Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
b) ( 2x + 3)2 – (2x + 1)(2x – 1) = 22
<=> 4x2 + 12x + 9 - 4x2 + 1 = 22
<=> 12x + 10 = 22
<=> 12x = 12
<=> x = 1
Vậy x = 1
c) (4x + 3)(4x – 3) – (4x - 5)2 = 16
<=> 16x2 - 9 - 16x2 + 40x - 25 = 16
<=> 40x - 34 = 16
<=> 40x = 50
<=> \(\frac{5}{4}\)
Vậy \(x=\frac{5}{4}\)
e) (x + 1)3 – x2(x + 3) = 2
<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2
<=> 3x + 1 = 2
<=> 3x = 2
<=> x = \(\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
f) (x – 2)3 – x(x – 1)(x + 1) + 6x2 = 5
<=> x3 - 6x2 + 12x - 8 - x3 + x + 6x2 = 5
<=> 13x - 8 = 5
<=> 13x = 13
<=> x = 1
Vậy x = 1
đề là gì
a)\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}3x-2=0\\x+6=0\\x^2+5=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x=2\\x=-6\\x^2=-5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{2}{3}\\x=-6\\x\in\varnothing\end{cases}}}\)
vậy x=2/3 hoặc x=-6
a, (3x-2) (x+6) (x^2 +5) = 0
<=> 3x - 2 = 0 hoặc x + 6 = 0 hoặc x2 + 5 = 0 (loại vì x2 \(\ge\)0 => x2 + 5 > 0)
<=> x = 2/3 hoặc x = -6
b, (2x+5)^2 = (3x-1)^2
<=> (2x + 5)2 - (3x - 1)2 = 0
<=> (2x + 5 - 3x + 1)(2x + 5 + 3x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-3x+6=0\\2x+3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}-x=-6\\5x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=6\\x=\frac{4}{5}\end{cases}}}\)
c, 4x2 (x-1) - x+1 = 0
<=> 4x2(x - 1) - (x - 1) = 0
<=> (x - 1)(4x2 - 1) = 0
<=> (x - 1)(2x - 1)(2x + 1) = 0
vậy x - 1 = 0 hoặc 2x - 1 = 0 hoặc 2x + 1 = 0
hay x = 1 hoặc x = 1/2 hoặc x = -1/2
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
a) ( x - 3 )2 - 4 = 0
<=> ( x - 3 )2 - 22 = 0
<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0
<=> ( x - 5 )( x - 1 ) = 0
<=> x = 5 hoặc x = 1
b( 2x + 3 )2 - ( 2x + 1 )( 2x - 1 ) = 22
<=> 4x2 + 12x + 9 - ( 4x2 - 1 ) = 22
<=> 4x2 + 12x + 9 - 4x2 + 1 = 22
<=> 12x + 10 = 22
<=> 12x = 12
<=> x = 1
c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )2 = 16
<=> 16x2 - 9 - ( 16x2 - 40x + 25 ) = 16
<=> 16x2 - 9 - 16x2 + 40x - 25 = 16
<=> 40x - 34 = 16
<=> 40x = 50
<=> x = 50/40 = 5/4
d) x3 - 9x2 + 27x - 27 = -8
<=> ( x - 3 )3 = -8
<=> ( x - 3 )3 = (-2)3
<=> x - 3 = -2
<=> x = 1
e) ( x + 1 )3 - x2( x + 3 ) = 2
<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2
<=> 3x + 1 = 2
<=> 3x = 1
<=> x = 1/3
f) ( x - 2 )3 - x( x - 1 )( x + 1 ) + 6x2 = 5
<=> x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 = 5
<=> x3 + 12x - 8 - x3 + x = 5
<=> 13x - 8 = 5
<=> 13x = 13
<=> x = 1
a) \(\left(x-3\right)^2-4=0\)
=> \(\left(x-3\right)^2-2^2=0\)
=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)
=> \(\left(x-5\right)\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)
=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)
=> \(4x^2+12x+9-4x^2+1=22\)
=> \(12x+9+1=22\)
=> \(12x+10=22\)
=> 12x = 12
=> x = 1
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)
=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)
=> \(16x^2-9-16x^2+40x-25=16\)
=> \(-9+40x-25=16\)
=> \(40x=16+25-\left(-9\right)=16+25+9=50\)
=> x = 50/40 = 5/4
d) \(x^3-9x^2+27x-27=-8\)
=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)
=> \(\left(x-3\right)^3=-8\)
=> \(\left(x-3\right)^3=\left(-2\right)^3\)
=> x - 3 = -2 => x = 1
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)
=> \(3x+1=2\)
=> \(3x=1\)=> x = 1/3
f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)
=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)
=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)
=> \(\left(12x+x\right)-8=5\)
=> 13x = 13
=> x = 1
(x+1).(x+2).(x+5)-x^2.(x+8)= 27
\(\Leftrightarrow\left(x^2+2x+x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+2x^2+10x+x^2+5x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x-17=0\)
\(\Leftrightarrow x=1\)
(x + 1).(x + 2).(x + 5) - x2.(x + 8) = 27
<=> x3 + 8x2 + 17x + 10 - x2 - 8x2 = 27
<=> 17x + 10 = 27
<=> 17x = 27 - 10
<=> 17x = 17
<=> x = 1
=> x = 1
\(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^2\left(x+8\right)=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-\left(x^3+8x^2\right)-27=0\)
\(\Leftrightarrow17x-17=0\\ \Leftrightarrow x=1\)
Trả lời:
( x + 1 ) ( x + 2 ) ( x + 5 ) - x2 ( x + 8 ) = 27
<=> ( x2 + 3x + 2 ) ( x + 5 ) - x3 - 8x2 = 27
<=> x3 + 5x2 + 3x2 + 15x + 2x + 10 - x3 - 8x2 = 27
<=> 17x + 10 = 27
<=> 17x = 17
<=> x = 1
Vậy x = 1 là nghiệm của pt.