tìm x, biết:
1, x+2/5=1/x-2
2, 3/x-4= x+4/3
3, x+2/x+6=3/x=1
K
Khách
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19 tháng 2 2022
\(x+\dfrac{1}{2}=\dfrac{33}{4}\\ \Rightarrow x=\dfrac{33}{4}-\dfrac{1}{2}\\ \Rightarrow x=\dfrac{31}{4}\\ \dfrac{5}{6}-x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{5}{6}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{2}\\ x+\dfrac{4}{5}=\dfrac{-2}{3}\\ \Rightarrow x=\dfrac{-2}{3}-\dfrac{4}{5}\\ \Rightarrow x=\dfrac{-22}{15}\)
5 tháng 8 2015
- 2(x+5)(x-5)-(x+2)(2x-3)+x(x^2-8)=(x+1)(x^2-x+1)
<=> 2(x^2-25) - 2x^2+3x-4x+6 + x^3-8x = x^3+1
=>2x^2-50 - 2x^2 -9x+6+x^3-x^3-1 = 0
<=>-9x - 45 =0
<=>-9x=45
<=>x=-5
Còn phần b và c bạn cứ khai triển ra,mình phải đi học nên không có thời gian giải cho bạn
TT
Tìm x biết :
a) 3(5/3x-7)-2(1.5x+6)-(5-x)(x+4)=80+x^2
b) 4/5x^2(x/3-1/2)-(1/5x-2/3)(4x^2/3+1)=22/45x^2
1
YN
10 tháng 3 2022
`Answer:`
\(3\left(\frac{5}{3}x-7\right)-2\left(1.5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)
\(\Leftrightarrow3\left(\frac{5x}{3}-7\right)-2\left(5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)
\(\Leftrightarrow5x-21-10x-12-5x-20+x^2+4x=80+x^2\)
\(\Leftrightarrow5x-21-10x-12-5x-20+4x=80\)
\(\Leftrightarrow-6x-53=80\)
\(\Leftrightarrow-6x=133\)
\(\Leftrightarrow x=-\frac{133}{6}\)
\(\frac{4}{5}x^2\left(\frac{x}{3}-\frac{1}{2}\right)-\left(\frac{1}{5}x-\frac{2}{3}\right)\left(4\frac{x^2}{3}+1\right)=\frac{22}{45}x^2\)
\(\Leftrightarrow36x^2\left(\frac{x}{3}-\frac{1}{2}\right)-45\left(\frac{x}{5}-\frac{2}{3}\right)\left(\frac{4x^2}{3}+1\right)=22x^2\)
\(\Leftrightarrow12x^3-18x^2-12x^3-9x+40x^2+30=22x^2\)
\(\Leftrightarrow22x^2-9x+30=22x^2\)
\(\Leftrightarrow-9x+30=0\)
\(\Leftrightarrow-9x=-30\)
\(\Leftrightarrow x=\frac{10}{3}\)
TT
Tìm x biết :
a) 3(5/3x-7)-2(1.5x+6)-(5-x)(x+4)=80+x^2
b) 4/5x^2(x/3-1/2)-(1/5x-2/3)(4x^2/3+1)=22/45x^2
0
7 tháng 6 2017
1).( 27,56 x 35 ) + ( 27,56 x 67 ) - ( 27,56 x 2)
= (964 + 1846,52) - 55,12
=2810,52 - 55,12
= 2755,4
2).( 4x 35 ) x ( 25 x 5 ) x 2
= ( 140 x 125 ) x2
= 17500 x 2
=35000
4). 3/10
5). 1188
6). 61/6
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
1.
$x(x+2)(x+4)(x+6)+8$
$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$
$=a(a+8)+8$ (đặt $x^2+6x=a$)
$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$
Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
2.
$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$
$=5-(x^2+5x-6)(x^2+5x+6)$
$=5-[(x^2+5x)^2-6^2]$
$=41-(x^2+5x)^2\leq 41$
Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$
NP
0
H9
HT.Phong (9A5)
CTVHS
14 tháng 7 2023
\(6-2\left(x-1\right)=4\)
\(\Rightarrow2\left(x-1\right)=6-4\)
\(\Rightarrow2\left(x-1\right)=2\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=1+1=2\)
________________
\(2\cdot\left(x-2\right)+1=7\)
\(\Rightarrow2\cdot\left(x-2\right)=7-1\)
\(\Rightarrow2\cdot\left(x-2\right)=6\)
\(\Rightarrow x-2=3\)
\(\Rightarrow x=3+2=5\)
_______________
\(\left(2\cdot x-3\right)+4=9\)
\(\Rightarrow2\cdot x-3=5\)
\(\Rightarrow2\cdot x=3+5\)
\(\Rightarrow2\cdot x=8\)
\(\Rightarrow x=\dfrac{8}{2}=4\)
________________
\(\left(3\cdot x-2\right)-1=3\)
\(\Rightarrow3\cdot x-2=3+1\)
\(\Rightarrow3\cdot x-2=4\)
\(\Rightarrow3\cdot x=6\)
\(\Rightarrow x=\dfrac{6}{3}=2\)
14 tháng 7 2023
a: =>2(x-1)=2
=>x-1=1
=>x=2
b: =>2(x-2)=6
=>x-2=3
=>x=5
c; =>2x-3=5
=>2x=8
=>x=4
d: =>3x-2=4
=>3x=6
=>x=2
e: =>2(6-x)=4
=>6-x=2
=>x=4
f: =>x-2=5
=>x=7
g: =>10-2x=4
=>2x=6
=>x=3
h: =>2x+4=3
=>2x=-1
=>x=-1/2
j: =>x+2=12
=>x=10
l: =>2x+3=3
=>2x=0
=>x=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) Ta có\(\frac{x+2}{5}=\frac{1}{x-2}\)
=> (x + 2)(x - 2) = 5
=> x2 + 2x - 2x - 4 = 5
=> x2 - 4 = 5
=> x2 = 9
=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
2) \(\frac{3}{x-4}=\frac{x+4}{3}\)
=> (x - 4)(x + 4) = 9
=> x2 + 4x - 4x - 16 = 9
=> x2 - 16 = 9
=> x2 = 25
=> \(\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
a, \(\frac{x+2}{5}=\frac{1}{x-2}ĐK:x\ne2\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{5\left(x-2\right)}=\frac{5}{5\left(x-2\right)}\Leftrightarrow\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^2-2x+2x-4=5\Leftrightarrow x^2=9\Leftrightarrow x\pm3\)
b, \(\frac{3}{x-4}=\frac{x+4}{3}ĐK:x\ne4\)
\(\Leftrightarrow\frac{9}{\left(x-4\right)3}=\frac{\left(x+4\right)\left(x-4\right)}{3\left(x-4\right)}\Leftrightarrow9=x^2-4x+4x-16\)
\(\Leftrightarrow x^2-16=9\Leftrightarrow x^2=25\Leftrightarrow x=\pm5\)
c, \(\frac{x+2}{x+6}=\frac{3}{x}=1ĐK:x\ne0;-6\)
Xét : \(\frac{x+2}{x+6}=1\Leftrightarrow x+2=x+6\Leftrightarrow-4\ne0\)
Xét : \(\frac{3}{x}=1\Leftrightarrow3=x\)