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Khách
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15 tháng 11 2019
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
KN
15 tháng 11 2019
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
2T
24 tháng 8 2019
a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
21 tháng 10 2020
Tìm x :)
a) x2( x2 + 4 ) - x2 = 4
⇔ x2( x2 + 4 ) - x2 - 4 = 0
⇔ x2( x2 + 4 ) - ( x2 + 4 ) = 0
⇔ ( x2 + 4 )( x2 - 1 ) = 0
⇔ ( x2 + 4 )( x - 1 )( x + 1 ) = 0
⇔ x2 + 4 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0
⇔ x = ±1 ( do x2 + 4 ≥ 4 > 0 ∀ x )
b) x4 - x3 + x2 - x = 0
⇔ x3( x - 1 ) + x( x - 1 ) = 0
⇔ ( x - 1 )( x3 + x ) = 0
⇔ ( x - 1 )x( x2 + 1 ) = 0
⇔ x - 1 = 0 hoặc x = 0 hoặc x2 + 1 = 0
⇔ x = 1 hoặc x = 0 ( do x2 + 1 ≥ 1 > 0 ∀ x )
c) x3 - 8 = ( x - 2 )( x - 12 )
⇔ ( x - 2 )( x2 + 2x + 4 ) - ( x - 2 )( x - 12 ) = 0
⇔ ( x - 2 )( x2 + 2x + 4 - x + 12 ) = 0
⇔ ( x - 2 )( x2 + x + 16 ) = 0
⇔ x - 2 = 0 hoặc x2 + x + 16 = 0
⇔ x = 2 ( do x2 + x + 16 = ( x2 + x + 1/4 ) + 63/4 = ( x + 1/2 )2 + 63/4 ≥ 63/4 > 0 ∀ x )
NC
28 tháng 8 2020
a) Ta có: \(\left(3x+5\right)^2-\left(x+3\right)^2-8x\left(x+3\right)=12\)
\(\Leftrightarrow9x^2+30x+25-x^2-6x-9-8x^2-24x-12=0\)
\(\Leftrightarrow4=0\) (vô lý)
=> pt vô nghiệm
b) \(\left(2x-5\right)^2-\left(x-2\right)^2-\left(x-1\right)\left(3x+2\right)=8\)
\(\Leftrightarrow4x^2-20x+25-x^2+4x-4-3x^2+x+2-8=0\)
\(\Leftrightarrow-15x=-13\)
\(\Rightarrow x=\frac{13}{15}\)
NC
28 tháng 8 2020
c) \(-2x\left(x+3\right)+\left(2x-5\right)^2=-3\left(x+2\right)\)
\(\Leftrightarrow-2x^2-6x+4x^2-20x+25+3x+6=0\)
\(\Leftrightarrow2x^2-23x+31=0\)
\(\Leftrightarrow2\left(x^2-\frac{23}{2}x+\frac{529}{16}\right)-\frac{281}{8}=0\)
\(\Leftrightarrow\left(x-\frac{23}{4}\right)^2-\left(\frac{\sqrt{281}}{4}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{23+\sqrt{281}}{4}\right)\left(x-\frac{23-\sqrt{281}}{4}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{23+\sqrt{281}}{4}=0\\x-\frac{23-\sqrt{281}}{4}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{23+\sqrt{281}}{4}\\x=\frac{23-\sqrt{281}}{4}\end{cases}}\)
6 tháng 9 2020
1. (x + 2)(x2 - 2x + 4) - (x3 + 2x2) = 5
=> x(x2 - 2x + 4) + 2(x2 - 2x + 4) - x3 - 2x2 - 5 = 0
=> x3 - 2x2 + 4x + 2x2 - 4x + 8 - x3 - 2x2 - 5 = 0
=> (x3 - x3) + (-2x2 + 2x2 - 2x2) + (4x - 4x) + (8 - 5) = 0
=> -2x2 + 3 = 0
=> -2x2 = -3
=> x2 = 3/2
=> x = \(\pm\sqrt{\frac{3}{2}}\)
2. \(\left(x+5\right)^2-6=0\)
=> x2 + 10x + 25 - 6 = 0
=> x2 + 10x + 19 = 0
=> x vô nghiệm(do mình không để căn nên ghi vô nghiệm thôi nhá)
3. \(\left(x+3\right)\left(x^2-3x+9\right)-x^3=2x\)
=> x(x2 - 3x + 9) + 3(x2 - 3x + 9) - x3 - 2x = 0
=> x3 - 3x2 + 9x + 3x2 - 9x + 27 - x3 - 2x = 0
=> (x3 - x3) + (-3x2 + 3x2) + (9x - 9x - 2x) + 27 = 0
=> -2x + 27 = 0
=> -2x = -27
=> x = 27/2
4. \(\left(x-2\right)^3-x^3+6x^2=7\)
=> x3 - 6x2 + 12x - 8 - x3 + 6x2 = 7
=> (x3 - x3) + (-6x2 + 6x2) + 12x - 8 = 7
=> 12x - 8 = 7
=> 12x = 15
=> x = 5/4
5. \(3\left(x-2\right)^2+9\left(x-1\right)-3\left(x^2+x-3\right)=12\)
=> 3x2 - 12x + 12 + 9x - 9 - 3x2 - 3x + 9 = 12
=> (3x2 - 3x2) + (-12x + 9x - 3x) + (12 - 9 + 9) = 12
=> -6x + 12 = 12
=> -6x = 0
=> x = 0
6. \(\left(4x+3\right)^2-\left(4x-3\right)^2-5x-2=0\)
=> 48x - 5x - 2 = 0
=> 43x - 2 = 0
=> 43x = 2
=> x = 2/43
Còn bài cuối tự làm :>
6 tháng 9 2020
Anh Sang làm cầu kì quá ;-;
1. ( x + 2 )( x2 - 2x + 4 ) - ( x3 + 2x2 ) = 5
<=> x3 + 8 - x3 - 2x2 = 5
<=> 8 - 2x2 = 5
<=> 2x2 = 3
<=> x2 = 3/2
<=> \(x^2=\left(\pm\sqrt{\frac{3}{2}}\right)^2\)
<=> \(x=\pm\sqrt{\frac{3}{2}}\)
2. ( x + 5 )2 - 6 = 0
<=> ( x + 5 )2 - ( √6 )2 = 0
<=> ( x + 5 - √6 )( x + 5 + √6 ) = 0
<=> \(\orbr{\begin{cases}x+5-\sqrt{6}=0\\x+5+\sqrt{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{6}-5\\x=-\sqrt{6}-5\end{cases}}\)
3. ( x + 3 )( x2 - 3x + 9 ) - x3 = 2x
<=> x3 + 27 - x3 = 2x
<=> 27 = 2x
<=> x = 27/2
4. ( x - 2 )3 - x3 + 6x2 = 7
<=> x3 - 6x2 + 12x - 8 - x3 + 6x2 = 7
<=> 12x - 8 = 7
<=> 12x = 15
<=> x = 15/12 = 5/4
5. 3( x - 2 )2 + 9( x - 1 ) - 3( x2 + x - 3 ) = 12
<=> 3( x2 - 4x + 4 ) + 9x - 9 - 3x2 - 3x + 9 = 12
<=> 3x2 - 12x + 12 + 6x - 3x2 = 12
<=> -6x + 12 = 12
<=> -6x = 0
<=> x = 0
6. ( 4x + 3 )2 - ( 4x - 3 )2 - 5x - 2 = 0
<=> 16x2 + 24x + 9 - ( 16x2 - 24x + 9 ) - 5x - 2 = 0
<=> 16x2 + 24x + 9 - 16x2 + 24x - 9 - 5x - 2 = 0
<=> 43x - 2 = 0
<=> 43x = 2
<=> x = 2/43
7, ( 4x + 7 )( 2 - 3x ) - ( 6x + 2 )( 5 - 2x ) = 0
<=> -12x2 - 13x + 14 - ( -12x2 + 26x + 10 ) = 0
<=> -12x2 - 13x + 14 + 12x2 - 26x - 10 = 0
<=> -39x + 4 = 0
<=> -39x = -4
<=> x = 4/39
16 tháng 8 2018
Bài 1:
a) (3x-2).(4x+5)-6x.(2x-1) = 12x^2 +15x - 8x -10 - 12x^2 + 6x = 13x - 10
b) (2x-5)^2 - 4.(x+3).(x-3) = 4x^2 - 20x + 25 - 4x^2 + 12x -12x + 36 = -20x + 61
Bài 2:
a)(2x-1)^2-(x+3)^2 = 0
<=> (2x-1-x-3).(2x-1+x+3) =0
<=>(x-4).(3x+2) = 0
<=> x-4 = 0 hoặc 3x+2=0
*x-4=0 => x=4
*3x+2 = 0 => 3x=-2 => x=-2/3
b)x^2(x-3)+12-4x=0 <=> x^2(x-3) - 4(x-3) =0 <=> (x-3).(x-2)(x+2) <=> x-3=0 hoặc x-2=0 hoặc x+2 =0
*x-3=0 => x=3
*x-2=0 =>x=2
*x+2=0 =>x=-2
c) 6x^3 -24x =0 <=> 6x(x^2 -4)=0 <=> 6x(x-2)(x+2)=0 <=> x=0 hoặc x-2 =0 hoặc x+2=0 <=> x=0 hoặc x=2 hoặc x=-2
VM
1
2 tháng 10 2015
a)(x2-x+1)(x2-x+2)-12 (1)
Đặt x2-x+1=a thì (1) <=> a(a+1)-12=a2+a-12
=(a2-3a)+(4a-12)
=a(a-3)+4(a-3)
=(a-3)(a+4)
=(x2-x+1-3)(x2-x+1+4)
=(x2-x-2)(x2-x+5)
Vậy......
b) Đặt x2+x=a thì a2 + 4a-12 = (a2-2a)+(6a-12)
= a(a-2) + 6(a-2)
= (a+6)(a-2)
= (x2+x+6)(x2+x-2)
Vậy....
a) x3 - 8 = ( x - 2 )( x - 12 )
<=> ( x - 2 )( x2 + 2x + 4 ) - ( x - 2 )( x - 12 ) = 0
<=> ( x - 2 )( x2 + 2x + 4 - x + 12 ) = 0
<=> ( x - 2 )( x2 + x + 16 ) = 0
<=> \(\orbr{\begin{cases}x-2=0\\x^2+x+16=0\end{cases}}\Leftrightarrow x=2\)( vì x2 + x + 16 = ( x2 + x + 1/4 ) + 63/4 = ( x + 1/2 )2 + 63/4 ≥ 63/4 > 0 ∀ x )
b) x2( x2 + 4 ) - x2 = 4
<=> x2( x2 + 4 ) - x2 - 4 = 0
<=> x2( x2 + 4 ) - ( x2 + 4 ) = 0
<=> ( x2 + 4 )( x2 - 1 ) = 0
<=> \(\orbr{\begin{cases}x^2+4=0\\x^2-1=0\end{cases}}\Leftrightarrow x=\pm1\)( vì x2 + 4 ≥ 4 > 0 ∀ x )
Ko khó nè :3, đừng tách ra nhé !
a, \(x^3-8=\left(x-2\right)\left(x-12\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)=\left(x-2\right)\left(x-12\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-x+12\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\Leftrightarrow x=2\)
b, \(x^2\left(x^2+4\right)-x^2=4\Leftrightarrow-x^2\left(-x^2-4\right)-x^2=4\)
\(\Leftrightarrow-x^2\left(4-x^2\right)-x^2=4\Leftrightarrow-x^2\left(2-x\right)\left(2+x\right)-x^2-4=0\)
\(\Leftrightarrow-x^2\left(2-x\right)\left(2+x\right)+\left(-x^2-4\right)=0\)
\(\Leftrightarrow-x^2\left(2-x\right)\left(2+x\right)+\left(2-x\right)\left(2+x\right)=0\)
\(\Leftrightarrow\left(-x^2+1\right)\left(2-x\right)\left(2+x\right)=0\Leftrightarrow x=\pm1;\pm2\)
Check hộ dáp án nhá :), ko chắc lắm nếu khai triển sẽ dễ nhìn hơn đấy.