tìm x biết : a)11x-(9x+12)=x-14
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Vì cứ 2 số tự nhiên thì tổng của chúng bằng -2x
Vậy dãy số trên có số số hạng là:
( 2015 - 1 ) : 2 + 1 = 1008 ( số hạng )
Dãy trên có số cặp là:
1008 : 2 = 504 ( cặp )
Do đó dãy trên có 504 số -2x
Ta được:\(x-3x+5x-7x+9x-11x+...+2013x-2015x=3024\)
\(\Leftrightarrow\left(-2x\right)+\left(-2x\right)+\left(-2x\right)+....+\left(-2x\right)=3024\)
Mà dãy này có 504 số -2x
\(\Leftrightarrow504.\left(-2x\right)=3024\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\)
Vậy x=-3
Bài 1:
a: \(x^3-6x^2+11x-6\)
\(=x^3-x^2-5x^2+5x+6x-6\)
\(=\left(x-1\right)\left(x^2-5x+6\right)\)
\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
b: \(x^3-6x^2-9x+14\)
\(=x^3-7x^2+x^2-7x-2x+14\)
\(=\left(x-7\right)\left(x^2+x-2\right)\)
\(=\left(x-7\right)\left(x+2\right)\left(x-1\right)\)
c: \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
a) 4(x+1)-(3x+1)=14
<=>4x+4-3x-1=14
<=>4x-3x = 14-4
<=>x=10
b) -12(x-5) +7.(3x)=5
<=> -12x +60 + 21+7x=5
<=> -12x+7x=5-60-21
<=> -5x=-76
<=> x=76/5
c) 17x + 34 -4x+28-9x=82
<=> 17x-9x-4x=82-34-28
<=> 4x=20
<=> x=5
\(a,2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Leftrightarrow2x-10-3x-21=14\)
\(\Leftrightarrow-x-31=14\Leftrightarrow-x=45\Leftrightarrow x=-45\)
\(c,5\left(x-6\right)-2\left(x+3\right)=12\)
\(\Leftrightarrow5x-30-2x-6=12\)
\(\Leftrightarrow3x-36=12\Leftrightarrow3x=48\Leftrightarrow x=16\)
Các câu còn lại tương tự
a, 2(x - 5) - 3(x + 7) = 14\(\Rightarrow2\chi-10-3\chi+21=14\Rightarrow2\chi-3\chi=14+10-21\)
\(\Rightarrow-\chi=3\Rightarrow\chi=3\)
a) x3 - 6x2 + 11x - 6
= ( x3 - 2x2 ) - ( 4x2 - 8x ) + ( 3x - 6 )
= x2( x - 2 ) - 4x( x - 2 ) + 3( x - 2 )
= ( x - 2 )( x2 - 4x + 3 )
= ( x - 2 )( x2 - x - 3x + 3 )
= ( x - 2 )[ x( x - 1 ) - 3( x - 1 ) ]
= ( x - 2 )( x - 1 )( x - 3 )
b) x3 - 6x2 - 9x + 14
= ( x3 - x2 ) - ( 5x2 - 5x ) - ( 14x - 14 )
= x2( x - 1 ) - 5x( x - 1 ) - 14( x - 1 )
= ( x - 1 )( x2 - 5x - 14 )
= ( x - 1 )( x2 + 2x - 7x - 14 )
= ( x - 1 )[ x( x + 2 ) - 7( x + 2 ) ]
= ( x - 1 )( x + 2 )( x - 7 )
c) x3 + 6x2 + 11x + 6
= ( x3 + 2x2 ) + ( 4x2 + 8x ) + ( 3x + 6 )
= x2( x + 2 ) + 4x( x + 2 ) + 3( x + 2 )
= ( x + 2 )( x2 + 4x + 3 )
= ( x + 2 )( x2 + x + 3x + 3 )
= ( x + 2 )[ x( x + 1 ) + 3( x + 1 ) ]
= ( x + 2 )( x + 1 )( x + 3 )
e) x6 - 9x3 + 8
Đặt t = x3
bthuc <=> t2 - 9t + 8
= t2 - t - 8t + 8
= t( t - 1 ) - 8( t - 1 )
= ( t - 1 )( t - 8 )
= ( x3 - 1 )( x3 - 8 )
= ( x - 1 )( x2 + x + 1 )( x - 2 )( x2 + 2x + 4 )
1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)
\(\Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{x+6-x-3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow x^2+9x+18=54\)
\(\Leftrightarrow x^2+9x-36=0\)
=>(x+12)(x-3)=0
=>x=-12 hoặc x=3
\(ĐKXĐ:x\ne-3,-4,-5,-6\)
\(\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{x+6-x-3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{3}{x^2+9x+18}=\dfrac{1}{18}\\ \Leftrightarrow x^2+9x+18=54\)
\(\Leftrightarrow x^2+9x-36=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-12\left(tm\right)\end{matrix}\right.\)
A) 3x² - x(3x - 5) = 9
3x² - 3x² + 5x = 9
5x = 9
x = 9/5
--------------------
B) 5x² + 9x - 2 = 0
5x² + 10x - x - 2 = 0
(5x² + 10x) - (x + 2) = 0
5x(x + 2) - (x + 2) = 0
(x + 2)(5x - 1) = 0
x + 2 = 0 hoặc 5x - 1 = 0
*) x + 2 = 0
x = -2
*) 5x - 1 = 0
5x = 1
x = 1/5
Vậy x = -2; x = 1/5
---------------------
D) 4(5 - 3x) = 5x - 5
20 - 12x = 5x - 5
-12x - 5x = -5 - 20
-17x = -25
x = 25/17
--------------------
E) 2x² - 11x + 14 = 0
2x² - 4x - 7x + 14 = 0
(2x² - 4x) - (7x - 14) = 0
2x(x - 2) - 7(x - 2) = 0
(x - 2)(2x - 7) = 0
x - 2 = 0 hoặc 2x - 7 = 0
*) x - 2 = 0
x = 2
*) 2x - 7 = 0
2x = 7
x = 7/2
Vậy x = 2; x = 7/2
\(11x-\left(9x+12\right)=x-14\)
\(11x-9x-12=x-14\)
\(2x-12=x-14\)
\(2x-x=-14+12\)
\(\Rightarrow x=-2\)