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Bài 2:
a: \(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)
=>(x+5)(x-6)=0
=>x=-5 hoặc x=6
b: \(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
=>-4x+2=0
hay x=1/2
c: \(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)
=>x=1 hoặc x=-1
\(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)
\(=\left(3^{60}-3^{57}\right):3^{57}+\left(5^{27}-5^{24}\right):5^{24}\)
\(=3^{57}\left(3^3-1\right):3^{57}+5^{24}\left(5^3-1\right):5^{24}\)
\(=3^3-1+5^3-1\)
\(=27-1+125-1\)
\(=150\)
2 )
\(x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
Vậy ...
b )
\(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
\(\Leftrightarrow2-4x=0\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
c )
\(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow x^2\left(x^2+4\right)-\left(4+x^2\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=-4\left(L\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy ...
a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)
\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=2^4.5+2-5^2\)
\(=57\)
b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)
\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)
c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)
\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)
\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)
a) 25x² - 16
= (5x)² - 4²
= (5x - 4)(5x + 4)
b) 16a² - 9b²
= (4a)² - (3b)²
= (4a - 3b)(4a + 3b)
c) 8x³ + 1
= (2x)³ + 1³
= (2x + 1)(4x² - 2x + 1)
d) 125x³ + 27y³
= (5x)³ + (3y)³
= (5x + 3y)(25x² - 15xy + 9y²)
e) 8x³ - 125
= (2x)³ - 5³
= (2x - 5)(4x² + 10x + 25)
g) 27x³ - y³
= (3x)³ - y³
= (3x - y)(9x² + 3xy + y²)
a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)
b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)
c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)
d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)
e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)
g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
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
1.(x -5)^2 - 25 =0
=> (x - 5)^2 = 25
=> x - 5 = 5 hoặc x - 5 = -5
=> x = 10 hoặc x = 0
vậy_
2. (x -2)^3 =27
=> x - 2 = 3
=> x = 5
vậy_
3. 3(x -7) + 2x(x+2) = 2x^2
=> 3x - 21 + 2x^2 + 4x = 2x^2
=> 7x - 21 = 0
=> 7x = 21
=> x = 3
vậy_
4. (x^2 - 4) (x +8) =0
=> x^2 - 4 = 0 hoặc x + 8 = 0
=> x^2 = 4 hoặc x = -8
=> x = 2 hoặc x = -2 hoặc x = -8
vậy_
5. x^ 2 + 3x = 0
=> x(x + 3) = 0
=> x = 0 hoặc x + 3 = 0
=> x = 0 hoặc x = -3
vậy_
6. 3x^3 - 3x = 0
=> 3x(x^2 - 1) = 0
=> 3x(x - 1)(x + 1) = 0
=> x = 0 hoặc x = 1 hoặc x = -1
vậy_
7. (x +1)^2 = ( 2x +3)^2
=> (x + 1 + 2x + 3)(x + 1 - 2x - 3) = 0
=> (3x + 3)(-x - 2) = 0
=> x = -1 hoặc x = -2
vậy_
Bài làm
1) ( x - 5 )2 - 25 = 0
<=> ( x - 5 - 5 )( x - 5 + 5 ) = 0
<=> x( x - 10 ) =
<=> \(\orbr{\begin{cases}x=0\\x-10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)
Vậy S = { 0; 10 }
2) \(\left(x-2\right)^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=5\)
Vậy x = 5 là nghiệm phương trình.
3) \(3\left(x-7\right)+2x\left(x+2\right)=2x^2\)
\(\Leftrightarrow3x+2x^2+4x-2x^2=21\)
\(\Leftrightarrow7x=21\)
\(\Leftrightarrow x=\frac{21}{7}=3\)
Vậy x = 3 là nghiệm phương trình
4) \(\left(x^2-4\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=\pm2\\x=-8\end{cases}}}\)
Vậy S = { 2; -2; -8 }
5) \(x^2+3x=0\)
\(\Leftrightarrow x\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)
Vậy S = { 0; -3 }
6) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)
Vậy S = { +1; 0 }
7) \(\left(x+1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-\left(2x+3\right)^2=0\)
\(\Leftrightarrow\left(x+1-2x-3\right)\left(x+1+2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x-2=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}}\)
Vậy S = { -2; -4/3 }
# Học tốt #
a. x mũ 2 - 2x + 1 = 25
= x^2 + 2.x.1 + 1^2
= ( x + 1 ) ^2
ko bt có đúng ko nữa, mấy câu kia tui ko bt lm
a)
\(\text{( 25 – 2x )³ : 5 – 3^2 = 4^2}\)
\(\text{( 25 – 2x )³ : 5 – 9 = 16}\)
\(\text{( 25 – 2x )³ : 5 = 16 + 9}\)
\(\text{( 25 – 2x )³ : 5 = 25}\)
\(\text{( 25 – 2x )³ = 25 . 5}\)
\(\text{( 25 – 2x )³ = 125}\)
\(\text{( 25 – 2x )³ = 5³}\)
\(\text{25 – 2x = 5}\)
\(\text{2x = 25 – 5}\)
\(\text{2x = 20}\)
\(\text{x = 10}\)
\(\text{________________________________________}\)
b)
\(\text{2.3^x = 10.3^12 + 8.27^4}\)
\(\text{2.3^x = 10.3^12 + 8.(3^3)^4}\)
\(\text{2.3^x = 3^12 . (10+8)}\)
\(\text{2.3^x = 3^12 . 18}\)
\(\text{3^x = 3^12 . 18:2}\)
\(\text{3^x = 3^12 . 9}\)
\(\text{3^x = 3^12 . 3^2}\)
\(\text{3^x = 3^14}\)
\(\text{=> x=14}\)