Tìm x biết
|x2+x|+|(x+1)*(x-2)|=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.
Bài 7. Tìm x,biết:
a) x-3x2=0 e) 5x(3x-1)+x(3x-1)-2(3x-1)=0
b) (x+3)2-x(x-2)=13 c) (x-4)2-36=0
d) x2-7x+12=0 g) x2-2018x-2019=0
Bài 8. Tìm x, biết
a) (2x-1)2=(x+5)2 b) x2-x+1/4
c) 4x4-101x2+25=0 d) x3-3x2+9x-91=0
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.
Tìm x, biết:
a) 8x(x - 2017) - 2x + 4034 = 0; b) x 2 + x 2 8 = 0;
c) 4 - x = 2 ( x - 4 ) 2 ; d) ( x 2 + 1)(x - 2) + 2x = 4.
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
(x-1)2-1+x2-(1-x)(x+3)=0
(x-1)2-1+x2-(1-x)(x+3)=0
⇒x2-2x+1-1+x2-x(1-x)+3(1-x)=0
⇒x2-2x+1-1+x2-x+x2+3-3x=0
⇒3x2-6x+3=0
⇒3(x2-2x+1)=0
⇒x2-2x+1=0
⇒(x-1)2=0
⇒x-1=0
⇒x=1
Lời giải:
$(x-1)^2-1+x^2-(1-x)(x+3)=0$
$\Leftrightarrow (x^2-2x+1)-1+x^2-(3-x^2-2x)=0$
$\Leftrightarrow x^2-2x+1-1+x^2-3+x^2+2x=0$
$\Leftrightarrow 3x^2-3=0$
$\Leftrightarrow x^2-1=0$
$\Leftrightarrow (x-1)(x+1)=0$
$\Leftrightarrow x=1$ hoặc $x=-1$
\(\left(x-1\right)^2-1+x^2+\left(1-x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(x^2-1\right)+\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1+x+1+x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)=0\)
\(\Leftrightarrow3\left(x^2-1\right)=0\) \(\Leftrightarrow x^2-1=0\) \(\Leftrightarrow x^2=1\) \(\Leftrightarrow x=\pm1\)
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 x biết:
a) x 6 + 2 x 3 +1 = 0; b) x(x - 5) = 4x - 20;
c) x 4 -2 x 2 =8-4 x 2 ; d) ( x 3 - x 2 ) - 4 x 2 + 8x-4 = 0.
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
Bài 11: Tìm x biết:
a) (x+2)(x2-2x+4) - x(x2+2) =15
b) (x+3)2 –x(3x+1)2 +(2x+1)(4x2 -2x+1) =28
c) (x2-1)3 - (x4+x2+1)(x2-1) = 0
d) (x-2)3 –(x-3)(x2 + 3x+9) +6(x+1)2 = 49
1 thưc hiện phép tính
a, 7x2.(2x3+3x5 ) b,(x3-x2+x-1):(x-1)
2 tìm x biết x : x2-8x+7=0
1. a) \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=14x^5+21x^7\)
b) \(\left(x^3-x^2+x-1\right):\left(x-1\right)=\dfrac{x^3-x^2+x-1}{x-1}\)
\(=\dfrac{x^2\left(x-1\right)+\left(x-1\right)}{x-1}=\dfrac{\left(x-1\right)\left(x^2+1\right)}{x-1}=x^2+1\)
2: \(x^2-8x+7=0\)
=>\(x^2-x-7x+7=0\)
=>\(x\left(x-1\right)-7\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x-7\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
1:
a: \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=21x^7+14x^5\)
b: \(\dfrac{x^3-x^2+x-1}{x-1}=\dfrac{x^2\left(x-1\right)+\left(x-1\right)}{\left(x-1\right)}\)
\(=x^2+1\)
1)
\(a,7x^2\cdot(2x^3+3x^5)\\=7x^2\cdot2x^3+7x^2\cdot3x^5\\=14x^5+21x^7\\---\\b,(x^3-x^2+x-1):(x-1)(dkxd:x\ne 1)\\=[x^2(x-1)+(x-1)]:(x-1)\\=(x-1)(x^2+1):(x-1)\\=x^2+1\)
2)
\(x^2-8x+7=0\\\Leftrightarrow x^2-x-7x+7=0\\\Leftrightarrow x(x-1)-7(x-1)=0\\\Leftrightarrow (x-1)(x-7)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
\(\text{#}Toru\)