(2x-1\right)^2-5^2=0\)
⇔ \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)
⇒ \(2x-1-5=0\) hoặc \(2x-1+5=0\)
⇔ \(x=3\) hoặc \(x=-2\)
Bài 1: Tìm x
a) (2x-1) ² - 25 = 0
<=> (2x-1)2 = 25
<=> 2x-1 = 5 hay 2x-1 =-5
<=> 2x= 6 hay 2x=-4
<=> x=3 hay x= -2
Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0
<=> (x-1)(3x+1)=0
<=> x-1=0 hay 3x+1=0
<=> x=1 hay 3x=-1
<=> x=1 hay x=\(\dfrac{-1}{3}\)
Vậy S={1;\(\dfrac{-1}{3}\)}
c) 2(x+3) - x ² - 3x = 0
<=> 2(x+3)- x(x+3)=0
<=> (x+3)(2-x)=0
<=> x+3=0 hay 2-x=0
<=> x=-3 hay x=2
Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0
<=> x(x-2)+3(x-2)=0
<=> (x-2)(x+3)=0
<=> x-2=0 hay x+3=0
<=> x=2 hay x=-3
Vậy S={2;-3}
e) 4x ² - 4x +1 = 0
<=> (2x-1)2=0
<=> 2x-1=0
<=> 2x=1
<=> x=\(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
f) x +5x2 = 0
<=> x(1+5x)=0
<=>x=0 hay 1+5x=0
<=> x=0 hay 5x=-1
<=> x=0 hay x= \(\dfrac{-1}{5}\)
Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0
<=> x2-x+3x-3=0
<=> x(x-1)+3(x-1)=0
<=> (x-1)(x+3)=0
<=> x-1=0 hay x+3=0
<=> x=1 hay x=-3
Vậy S={1;-3}
Thời gian có hạn copy cái này hộ mình vào google xem nha :
https://lazi.vn/quiz/d/16491/nhac-edm-la-loai-nhac-the-loai-gi
Vào xem xong các bạn nhận được 1 thẻ cào mệnh giá 100k nhận thưởng bằng cách nhắn tin vs mình và 1 phần thưởng bí mật là chiếc áo đá bóng,....
Có 500 giải nhanh nha đã có 401 người nhận rồi
OK
\(\left(3x+1\right)\left(x-2\right)< 0.\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1>0,x-2< 0\\3x+1< 0,x-2>0\end{cases}}\)
\(Th1\hept{\begin{cases}3x+1>0\\x-2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}3x>-1\\x< 2\end{cases}\Leftrightarrow\hept{\begin{cases}x>\frac{-1}{3}\\x< 2\end{cases}\Rightarrow}}}\frac{-1}{x}< x< 2\)
\(Th2:\hept{\begin{cases}3x+1< 0\\x-2>0\end{cases}\Leftrightarrow\hept{\begin{cases}3x< -1\\x>2\end{cases}\Leftrightarrow\hept{\begin{cases}x< \frac{-1}{2}\\x>2\end{cases}\left(loại\right)}}}\)
Vậy \(\frac{-1}{x}< x< 2\)
bạn đăng tách ra nhé
a, \(\left(2x+1\right)\left(x-4\right)=\left(2x+1\right)^2\)
\(\Leftrightarrow2x^2-7x-4=4x^2+4x+1\Leftrightarrow2x^2+11x+5=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)=0\Leftrightarrow x=-5;x=-\frac{1}{2}\)
b, sửa đề : \(\left(x-4\right)\left(x^2+4x+16\right)-\left(x^2-6\right)=2\)
\(\Leftrightarrow x^3-64-x^2+6=2\Leftrightarrow x^3-x^2-60=0\Leftrightarrow x=4,27...\)
c, \(\left(2x-1\right)^2-\left(3x+4\right)^2=0\Leftrightarrow\left(2x-1+3x+4\right)\left(2x-1-3x-4\right)=0\)
\(\Leftrightarrow\left(5x+3\right)\left(-x-5\right)=0\Leftrightarrow x=-\frac{3}{5};x=-5\)
d, \(\left(9x+2\right)\left(x-1\right)-\left(3x-1\right)^2=0\)
\(\Leftrightarrow9x^2-7x-2-9x^2+6x-1=0\Leftrightarrow-x-3=0\Leftrightarrow x=-3\)
e, \(\left(2x+3\right)^2-4\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow4x^2+12x+9-4\left(x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow4x^2+12x+9-4\left(x^3-x-x^2+1\right)=0\)
\(\Leftrightarrow4x^2+12x+9-4x^3+4x+4x^2-4=0\)
\(\Leftrightarrow-4x^3+8x^2+16x+5=0\Leftrightarrow x=-0,9...;x=-0,41...;x=3,31...\)
f, \(15x\left(x+4-6x-24\right)=0\Leftrightarrow15\left(-5x-20\right)=0\)
\(\Leftrightarrow-75x-300=0\Leftrightarrow x=-4\)
g, \(\left(4x-10\right)\left(2-3x\right)-30^2=0\)
\(\Leftrightarrow8x-12x^2-20+30x-900=0\Leftrightarrow-12x^2+38x-920=0\)
vô nghiệm
a) 0 : x = 0
=> x \(\in\)N ( x \(\ne\)0 )
b) 3x = 9
=> x = 2
c) 4x = 64
=> x = 3
d) 2x = 16
=> x = 4
e) 9x-1 = 9
9 = 91
=> x = 1 + 1 = 2
x4 = 16
=> x = 2
g) 2x : 25 = 1
Quy ước a0 = 1 , ta suy ra : x = 5
Giải ko giải thích , bỏ qua nhá :))
Mình chưa có học lớp 6 nên mk ko biết.Sorry nha