20+[4^2+(2×x-6)^2]=72
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TN
0
10 tháng 1 2023
= [(2+18)+(3+17)+(4+16)+(5+15)+(1+19)+20+10].(168−8.21)
= (20 x 5 + 10) x 0
= 110 x 0
= 0
9 tháng 1 2023
(2+4+6+......+20)x(56x3-72:9x21)
=110x168-8x21
=18480-168
=18312
9 tháng 1 2023
\(=\left[\left(2+18\right)+\left(3+17\right)+\left(4+16\right)+\left(5+15\right)+\left(1+19\right)+20+10\right].\left(168-8.21\right)\)= (20 x 5 + 10) x 0
= 110 x 0
= 0
LT
1
LG
12 tháng 10 2023
(1/2+1/6+1/12+1/20+...+1/72) : X = 3/4
(1/1x2+1/2x3+1/3x4+1/4x5+...+1/7x8) : X = 3/4
(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/7-1/8) : X = 3/4
(1/1-1/8) : X = 3/4
7/8 : X = 3/4
X = 7/8 : 3/4
X = 7/6
LC
1
TV
19 tháng 3 2018
\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\)ĐKXĐ là \(x\ne-2;x\ne-8;x\ne-4;x\ne-6\)
\(\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+16x+64+8}{x+8}=\dfrac{x^2+8x+16+4}{x+4}+\dfrac{x^2+12x+36+6}{x+6}\)\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)
\(\Leftrightarrow\left(\dfrac{2}{x+2}-1\right)+\left(\dfrac{8}{x+8}-1\right)=\left(\dfrac{4}{x+4}-1\right)+\left(\dfrac{6}{x+6}-1\right)\)\(\Leftrightarrow\dfrac{-x}{x+2}+\dfrac{-x}{x+8}=\dfrac{-x}{x+4}+\dfrac{-x}{x+6}\)
\(\Leftrightarrow\dfrac{x}{x+2}+\dfrac{x}{x+8}-\dfrac{x}{x+4}-\dfrac{x}{x+6}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)=0\)
Do \(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\ne0\)
=> x=0
Vậy ....
14 tháng 7 2019
Sửa đề:
\(\frac{x^2+4x+6}{x+2}+\frac{x^2+16x+72}{x+8}=\frac{x^2+8x+20}{x+4}+\frac{x^2+12x+20}{x+6}\)
\(\Leftrightarrow\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}=\frac{\left(x+4\right)^2+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow\frac{2}{x+2}+\frac{8}{x+8}=\frac{4}{x+4}+\frac{6}{x+6}\)
Quy đồng giải tiếp nhé
3 tháng 3 2018
\(\Leftrightarrow1+\dfrac{2}{x+2}+1+\dfrac{8}{x+8}=1+\dfrac{4}{x+4}+1+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{1}{x+2}+\dfrac{4}{x+8}=\dfrac{2}{x+4}+\dfrac{3}{x+6}\)
\(\Leftrightarrow\dfrac{4}{x+8}-\dfrac{3}{x+6}=\dfrac{2}{x+4}-\dfrac{1}{x+2}\)
\(\Leftrightarrow\dfrac{4x+24-3\left(x+8\right)}{\left(x+8\right)\left(x+6\right)}=\dfrac{2x+4-\left(x+4\right)}{\left(x+4\right)\left(x+2\right)}\)
\(\dfrac{x}{\left(x+8\right)\left(x+6\right)}=\dfrac{x}{\left(x+4\right)\left(x+2\right)}\)
x=0 là nghiệm
x khác 0
\(\left\{{}\begin{matrix}x\ne\left\{-8;-6;-4;-2\right\}\\\left(x+4\right)\left(x+2\right)=\left(x+8\right)\left(x+6\right)\end{matrix}\right.\)<=>x^2 +6x+8 =x^2 +14x+48
-40 =8x=> x =-5 nhận
x={-5;0}
P
2 tháng 1 2018
=>\(\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}\)=\(\frac{\left(x+4\right)+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)
=>2x+10+\(\frac{2}{x+2}+\frac{8}{x+8}\)=2x+10+\(\frac{4}{x+4}+\frac{6}{x+6}\)
=>-x\(\left(\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}+\frac{1}{x+8}\right)\)=0
=>\(\orbr{\begin{cases}x=0\\\frac{1}{x+2}-.....+\frac{1}{x+8}=0\end{cases}}\)
Voi \(\frac{1}{x+2}-....\)=0 ta co
Dat x+5=t
=>\(\frac{1}{t-3}-\frac{1}{t-1}-\frac{1}{t+1}+\frac{1}{t+3}\)=0
=> \(2t\left(\frac{1}{t^2-1}+\frac{1}{t^2-9}\right)=0\)
=>t=0
=>x=-5
Vay phuong trinh co nghiem x=0;-5
11 tháng 3 2018
=> \(\frac{(x+2)^2+2}{x+2}+\frac{(x+8)^2+8}{x+8}=\frac{(x+4)+4}{x+4}+\frac{(x+6)^2+6}{x+6}\)
=> 2x + 10 + \(\frac{2}{x+2}+\frac{8}{x+8}=2x+10+\frac{4}{x+4}+\frac{6}{x+6}\)
=>-x \((\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}-\frac{1}{x+8})=0\)
\(x=0\)
\(=>\orbr{\frac{1}{x+2}}-.....+\frac{1}{x+8}=0\)
Với \(\frac{1}{x+2}-...=0\). Ta có :
Đặt x + 5 = t
=> \(\frac{1}{t-3}-\frac{1}{t-1}-\frac{1}{t+1}+\frac{1}{t+3}=0\)
\(=>2t(\frac{1}{t^2-1}+\frac{1}{t^2-9})=0\)
=> t = 0
=> x = -5
Vậy phương trình có nghiệm x= 0 ; - 5
20 + [ 42 + ( 2x - 6 )2 ] = 72
16 + ( 2x - 6 )2 = 72 - 20 = 52
( 2x - 6 )2 = 52 - 16 = 36 = 62
2x - 6 = 6
2x = 6 + 6 = 12
x = 12 : 2 = 6