Giải Phương trình:
1) (x+1)4+(x-3)4=82.
2) x4-2x3+5x2-2x+1=0.
3) ((29-x)/21)+((27-x)/23)+((25-x)/25)+((23-x)/27)+((21-x)/29)=0
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21 tháng 7 2019
a) \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)
\(=>\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(=>\left(4x+2\right)\left(2x-4\right)=0\)
\(=>4\left(2x+1\right)\left(x-2\right)=0\)
\(=>\orbr{\begin{cases}2x+1=0\\x-2=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=-\frac{1}{2}\\x=2\end{cases}}\)
b)\(x^3-\frac{x}{49}=0=>x\left(x^2-\frac{1}{49}\right)=0=>x\left(x-\frac{1}{7}\right)\left(x+\frac{1}{7}\right)=0\)
\(=>x=0\)hoặc \(x=\frac{1}{7}\) hoặc \(x=-\frac{1}{7}\)
21 tháng 7 2019
a)\(\(\left(3x-1\right)^2-\left(x+3\right)^2=0\)\)
\(\(\Leftrightarrow\left(3x-1-x-3\right)\left(3x-1+x+3\right)=0\)\)
\(\(\Leftrightarrow\left(2x-4\right)\left(4x+2\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}2x-4=0\\4x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}}\)\)
b)\(\(x^3-\frac{x}{49}=0\)\)
\(\(\Leftrightarrow\frac{49x^3-x}{49}=0\)\)
\(\(\Leftrightarrow x\left(49x^2-1\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}x=0\\49x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(7x-1\right)\left(7x+1\right)=0\end{cases}}}\)\)\
\(\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7};x=-\frac{1}{7}\end{cases}}\)\)
c)\(\(x^2-7x+12=0\)\)
\(\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}}\)\)
d) \(\(4x^2-3x-1=0\)\)
\(\(\Leftrightarrow4x^2-4x+x-1=0\)\)
\(\(\Leftrightarrow4x\left(x-1\right)+\left(x-1\right)=0\)\)
\(\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\4x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{4}\end{cases}}}\)\)
e) Tham khảo tại : [Toán 8]Giải phương trình | Cộng đồng học sinh Việt Nam - HOCMAI Forum
https://diendan.hocmai.vn/threads/toan-8-giai-phuong-trinh.290061/
_Y nguyệt_
18 tháng 1 2016
a, <=> (x-5/100) -1 +(x-4/101) -1 +(x-3/102) -1= (x-100/5) -1+(x-101/4) -1 +(x-102/3) -1
<=> (x-105)(1/100 +1/101 +1/102)= (x-105)(1/5+1/4+1/3)
<=> (x-105)(1/100+1/101+1/102-1/5-1/4-1/3)=0
vì 1/100+1/101+1/102-1/5-1/4-1/3 khác 0 <=> x-105=0
<=> x=105
18 tháng 1 2016
b, 29-x/21 +1+27-x/23 +1+25-x/25 +1+23-x/27 +1+21-x/29 +1=0
<=> 50-x/21 +50-x/23 +50-x/25 +50-x/27 +50-x/29=0
<=> (50-x)(1/21 +1/23 +1/25 +1/27 +1/29)=0
vì 1/21+1/23+1/25+1/27+1/29 lớn hơn 0
nên 50-x=0
<=> x=50
DN
25 tháng 12 2017
a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x=105\)
b) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)
CH
Cô Hoàng Huyền
Admin
VIP
8 tháng 8 2016
\(pt\Leftrightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+...=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
Do \(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}>0\) nên 50 - x = 0 hay x = 50.
9 tháng 7 2017
pt<=>29-x/21+1+27-x/23+1+...=0
<=>50-x/21+50-x/23+50-x/25+50-x/27+50-x/29=0
<=>(50-x).(1/21+1/23+1/25+1/27+1/29)=0
Do 1/21+1/23+1/25+1/27+1/29>0 nên 50-x=0 hay x=50
H
19 tháng 1 2019
a, \(\Leftrightarrow3x^2-3+5=3x^2+2x-3x-2\)
\(\Leftrightarrow3x^2-3x-2x+3x=-2+3-5\)
<=>x=-4
b, \(\Leftrightarrow\dfrac{x+4}{5}-\dfrac{5x}{5}+\dfrac{20}{5}=\dfrac{2x}{6}-\dfrac{3\left(x-2\right)}{6}\)
\(\Leftrightarrow\dfrac{x+4-5x+20}{5}=\dfrac{2x-3x+6}{6}\)
\(\Leftrightarrow\dfrac{6\left(-4x+24\right)}{30}=\dfrac{5\left(-x+6\right)}{30}\)
<=>-24x+144=-5x+30
<=>-5x+24x=144-30
<=>19x=114
<=>x=6
BL
19 tháng 3 2020
a. \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Rightarrow\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)
\(\Rightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-105}{4}-\frac{x-105}{3}=0\)
\(\Rightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Rightarrow x-105=0\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)
\(\Rightarrow x=105\)
b. \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Rightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+\frac{25-x}{25}+1+\frac{23-x}{27}+1+\frac{21-x}{29}+1=0\)
\(\Rightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Rightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Rightarrow50-x=0\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\ne0\right)\)
\(\Rightarrow x=50\)
19 tháng 3 2020
a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
Dễ dàng thấy nhân tử thứ hai luôn bé thua 0 nên \(x-105=0\)\(\Leftrightarrow x=105\)
b) Kĩ thuật làm tương tự câu a cộng mỗi phân số VT với 1 thì VP=0 và ta có nhân tử chung 50-x
3)
\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)
có vấn đề thì phải
= 5 mới đúng chứ-.-
p/s: câu c = -5 chứ-.-