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YY
5
TQ
2 tháng 7 2016
(x+1/5)^2 =26/25-17/25
<=> (x +1/5)^2 =(3/5)^2
<=> x+1/5=3/5
=> x= 2/5
KT
2
2 tháng 7 2016
1.
(x + 1/5)² = 26/25 - 17/25
(x + 1/5)² = 9/25
Rút căn hai vế :
|(x + 1/5)| = 3/5
x = -4/5
hoặc
x = 2/5
2.
(x + 2) / 327 + (x + 3) / 326 + (x + 4) / 325 + (x + 5) / 324 + (x + 349) / 5 = 0
<=> (x + 2) / 327 +1+ (x + 3) / 326 +1+ (x + 4) / 325 +1+ (x + 5) / 324 +1+ (x + 349) / 5 -4 = 0
<=> (x+ 329)/327 + (x+ 329)/326 + (x+ 329)/325 + (x+ 329)//324 + (x+ 329)/5 =0
<=> (x+ 329).(1/327 + 1/ 326 + 1/325 + 1/324 +1/5) =0
Do (1/327 + 1/ 326 + 1/325 + 1/324 +1/5) >0 nên x+ 329 =0 => x= -329
Câu 1 chưa chắc đã đúng ( quên hết kiến thức lớp 6 rùi ) hihi
2 tháng 7 2016
aaaaaaaa . chết rồi . cho mình sủa câu thứ nhất :
(x+1/5)2 + 17/25=26/25
( x + 1/5 ) 2 = 26/25 - 17/25
( x + 1/5 ) 2 = 3/52
x + 1/5 = 3/5
x = 2/5.
NN
13 tháng 8 2023
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)
\(=>x+1=0\)
\(=>x=-1\)
b,
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)
\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)
\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)
\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)
Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)
\(=>x+2021=0\)
\(=>x=-2021\)
c,
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)
\(=>x+329=0\)
\(=>x=-329\)
11 tháng 3 2018
b, |5x-3| >= 7
=> 5x-3 < = -7 hoặc 5x-3 >= 7
=> x < = -4/5 hoặc x >= 2
Vậy ..........
Tk mk nha
NC
0
15 tháng 9 2017
a) \(234+\left(345-x\right)=500\)
\(345-x=500-234\)
\(345-x=266\)
\(x=79\)
vay \(x=79\)
b) \(456-\left(x+23\right)=326\)
\(x+23=456-326\)
\(x+23=130\)
\(x=107\)
vay \(x=107\)
c) \(\left(5x-15\right):5=0\)
\(5x-15=0\)
\(5x=15\)
\(x=3\)
vay \(x=3\)
d) \(84-4\left(2x+1\right)=48\)
\(8x+4=84-48\)
\(8x+4=36\)
\(8x=32\)
\(x=4\)
vay \(x=4\)
e) \(\left(x-76\right).54=0\)
\(x-76=0\)
\(x=76\)
vay \(x=76\)
f) \(\left(x-3\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
vay \(\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
g) \(\left(x-6\right)\left(x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-6=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}\)
vay \(\orbr{\begin{cases}x=6\\x=7\end{cases}}\)
h) \(\left(x-2\right)\left(x-3\right)\left(x-5\right)=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)hoac \(\orbr{\begin{cases}x-3=0\\x-5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=5\end{cases}}\)
vay \(x=2\)hoac \(\orbr{\begin{cases}x=3\\x=5\end{cases}}\)
TN
25 tháng 7 2018
\(\left(x^2-5\right)\left(x^2+1\right)=0\)
<=> \(\hept{\begin{cases}x^2-5=0\\x^2+1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x^2=5\\x^2=-1\end{cases}}\)
<=> \(\hept{\begin{cases}x=\sqrt{5};x=-\sqrt{5}\\x\in\varnothing\end{cases}}\)
câu còn lại tương tự nha
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(=\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5-4}=0\)
\(=\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
=x+329.1/327+1/326+1/325+1/324+1/5=0
Do 1/327+1/326+1/325+1/324+1/5 # 0 nên x+329=0
=>x=-329