Giải các phương trình sau
a ( 3x-1)^2 - (x+3)^2
b x^3-x/49 = 0
c x^2 -7x+12
d 4x^2 -3x -1 =0
e . 29-x/21 + 27-x/23 + 25-x/25 + 23-x/28 + 21-x/29
giải các phương trình sau
a) \(3\left(x+1\right)\left(x-1\right)+5=\left(x-1\right)\left(3x+2\right)\)
b) \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
c) \(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5\)
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
a ) <=> 3x2 - 3 + 5 = 3x2 + 2x - 3x - 2
<=> 3x2 - 3x2 - 2x + 3x = -2 - 5 + 3
<=> x = - 4
Vậy s = \(\left\{-4\right\}\)
b)<=> \(\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x+4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
<=> 6x + 24 - 30x - 120 = 10x - 15x + 30
<=> 6x -30x - 10x + 15x = 30 - 24 + 120
<=> -19x = 126
<=> x =-6,6
Vậy s = \(\left\{-6,6\right\}\)
a ) <=> 3x2 - 3 + 5 = 3x2 + 2x - 3x - 2
<=> 3x2 - 3x2 - 2x + 3x = -2 - 5 + 3
<=> x = - 4
Vậy s = \(\left\{-4\right\}\)
b)<=> 6x + 24 - 30x - 120 = 10x - 15x + 30
<=> 6x + 24 - 30x - 120 = 10x - 15x + 30
<=> -19x = 126
<=> x =-6,6
ai giải cụ thể giùm em với
Giải phương trình sau
a) x-5/100+x-4/101+x-3/102=x-100/5+x-101/4+x-102/3
b) 29-x/21+27-x/23+25-x/25+23-x/27+21-x/29=-5
ai giải cụ thể giùm em với
Giải phương trình sau
a) x-5/100+x-4/101+x-3/102=x-100/5+x-101/4+x-102/3
b) 29-x/21+27-x/23+25-x/25+23-x/27+21-x/29=-5
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
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
Giải phương trình:
1) (3x-1)^2-5(2x+1)^2+96x-3)(2x+1)=(x-1)^2
2) (x+2)^3-(x-2)^3=12(x-1)-8
3) x-1/4-5-2x/9=3x-2/3
4) 25x-655/95-5(x-12)/209=[89-3x-2(x-13)/5]/11
5) 29-x/21+27-x/23+25-x/25+23-x/27=-4
6) x-69/30+x-67/32=x-63/36+x-61/38
7)x+117/19+x+4/28+x+3/57=0
8) 59-x/41+57-x/43+2=x-55?45+x-53/47-2
9) Cho phương trình: mx+x-m^2=2x-2 (x là ẩn). Tìm m để phương trình:
a) Có nghiệm duy nhất
b) Vô số nghiệm
c) Vô nghiệm
\(\dfrac{29-x}{21}\)+\(\dfrac{27-x}{23}\)+\(\dfrac{25-x}{25}\)+\(\dfrac{23-x}{27}\)+\(\dfrac{21-x}{29}\)=\(\dfrac{(29-x+1}{21}\)+\(\dfrac{(27-x+1)}{23}\)+\(\dfrac{(25-x+1)}{25}\)+\(\dfrac{(23-x+1)}{21}\)=-5 +5
GIẢI nốt hộ mình với ạ
Giải phương trình sau : \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(pt\Leftrightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+...=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
Do \(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}>0\) nên 50 - x = 0 hay x = 50.
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
1) giải pt :
a) \(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
b) \(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5\)
2) giải pt :
a) \(\left(5x+1\right)^2=\left(3x-2\right)^2\)
b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)
c) \(\left(x+3\right)^4+\left(x+5\right)^4=2\)
d) \(x^4-3x^3+4x^2-3x+1=0\)
1)
\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
\(\Leftrightarrow\dfrac{x-5}{100}+1+\dfrac{x-4}{101}+1+\dfrac{x-3}{102}+1=\dfrac{x-100}{5}+1+\dfrac{x-101}{4}+1+\dfrac{x-102}{3}+1\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)
\(\Leftrightarrow x=105\)
b)
\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)
\(\Leftrightarrow50-x=0\)
\(\Leftrightarrow x=50\)
2)
\(\left(5x+1\right)^2=\left(3x-2\right)^2\)
\(\Leftrightarrow\left|5x+1\right|=\left|3x-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=3x-2\\5x+1=-3x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)
b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)
\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)
\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)
\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)
\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
c. \(\left(x+3\right)^4+\left(x+5\right)^4=2\)
Đặt: \(y=x+4\), ta có:
\(\left(y-1\right)^4+\left(y+1\right)^4=2\)
\(\Leftrightarrow y^4-4y^3+6y^2-4y+1+y^4+4y^3+6y^2+4y+1=2\)
\(\Leftrightarrow2y^4+12y^2=0\)
\(\Leftrightarrow2y^2\left(y^2+6\right)=0\)
\(\Leftrightarrow y=0\)
\(\Leftrightarrow x=-4\)
d) \(x^4-3x^3+4x^2-3x+1=0\)
\(\Leftrightarrow x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)
\(\Leftrightarrow x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x=1\)
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
MONG M.N GIÚP MIK NHA. MIK CẢM ƠN NHÌU...
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-.-
3)
\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=-5+5\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)
\(\Leftrightarrow50-x=0\) Vì \(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\ne0\)
\(\Leftrightarrow x=50\)
Giải phương trình
a,\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
b, \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
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) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)