Tìm x biết:
a) x^2 > 0
b) x^2 < = 0
c) |x| = -|x+1|
d) |x| = -x^2
Bài 13: Tìm x biết: a) (x-2)(x-3)-D0. b) (x-3)(x-4)-0. c) (x-7)(6-x)=0. d) (x-3)(x-13)=0. The Bài 14: Tìm x biết: a) (12-x)(2-x)=0. b) (x-33)(11-x)=0. c) (21-x)(12-x)=0. d) (50-x)(x-150) =0. Bài 15: Tìm x biết: a) 2x +x = 45. b) 2x +7x = 918. c) 2x+3x 60+5. d) 11x+22x 33.2.
Tìm x, biết 25 x - 2. 10 x + 4 x = 0
A. x = 1 B. x = -1
C. x = 2 D. x = 0
Tìm x, biết:
a) 3x(x - 1) + x - 1 = 0;
b) (x - 2)( x 2 + 2x + 7) + 2( x 2 - 4) - 5(x - 2) = 0;
c) ( 2 x - 1 ) 2 - 25 = 0;
d) x 3 + 27 + (x + 3)(x - 9) = 0.
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
Tìm x biết a) x(x-25)=0 b)2x(x-4)-x(2x-1)=-28 c)x^2 -5x=0 d)(x-2)^2-(x+1)(x+3)=-7 e)(3x+5).(4-3x)=0 f)x^2-1/4=0
a: \(x\in\left\{0;25\right\}\)
c: \(x\in\left\{0;5\right\}\)
tìm số nguyên x biết
a, (x-1)^2=0
b, x.(x-1)=0
c, (x+1) .(x-2) =0
d,(x+5).(x^2+1)=0
a) \(\left(x-1\right)^2=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
b) \(x\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
c) \(\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
d) \(\left(x+5\right)\left(x^2+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x^2=-1\end{cases}}\Rightarrow x=-5\)
a) ( x - 1 )2 = 0
x - 1 = 0
x = 1
b) x . ( x - 1 ) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
c) ( x + 1 ) . ( x - 2 ) = 0
tương tự, xét 2 trường hợp như câu b
d) tương tự, xét 2 trường hợp như câu b
a) (x - 1)2 = 0
=> (x - 1)2 = 02
=> x - 1 = 0
=> x = 0 + 1
=> x = 1
Vậy x = 1
b) x(x - 1) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy x = 0 hoặc x = 1
Tìm x, biết:
a) 2-x = 2 ( x - 2 ) 3 ; b) 8 x 3 - 72x = 0;
c) ( x - 1 , 5 ) 6 + 2 ( 1 , 5 - x ) 2 = 0; d) 2 x 3 +3 x 2 +3 + 2x = 0;
e) x 3 - 4x- 14x(x - 2) = 0; g) x 2 (x + 1)- x(x + 1) + x(x - 1) = 0.
Bài 1: Tìm x biết a) x^3 - 4x^2 - x + 4= 0 b) x^3 - 3x^2 + 3x + 1=0 c) x^3 + 3x^2 - 4x - 12=0 d) (x-2)^2 - 4x +8 =0
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
Tìm x thuộc Z biết:
a, x (x + 2) = 0
b, (x - 1) (x - 2) = 0
c, (x - 2) (x^2 + 1) = 0
d, (x + 1) (x^2 - 4) = 0
e, x (x - 3) > 0
a) \(x\left(x+2\right)=0\)
=> \(\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
Vậy ...
b) (x - 1)(x - 2) = 0
=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
Vậy ...
còn lại tương tự
a) x(x+2) = 0
=> x=0 hoặc x+2 = 0
+ x + 2 = 0
x = 0 - 2
x = -2
Vậy x thuộc tập hợp 0 ; -2
b) (x-1)(x-2)=0
=> x-1 =0 hoặc x-2=0
+ x-1=0 + x-2=0
x=0+1 x=0+2
x=1 x=2
Vậy x thuộc tập hợp 1;2
ý c , d làm giống 2 ý đầu
e) x(x-3)>0
=> \(\hept{\begin{cases}x>0\\x-3>0\end{cases}}\)=> \(\hept{\begin{cases}x>0\\x>3\end{cases}}=>\hept{\begin{cases}\\\end{cases}}x>3\)
Vậy x > 3
Tìm x, biết : a, |x + 2| = x b, |x - 1|- 2x = 1212 c, |0,5x - 2|-|x + 3| = 0 d, |x - 3,5|+|x + 1,3| = 0
c: Ta có: \(\left|\dfrac{1}{2}x-2\right|-\left|x+3\right|=0\)
\(\Leftrightarrow\left|\dfrac{1}{2}x-2\right|=\left|x+3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-2=x+3\\\dfrac{1}{2}x-2=-x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{-1}{2}=5\\x\cdot\dfrac{3}{2}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Bài 2: Tìm x, biết: a) (x + 2)^2 – 2(x + 2)(x – 5) = 0. b) 2x^2 + 3x – 5 = 0. c) x + 2 ^2 x 2 + 2x^3 = 0. d) (3x-1)^2-4(x+5)^2=0
a: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)