NK
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
CB
1
16 tháng 8 2022
b: \(\Leftrightarrow x^3-4x-3\left(4x^2-4x+1\right)-2x-5=-6x^2-6x\)
\(\Leftrightarrow x^3-4x-12x^2+12x-3-2x-5=-6x^2-6x\)
\(\Leftrightarrow x^3-12x^2+6x-8+6x^2+6x=0\)
\(\Leftrightarrow x^3-6x^2+12x-8=0\)
=>x-2=0
hay x=2
c: \(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)
=>12x-4=-10
=>12x=-6
hay x=-1/2
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 ....
CN
27 tháng 12 2017
a) (x - 2)2 - (x - 3)(x + 3) = 17
⇔ (x2 - 4x + 4) - (x2 - 9) = 17
⇔ x2 - 4x + 4 - x2 + 9 = 17
⇔ 13 - 4x = 17
⇔ - 4x = -4
⇔ x = 1
b) 4(x - 3)2 - (2x - 1)(2x + 1) = 10
⇔ [2(x - 3)]2 - (4x2 - 1) = 10
⇔ (2x - 6)2 - 4x2 + 1 = 10
⇔ 4x2 - 24x + 36 - 4x2 + 1 = 10
⇔ - 24x = -27
⇔ x = \(\dfrac{9}{8}\)
c) (x - 4)2 - (x - 2)(x + 2) = 36
⇔ x2 - 8x + 16 - x2 + 4 = 36
⇔ -8x = 16
⇔ x = -2
d) (2x + 3)2 - (2x - 1)(2x + 1) = 10
⇔ 4x2 + 12x + 9 - 4x2 + 1 = 10
⇔ 12x = 0
⇔ x = 0
27 tháng 12 2017
Tìm x ,biết :
a, \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=17\)
\(\Rightarrow x^2-4x+4-x^2+9=17\)
\(\Rightarrow-4x+13=17\)
\(\Rightarrow-4x=4\)
\(\Rightarrow x=-1\)
b,\(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Rightarrow4\left(x^2-6x+9\right)-4x^2+1=10\)
\(\Rightarrow4x^2-24x+36-4x^2+1=10\)
⇒ \(-24x+37=10\)
\(\Rightarrow-24x=-27\)
\(\Rightarrow x=\dfrac{-27}{-24}=\dfrac{9}{8}\)
c,\(\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=36\)
⇒ \(x^2-8x+16-x^2+4=36\)
⇒ \(-8x+20=36\)
⇒ \(-8x=16\Rightarrow x=-2\)
d,\(\left(2x+3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Rightarrow4x^2+12x+9-4x^2+1=10\)
\(\Rightarrow12x+10=10\)
\(\Rightarrow12x=0\Rightarrow x=0\)
14 tháng 6 2022
a: \(B=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{10}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)
b: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)
\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)
\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)
c: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=\dfrac{-5}{2}\)
d: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)
\(=\dfrac{1-4x^2}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x+4}\cdot\dfrac{3}{2\left(1-2x\right)}=\dfrac{3\left(2x+1\right)}{x+4}\)
9 tháng 1 2021
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
PN
16 tháng 5 2021
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)