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Tính nhanh :
a) 252 - 152 = (25 + 15)(25 - 15) = 40 . 10 = 400
b) 872 + 732 - 272 - 132 = (872 - 132) + (732 - 272)
= (87 + 13)(87 - 13) + (73 + 27)(73 - 27)
= 100 . 74 + 100 . 26 = 100 . (74 + 26) = 100 . 100 = 10000
Bài 1:
a)\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x\cdot2y=2\left(x+y\right)\)
b) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\\ =\left(4x+2\right)\cdot2x=4x\left(2x+1\right)\)
Bài 2:
a) \(25^2-15^2=\left(25-15\right)\left(25+15\right)=10\cdot40=400\)
b) \(87^2+73^2-27^2-13^2=\left(87^2-27^2\right)+\left(73^2-13^2\right)\\ =\left(87-27\right)\left(87+27\right)+\left(73-13\right)\left(73+13\right)\)
\(=60\cdot114+60\cdot86=60\cdot\left(114+86\right)=60\cdot200=12000\)
Bài 2:
a) \(x^3-0,25\cdot x=0\)
\(\Leftrightarrow x^2\left(x-0,25\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-0,25=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=0,25\end{array}\right.\)
b) \(x^2-10=-25\)
\(\Leftrightarrow x^2=-15\) (vô nghiệm0
c) \(4x^2-4x=-1\)
\(\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)
d) \(8x^3+12x^2+6x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^3=0\)
\(\Leftrightarrow2x+1=0\Leftrightarrow x=-\frac{1}{2}\)
\(2.\left(x-4\right)-x+3=0\)
\(2x-8-x+3=0\)
\(x-5=0\)
\(x=5\)
\(x^2-25-x-5=0\)
\(\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(x-5-1\right)=0\)
\(\left(x+5\right)\left(x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=6\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=-5\\x=6\end{cases}}\)
Ta có : (x - 1)2 - 4 = 0
<=> (x - 1)2 - 22 = 0
=> (x - 1 - 2)(x - 1 + 2) = 0
=> (x - 3)(x + 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Ta có \(\left(x-1\right)^2-4=0\)
<=>\(\left(x-1\right)^2-2^2=0\)
<=>\(\left(x-1-2\right)\left(x-1+2\right)=0\)
<=>\(\left(x-3\right)\left(x+1\right)=0\)
<=>\(\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)
<=>\(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
a) \(x^3-5x^2+8x-4=0\)
\(\Leftrightarrow x^3-x^2-4x^2+4x+4x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy nghiệm của phương trình là: \(x=\left\{1;2\right\}\)
b: =>2x^3+2x^2-3x^2-3x+6x+6=0
=>(x+1)(2x^2-3x+6)=0
=>x+1=0
=>x=-1
c: =>(x^2+x)^2+(x^2+x)-6=0
=>(x^2+x-2)=0
=>(x+2)(x-1)=0
=>x=1 hoặc x=-2
d: =>(x^2-4x-3)(x^2-4x-5)=0
=>(x-5)(x+1)(x^2-4x-3)=0
hay \(x\in\left\{2+\sqrt{7};2-\sqrt{7};5;-1\right\}\)
a) ( 5 - 2x )( 2x + 7 ) - 4x2 + 25 = 0
<=> ( 5 - 2x )( 2x + 7 ) + ( 5 - 2x )( 5 + 2x ) = 0
<=> ( 5 - 2x )( 2x + 7 + 5 + 2x ) = 0
<=> ( 5 - 2x )( 4x + 12 ) = 0
<=> \(\orbr{\begin{cases}5-2x=0\\4x+12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
b) ( 5x2 + 3x - 2 )2 - ( 4x2 - x - 5 )2 = 0 ( như này chứ nhỉ ? )
<=> [ ( 5x2 + 3x - 2 ) - ( 4x2 - x - 5 ) ][ ( 5x2 + 3x - 2 ) + ( 4x2 - x - 5 ) ] = 0
<=> ( 5x2 + 3x - 2 - 4x2 + x + 5 )( 5x2 + 3x - 2 + 4x2 - x - 5 ) = 0
<=> ( x2 + 4x + 3 )( 9x2 + 2x - 7 ) = 0
<=> ( x2 + x + 3x + 3 )( 9x2 + 9x - 7x - 7 ) = 0
<=> [ x( x + 1 ) + 3( x + 1 ) ][ 9x( x + 1 ) - 7( x + 1 ) ] = 0
<=> ( x + 1 )( x + 3 )( x + 1 )( 9x - 7 ) = 0
<=> ( x + 1 )2( x + 3 )( 9x - 7 ) = 0
<=> x + 1 = 0 hoặc x + 3 = 0 hoặc 9x - 7 = 0
<=> x = -1 hoặc x = -3 hoặc x = 7/9
c) 15x4 - 8x3 - 14x2 - 8x + 15 = 0
<=> 15x4 + 22x3 - 30x3 + 15x2 + 15x2 - 44x2 - 30x + 22x + 15 = 0
<=> ( 15x4 + 22x3 + 15x2 ) - ( 30x3 + 44x2 + 30x ) + ( 15x2 + 22x + 15 ) = 0
<=> x2( 15x2 + 22x + 15 ) - 2x( 15x2 + 22x + 15 ) + ( 15x2 + 22x + 15 ) = 0
<=> ( 15x2 + 22x + 15 )( x2 - 2x + 1 ) = 0
<=> ( 15x2 + 22x + 15 )( x - 1 )2 = 0
<=> \(\orbr{\begin{cases}15x^2+22x+15=0\\\left(x-1\right)^2=0\end{cases}}\)
+) ( x - 1 )2 = 0 <=> x = 1
+) 15x2 + 22x + 15 = 15( x2 + 22/15x + 121/225 ) + 104/15 = 15( x + 11/25 )2 + 104/15 ≥ 104/15 > 0 ∀ x
Vậy phương trình có nghiệm duy nhất là x = 1
a, x^2 - 25 = 0
=> x^2 = 25
Mà : 5^2=25
=> x^2=25
=> x^2=5^2
=> x=5
b, x^2 - x - 6 = 0
=> x^2-x=6
Ta có : 3^2=6
=> x = 3
=> 3^2-3=6
cái này mk ko chắc lắm :v
c, x^2 - 8x + 15 lộn đề òi bạn
a) \(x^2-25=0\)
\(x^2=0+25=25\)
\(x^2=5^2=\left(-5\right)^2\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=5\\x=-5\end{array}\right.\)
b) \(x^2-x-6=0\)
\(x^2-\left(3x-2x\right)-6=0\)
\(x^2-3x+2x-6=0\)
\(\left(x^2+2x\right)-\left(3x+6\right)=0\)
\(x\left(x+2\right)-3\left(x+2\right)=0\)
\(\left(x-3\right)\left(x+2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-3=0\\x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\end{array}\right.\)
Câu c bằng bao nhiêu bn ơi!!!!!!