Tìm x, biết
a) ( x^2 -1/2.x ) :2x + ( 3x - 1 )^2 : ( 1-3x) =0
b) ( x-1)^3 + 1 + 3x( x - 4) = 0
Những câu hỏi liên quan
1Rút gọn biểu thức a) (3x+1)^2+(3x-1)^2-2(3x+1)(3x-1) b) 8(3^2+1)(3^4+1)...(2^16+1) c ) (2^2+1)(2^4+1)...(2^32+1) 2 Tìm x biết a) x(2x-1)-2x+1=0 b) 3x(x-1)=x-1 c) 3(x+2)-x^2-2x=0 d) x^3+x=0 3 Phân tích thành nhân tử a) 4x^3-x b) 6x^2-12xy+6y^2-24z^2
Bài 2:
a: Ta có: \(x\left(2x-1\right)-2x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
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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: \(x\in\left\{0;5\right\}\)
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tìm x biết
a/x^3+3x^2+3x+2=0
b/x^4-2x^3+2x-1=0
c/x^4-3x^3-6x^2+8x=0
a) \(x^3+3x^2+3x+2=0\)
<=> \(x^3+x^2+x+2x^2+2x+2=0\)
<=> \(x\left(x^2+x+1\right)+2\left(x^2+x+1\right)=0\)
<=> \(\left(x+2\right)\left(x^2+x+1\right)=0\)
tự làm
b) \(x^4-2x^3+2x-1=0\)
<=> \(\left(x^4-3x^3+3x^2-x\right)+\left(x^3-3x^2+3x-1\right)=0\)
<=> \(x\left(x^3-3x^2+3x-1\right)+\left(x^3-3x^2+3x-1\right)=0\)
<=> \(\left(x^3-3x^2+3x-1\right)\left(x+1\right)=0\)
<=> \(\left(x-1\right)^3\left(x+1\right)=0\)
tự làm
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c) \(x^4-3x^3-6x^2+8x=0\)
<=> \(x\left(x^3-3x^2-6x+8\right)=0\)
<=> \(x\left[\left(x^3+x^2-2x\right)-\left(4x^2+4x-8\right)\right]=0\)
<=>\(x\left[x\left(x^2+x-2\right)-4\left(x^2+x-2\right)\right]=0\)
<=> \(x\left(x-4\right)\left(x^2+x-2\right)=0\)
<=> \(x\left(x-4\right)\left(x-1\right)\left(x+2\right)=0\)
tự làm
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Bài 1 làm tính chia :
a,[5.(x-y)^4-3.(x-y)^3+4.(x-y)^2]:(y-x)^2
b,[(x+y)^5-2.(x+y)^4+3.(x+y)^3]:(3x-1)=0
Bài 2 tìm x biết :
(x^2-1/2x):2x-(3x-1)^2.(3x-1)=0
Tìm x
a.(x+2).(x+3)-(x-2).(x+5) = 0
b. (2x+3).(x-4)+(x-5)(x+2) = (3x-5)(x-4)
c. (3x+2)(2x+9)-(x+2)(6x+1) = x+1-(x-6)
d. 3( 2x-1).(3x-1)-(2x-3).(9x-1)=0
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
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Tìm x,biết:
a) (x-3)^2-4=0
b) x^2-9=0
c) x(x-2x)-y^2-8=0
d) 2x(x-1)-2x^2+x-5=0
e) x(x-3)-(x+1)(x-2)=0
f) x(3x-1)-3x^2-7x=0
a) ( x - 3 )2 - 4 = 0
<=> ( x - 3 )2 = 4
<=> \(\orbr{\begin{cases}\left(x-3\right)^2=2^2\\\left(x-3\right)^2=\left(-2\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy S = { 5 ; 1 }
b) x2 - 9 = 0
<=> x2 = 9
<=> \(\orbr{\begin{cases}x^2=3^2\\x^2=\left(-3\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy S = { 3 ; -3 }
c) x( x - 2x ) - x2 - 8 = 0
<=> x2 - 2x2 - x2 - 8 = 0
<=> -2x2 - 8 = 0
<=> -2x2 = 8
<=> x2 = -4 ( vô lí )
<=> x = \(\varnothing\)
Vậy S = { \(\varnothing\)}
d) 2x( x - 1 ) - 2x2 + x - 5 = 0
<=> 2x2 - 2x - 2x2 + x - 5 = 0
<=> -x - 5 = 0
<=> -x = 5
<=> x = -5
Vậy S = { -5 }
e) x( x - 3 ) - ( x + 1 )( x - 2 ) = 0
<=> x2 - 3x - ( x2 - x - 2 ) = 0
<=> x2 - 3x - x2 + x + 2 = 0
<=> - 2x + 2 = 0
<=> -2x = -2
<=> x = 1
Vậy S = { 1 }
f) x( 3x - 1 ) - 3x2 - 7x = 0
<=> 3x2 - x - 3x2 - 7x = 0
<=> -8x = 0
<=> x = 0
Vậy S = { 0 }
d) 2x(x - 1) - 2x2 + x - 5 = 0
=> 2x2 - 2x - 2x2 + x - 5 = 0
=> -x = 5
=> x = -5
e) x(x - 3) - (x + 1)(x - 2) = 0
=> x2 - 3x - (x2 - x - 2) = 0
=> x2 - 3x - x2 + x + 2 = 0
=> -2x = - 2
=> x = 1
f) x(3x - 1) - 3x2 - 7x = 0
=> 3x2 - x - 3x2 - 7x = 0
=> -8x = 0
=> x = 0
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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: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
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Tìm x biết :
a) x^2 - 3x + 2 (x-3) = 0
b) (x-1)(x+1) + x (x-9) = 2x^2 - 4
c) x (x-3) - 3x + 9 = 0
d) x (x+2) - (x-3)(x+3) = 5
đ) 2x (x+1) - (2x+1)(x-3) = 6
\(x^2-3x+2.\left(x-3\right)=0\)
\(x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(x.\left(x-3\right)-3x+9=0\)
\(x.\left(x-3\right)-3.\left(x-3\right)=0\)
\(\left(x-3\right)^2=0=>x=3\)
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a,\(x^2-3x+2\left(x-3\right)=0.\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
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B1. Tính giá trị: A=(x+2)^2 - (x-2)^2 với x=125
B2. Thu gọn:
a, 2x.(2x-1)^2 - 3x.(x+3).(x-3) - 4x.(x+1)^2
b. (3x+1)^2 - 2.(3x+1).(3x+5) + (3x+5)^2
B3: Tìm x biết (x+2)^2 - x^2 +4=0
b1
A=(125+2)2 - (125-2)2 = 1272 - 1232 = 1000
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