tìm x nguyên biết
( x2 + 3).(x2 - 17).(x2 - 26) < 0
tìm x biết:
a)x2 + 3x = 0 b) x3 – 4x = 0
c) 5x(x-1) = x-1 d) 2(x+5) - x2-5x = 0
e) 2x(x-5)-x(3+2x)=26 f) 5x.(x – 2012) – x + 2012 = 0
a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
Tìm x, biết:
a) ( x + 3 ) 2 + (4 - x)(x + 4) = 1;
b) (2 - x) 3 +(3 +x)(9 - 3x + x 2 ) + 6x(1 - x) = 17;
c) x 4 - 2 x 2 +1 = 0.
a) Tìm được x = -4.
b) Tìm được x = 3.
c) Tìm được x = ±1.
Tìm số nguyên x, biết:
(x2 + 3) . (x – 15) = 0
TH1: \(x^2+3=0\) (vô lý)
TH2: \(x-15=0\Leftrightarrow x=15\)
\(TH1:x^2+3=0\)
\(Do\) \(x^2\ge0\Rightarrow x^2+3\ge3\Rightarrow\) (\(vô\) \(lý\))
\(\Rightarrow x-15=0\\ \Rightarrow x=15\)
h*) (x + 3)(1 – x) > 0
i*) (x2 – 1)(x2 – 4) < 0
k*) (x2 – 20)(x2 – 30) < 0
Bài 4: Tìm các số nguyên x sao cho
a) –3 ⋮ (x – 2)
b) (3x + 7) ⋮ (x – 2)
c*) (x2 + 7x + 2) ⋮ (x + 7)
a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
x-2 | 1 | -1 | 3 | -3 |
x | 3 | 1 | 5 | -1 |
b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
x-2 | 1 | -1 | 13 | -13 |
x | 3 | 1 | 15 | -11 |
c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+7 | 1 | -1 | 2 | -2 |
x | -6 | -8 | -5 | -9 |
Tìm x, biết:
a) x 2 (x - 5) + 5 - x = 0; b) 3 x 4 - 9 x 3 = -9 x 2 + 27x;
c) x 2 (x + 8) + x 2 = -8x; d) (x + 3)( x 2 -3x + 5) = x 2 + 3x.
cho phương trình x*2-4x+m=0
tìm m đề pt có hai nghiệm x1, x2 thõa mãn
x1*3+x2*3-5(x1*2+x2*2)=26
Tìm số nguyên x, biết:
(x2 - 1) . (x2 - 4) <0
x2 là x mũ 2
Mọi người ơi help me
a) (x2-1)(x2-4)<0
=> x2-1 và x2-4 trái dấu nhau
Ta thấy: x2 >=0 với mọi x => x2-1 > x2-4
=> \(\hept{\begin{cases}x^2-1>0\\x^2-4< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2>1\\x^2< 4\end{cases}\Leftrightarrow}\hept{\begin{cases}x>\pm1\\x< \pm2\end{cases}}}\)
=> Không có giá trị củ x thỏa mãn đề bài
Câu 2.(1,5 điểm) Tìm x, biết:
a) 5x(x2 – 9) = 0. b) 3(x+3) - x2 - 3x =0. c) x2 – 9x – 10 = 0
\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
a, 5\(x\)(\(x^2\) - 9) = 0
\(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -3; 0; 3}
b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0
3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0
(\(x+3\))( 3 - \(x\)) = 0
\(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -3; 3}
c, \(x^2\) - 9\(x\) - 10 = 0
\(x^2\) + \(x\) - 10\(x\) - 10 = 0
\(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0
(\(x+1\))(\(x-10\)) = 0
\(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -1; 10}
a) 5x(x2-9)=0
=> TH1 5x=0
<=> x= 0
TH2: 2x-9=0
<=> 2x=9
<=> x = \(\dfrac{9}{2}\)
b, 3(x+3) - x2- 3x = 0
<=> 3x + 9 - x2 -3x = 0
<=> - x2 +9 = 0
<=> - x2 = -9
<=> x = 3
c, x2 -9x -10 = 0
<=> x2 -x + 10x -10 = 0
<=> x(x-1)+10(x-1)=0
<=> (x-1)(x+10)=0
=> TH1: x-1=0
<=> x=1
TH2: x +10=0
<=> x=-10
Tìm các số nguyên x, biết:
a, (22 + 5)(x2 + 25) = 0
b, (x2 + 7)(x2 - 49) < 0
c, (x2 - 7)(x2 - 49) < 0
d, (x2 - 36)(x2 - 81) ≤ 0
Tìm x biết rằng:
a) ( x 2 + 2x + 4)(2 - x) + x(x - 3)(x + 4) - x 2 + 24 = 0;
b) x 2 + 3 ( 5 − 6 x ) + ( 12 x − 2 ) x 4 + 3 = 0 .