tìm x :
( 4x + 4 ) . 2 - 6 = 10
K
Khách
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12 tháng 6 2023
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
20 tháng 7 2018
Tìm x, biết:
1) 2x ( x - 5) - x ( 2x - 4 ) = 15
<=> 2x2 - 10x - 2x2 + 4x - 15 = 0
<=> -6x - 15 = 0
<=> -6x = 15
<=> x = -15/6
2) ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6
<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0
<=> -4x = -16
<=> x = 4
3) 4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x
<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0
<=> x + 4 = 0
<=> x = -4
4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5
<=> 2x2 + x + 6x + 3 - 2x2 - 4x + 5 = 0
<=> 3x + 8 = 0
<=> 3x = -8
<=> x = -8/3
5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0
<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0
.......
6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)
<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0
<=> -2x + 40 = 0
<=> -2x = -40
<=> x = 20
Còn lại tương tự ....
P
Phong
CTVHS
28 tháng 7 2023
a) \(3x\cdot\left(12x-4\right)-9x\cdot\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x=30\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=\dfrac{30}{15}\)
\(\Leftrightarrow x=2\)
b) \(\left(2x+1\right)-5\left(x-2\right)=10\)
\(\Leftrightarrow2x+1-5x+10=10\)
\(\Leftrightarrow-3x+11=10\)
\(\Leftrightarrow-3x=10-11\)
\(\Leftrightarrow-3x=-1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
NT
0
27 tháng 8 2021
\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)
KK
27 tháng 8 2021
a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)
ĐKXĐ: x ≠ -1
⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)
⇔ 65 + 52 = -3(x + 1)
⇔ 117 = -3x - 3
⇔ 117 + 3 = -3x
⇔ 120 = -3x
⇔ x = \(\dfrac{120}{-3}=-40\) (TM)
b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)
⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)
⇔ 4x = -2,75
⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)
c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)
⇔ \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)
⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x2 + 96x + 144x + 48
⇔ 936x + 1560x - 624x - 96x - 144x - 288x2 = 48 - 312 - 520 + 312
⇔ 1632x - 288x2 = -472
⇔ -288x2 + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)
⇔ x = 5,942459684 \(\approx\) 6
1 tháng 8 2015
1. x2-8x+1 = x2 -2x.4 + 42 - 42 +1 = ( x- 4 )2 - 15
mà ( x - 4 )2 > 0
=> ( x - 4 )2 -15 > 0
Vậy -15 là gt min của biểu thức khi x = 4
2. x2 - 4x + y2 - 6y + 2 = x2 - 2.2x + 22 + y2 - 2.3y + 32 -11 = (x-2)2 + ( y - 3)2 -11
mà ( x - 2)2 > 0
( y - 3)2 > 0
Vậy -11 là gt min của biểu thức khi x=2 và y = 3
Mình nghĩ là bài 3 là tìm gt lớn nhất chứ bạn ^^
20 tháng 10 2021
a: \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b: \(\left(x+1\right)^2-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
19 tháng 8
1: \(x^2+3x+2\)
\(=x^2+x+2x+2\)
=x(x+1)+2(x+1)
=(x+1)(x+2)
2: \(x^2+4x+3\)
\(=x^2+x+3x+3\)
=x(x+1)+3(x+1)
=(x+1)(x+3)
3: \(x^2+5x+4\)
\(=x^2+x+4x+4\)
=x(x+1)+4(x+1)
=(x+1)(x+4)
4: \(x^2-4x+3\)
\(=x^2-x-3x+3\)
=x(x-1)-3(x-1)
=(x-1)(x-3)
5: \(x^2-4x+4=x^2-2\cdot x\cdot2+2^2=\left(x-2\right)^2\)
6: \(x^2-5x+4\)
\(=x^2-x-4x+4\)
=x(x-1)-4(x-1)
=(x-1)(x-4)
7: \(x^2-5x+6\)
\(=x^2-2x-3x+6\)
=x(x-2)-3(x-2)
=(x-2)(x-3)
8: \(x^2+6x+5\)
\(=x^2+x+5x+5\)
=x(x+1)+5(x+1)
=(x+1)(x+5)
9: \(x^2-7x+10\)
\(=x^2-2x-5x+10\)
=x(x-2)-5(x-2)
=(x-2)(x-5)
10: \(x^2+8x+12\)
\(=x^2+2x+6x+12\)
=x(x+2)+6(x+2)
=(x+2)(x+6)
11: \(x^2-8x+16=x^2-2\cdot x\cdot4+4^2=\left(x-4\right)^2\)
12: \(x^2+8x+15\)
\(=x^2+3x+5x+15\)
=x(x+3)+5(x+3)
=(x+3)(x+5)
13: \(x^2-8x+7\)
\(=x^2-x-7x+7\)
=x(x-1)-7(x-1)
=(x-1)(x-7)
14: \(x^2+9x+8\)
\(=x^2+x+8x+8\)
=x(x+1)+8(x+1)
=(x+1)(x+8)
15: \(x^2-9x+14\)
\(=x^2-2x-7x+14\)
=x(x-2)-7(x-2)
=(x-2)(x-7)
16: \(x^2+9x+18\)
\(=x^2+3x+6x+18\)
=x(x+3)+6(x+3)
=(x+3)(x+6)
17: \(x^2-9x+20\)
\(=x^2-4x-5x+20\)
=x(x-4)-5(x-4)
=(x-4)(x-5)
18: \(2x^2-3x+1\)
\(=2x^2-2x-x+1\)
=2x(x-1)-(x-1)
=(x-1)(2x-1)
S
subjects
CTVHS
19 tháng 8
1. \(x^2+3x+2=\left(x+1\right)\left(x+2\right)\)
2. \(x^2+4x+3=\left(x+1\right)\left(x+3\right)\)
3. \(x^2+5x+4=\left(x+1\right)\left(x+4\right)\)
4. \(x^2-4x+3=\left(x-1\right)\left(x-3\right)\)
5. \(x^2-4x+4=\left(x-2\right)^2\)
6. \(x^2-5x+4=\left(x-1\right)\left(x-4\right)\)
7. \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)
8. \(x^2+6x+5=\left(x+1\right)\left(x+5\right)\)
9. \(x^2-7x+10=\left(x-2\right)\left(x-5\right)\)
10. \(x^2+8x+12=\left(x+2\right)\left(x+6\right)\)
11. \(x^2-8x+16=\left(x-4\right)^2\)
12. \(x^2+8x+15=\left(x+3\right)\left(x+5\right)\)
13. \(x^2-8x+7=\left(x-1\right)\left(x-7\right)\)
14. \(x^2+9x+8=\left(x+1\right)\left(x+8\right)\)
15. \(x^2-9x+14=\left(x-2\right)\left(x-7\right)\)
16. \(x^2+9x+18=\left(x+3\right)\left(x+6\right)\)
17. \(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)
\(18.2x^2-3x+1=2x^2-x-2x+1\)
\(=x\cdot\left(2x-1\right)-\left(2x-1\right)=\left(2x-1\right)\left(x-1\right)\)
\(\Rightarrow8x+8=16\)
\(\Rightarrow8x=8\)
\(\Rightarrow x=1\)
Vậy \(x=1.\)
(4x + 4) . 2 - 6 = 10
(4x + 4) . 2 = 10 + 6
(4x + 4) . 2 = 16
4x + 4 = 16 : 2
4x + 4 = 8
4x = 8 - 4
4x = 4
x = 4 : 4
x = 1