3^2x-4 -x^0=8
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2: \(3x\left(x-4\right)+2x-8=0\)
=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
3: 4x(x-3)+x2-9=0
=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(4x+x+3\right)=0\)
=>\(\left(x-3\right)\left(5x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)
4: \(x\left(x-1\right)-x^2+3x=0\)
=>\(x^2-x-x^2+3x=0\)
=>2x=0
=>x=0
5: \(x\left(2x-1\right)-2x^2+5x=16\)
=>\(2x^2-x-2x^2+5x=16\)
=>4x=16
=>x=4

\(2x+\left(1+2+3+...+100\right)=15150\)
\(2x+\left[\left(1+100\right)+\left(2+99\right)+...+\left(50+51\right)\right]=15150\)
\(2x+\left[101+101+...+101\right]=15150\)CÓ 50 SỐ 101
\(2x+\left[101\times50\right]=15150\)
\(2x=15150:5050\)
\(2x=3\)
\(x=3:2\)
\(x=1.5\)
a, 2x + (1+2+3+4+...+100) = 15150
=> 2x + \(\frac{\left(1+100\right).\left[\left(100-1\right)+1\right]}{2}\)= 15150
=> 2x + \(\frac{101.100}{2}\)= 15150
=> 2x + 5050 = 15150
=> 2x = 15150 - 5050
=> 2x = 10100
=> x = 10100 : 2
=> x = 5050
Vậy x = 5050
b, .(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8)=36
=> (x + x + x + x +x + x +x +x ) + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 36
=> 8x + 36 = 36
=> 8x = 0
=> x = 0
Vậy x = 0
c, 0+0+4+6+8+...+2x=110
Sửa đề :0 + 2 + 4 + 6 + 8 + ... + 2x = 110 = 2 + 4 + 6 + 8 + ... + 2x = 110
SSH : \(\frac{\left(2\text{x}-2\right)}{2}+1=x-1+1=x\)
Tổng : \(\frac{\left(2\text{x}+2\right).x}{2}=110\Leftrightarrow\frac{2.\left(x+1\right).x}{2}=110\)
\(\Leftrightarrow\left(x+1\right)x=110\)
\(\Leftrightarrow\left(10+1\right).10=110\)
=> x = 10
Vậy x = 10

1: =>(x+3)(x-5)=0
=>x=5 hoặc x=-3
2: =>(x-1)(5x-1)=0
=>x=1/5 hoặc x=1
5: =>(x-4)*x=0
=>x=0 hoặc x=4
10: =>(x+5)(x-3)=0
=>x=3 hoặc x=-5
9: =>(x-2)(x-4)=0
=>x=2 hoặc x=4
7: =>(x-6)(2x-1)=0
=>x=1/2 hoặc x=6
8: =>(2x-1)(3x-12)=0
=>x=4 hoặc x=1/2

Bài 1L
a) \(\left(x-7\right)\left(x+3\right)< 0\)
TH1:
\(\hept{\begin{cases}x-7>0\\x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>7\\x< -3\end{cases}}}\)( loại )
TH2:
\(\hept{\begin{cases}x-7< 0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 7\\x>-3\end{cases}\Leftrightarrow}-3< x< 7}\)( chọn )
Vậy \(-3< x< 7\)
Bài 2:
a) \(\left(5x+8\right)-\left(2x-15\right)+21=2x-5\)
\(\Leftrightarrow5x+8-2x+15+21=2x-5\)
\(\Leftrightarrow5x-2x-2x=-5-21-8-15\)
\(\Leftrightarrow x=-49\)
Vậy ...

3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

(2x - 7) + 17 = 6
=> 2x - 7 = 6 - 17
=> 2x - 7 = -11
=> 2x = -11 + 7
=> 2x = -4
=> x = -4 : 2
=> x = -2
+) 12 -2(3 - 3x)= -2
=> 2(3 - 3x) = 12 + 2
=> 2(3 - 3x) = 14
=> 3 - 3x = 14 : 2
=> 3 - 3x = 7
=> 3x = 3 - 7
=> 3x = -4
=> x = -4/3
\(\left(x+1\right)\left(x-3\right)=0\)
=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...

a: Ta có: \(4\left(x+1\right)^2+\left(2x+1\right)^2-8\left(x-1\right)\left(x+1\right)-11=0\)
\(\Leftrightarrow4x^2+8x+4+4x^2+4x+1-8x^2+8-11=0\)
\(\Leftrightarrow12x=-2\)
hay \(x=-\dfrac{1}{6}\)
b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)-1=0\)
\(\Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\)
\(\Leftrightarrow2x=-40\)
hay x=-20

3x(x - 10) = x - 10
(x - 10)(3x - 1) = 0
Th1:
x - 10 = 0
x = 10
TH2:
3x - 1 = 0
3x = 1
x = 1/3
Vậy x = 10 hoặc x = 1/3
x(x + 7) - (4x + 28) = 0
x(x + 7) - 4(x + 7) = 0
(x + 7)(x - 4) = 0
Th1:
x + 7 = 0
x = - 7
Th2:
x - 4 = 0
x = 4
Vậy x = - 7 hoặc x = 4
x(x - 4) = 2x - 8
x(x - 4) - 2(x - 4) = 0
(x - 2)(x - 4) = 0
Th1:
x - 2 = 0
x = 2
Th2:
x - 4 = 0
x = 4
Vậy x = 2 hoặc x = 4
(2x + 3)(x - 1) + (2x - 3)(x - 1) = 0
(x - 1)(2x + 3 + 2x - 3) = 0
4x(x - 1) = 0
Th1:
x = 0
Th2:
x - 1 = 0
x = 1
Vậy x = 0 hoặc x = 1
a)
\(3x\left(x-10\right)=x-10\)
\(\Rightarrow3x\left(x-10\right)-\left(x-10\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(x-10\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x-1=0\\x-10=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=10\end{array}\right.\)
b)
\(x\left(x+7\right)-\left(4x+28\right)=0\)
\(\Rightarrow x\left(x+7\right)-4\left(x+7\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x+7\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=-7\end{array}\right.\)
c)
\(x\left(x-4\right)=2x-8\)
\(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=2\end{array}\right.\)
d)
\(\left(2x+3\right)\left(x-1\right)+\left(2x+3\right)\left(x-1\right)=0\)
\(\Rightarrow2\left(2x+3\right)\left(x-1\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x+3=0\\x-1=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=1\end{array}\right.\)

a) ( 4 - x) (x + 3) = 0
=> \(\hept{\begin{cases}4-x=0\\x+3=0\end{cases}}\) => \(\hept{\begin{cases}x=4\\x=-3\end{cases}}\)
b) 7(2x - 8) = 0
=> 2x - 8 = 0
=> 2x = 8
=> x = 4
c) -x(8 + x) = 0
=> 8 + x = 0
=> x = -8
câu c) ko chéc nghen!!
565454785698532275475568679760351352636547765768843
a) ( 4 - x ) . ( x + 3 ) = 0
Ta có các trường hợp sau đây :
\(\orbr{\begin{cases}4-x=0\\x+3=0\end{cases}}\orbr{\begin{cases}x=4-0=4\\x=0-3=-3\end{cases}}\)
==> x = -3 ; 4
b) 7. ( 2x - 8 ) = 0
2x - 8 = 0 : 7
2x - 8 = 0
2x = 0 + 8 =8
==> x = 8 : 2 = 4
c) -x . ( 8 + x ) = 0
Ta có các trường hợp sau đây :
\(\orbr{\begin{cases}-x=0\\8+x=0\end{cases}}\orbr{\begin{cases}x=0\\x=-8\end{cases}}\)
MÌNH HK CHẮC CÂU c) ÂU NHA, MÌNH THẤY CÁI NÀO CŨNG LẠ MÀ 0 CÓ THỂ ĐỂ LÀ -0 ĐƯỢC HK CÁC BẠN
\(3^{2x-4}-x^0=8\)
\(3^{2x-4}-1=8\)
\(3^{2x-4}=8+1\)
\(3^{2x-4}=9\)
⇒ 2x - 4 = 2
⇒ 2x = 6
⇒ x = 3
vậy x = 3